Page 1 Dynamics Newton's Laws......................................................................................... 3 Newton’s First Law............................................................................. 3 Example 1.............................................................................................. 3 Newton’s Second Law......................................................................... 4 Example 2............................................................................................. 5 Questions A............................................................................................. 6 Vertical Motion........................................................................................ 7 Example 3............................................................................................. 7 Example 4............................................................................................. 9 Example 5............................................................................................ 10 Example 6............................................................................................ 13 Motion of Two Connected Particles.................................................. 15 Newton's Third Law......................................................................... 15 Example 7............................................................................................ 16 Pulleys...................................................................................................... 19 Example 8............................................................................................ 19 Example 9............................................................................................ 21 Example 10......................................................................................... 23 Example 11.......................................................................................... 26 Extension............................................................................................ 27 Momentum and Impulse...................................................................... 28 Momentum.......................................................................................... 28 Example 12......................................................................................... 28 Change in Momentum....................................................................... 28 Example 13......................................................................................... 28 Impulse............................................................................................... 29 Example 14......................................................................................... 30 Example 15......................................................................................... 30 The Principle of the Conservation of Momentum......................... 32
Transcript
Page 1
Dynamics
Newton's Laws.........................................................................................3Newton’s First Law............................................................................. 3Example 1.............................................................................................. 3Newton’s Second Law.........................................................................4Example 2............................................................................................. 5
Example 3............................................................................................. 7Example 4............................................................................................. 9Example 5............................................................................................ 10Example 6............................................................................................ 13
Motion of Two Connected Particles..................................................15Newton's Third Law......................................................................... 15Example 7............................................................................................ 16
Momentum and Impulse...................................................................... 28Momentum.......................................................................................... 28Example 12......................................................................................... 28Change in Momentum....................................................................... 28Example 13......................................................................................... 28Impulse............................................................................................... 29Example 14......................................................................................... 30Example 15......................................................................................... 30
The Principle of the Conservation of Momentum......................... 32
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Example 16......................................................................................... 32Example 17......................................................................................... 34Example 18......................................................................................... 35
Jerk in a String.................................................................................... 37Example 19......................................................................................... 37
Dynamics is the study of moving objects. In previous chapters we have considered the sum total of forces on a body (resultant) and our next consideration is the movement that these forces bring about. Forces of friction, thrust, gravity and tension will be considered and the subsequent motion will be measured by the application of constant acceleration equations and the equation of motion. As in most mechanics questions a certain amount of modeling will have to be used in our working.
Our first area of study of moving objects will involve the application of Newton’s laws.
Newton's Laws
Newton’s First Law
A body will remain at rest, or will continue to move with constant velocity, unless external forces force it to do otherwise.A change in state of motion of a body is caused by a force. The unit of force is the Newton, (N).
Example 1
A body of mass 3500Kg moves horizontally at a constant speed of 5ms-1 subject to the forces shown. Find P and S.
350N
S
3500g
S
5ms-1
P
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There is no vertical motion therefore:
2S = 3500g
S = 1750g
The horizontal acceleration is zero, therefore:
P =350N
Newton’s Second Law
The force F applied to a particle is proportional to the product of mass of the particle and the acceleration produced.A force of 1N produces an acceleration of 1ms-2 in a body of mass 1kg. Newton’s Second Law is summarized by the equation:
F = ma
this is often termed the equation of motion.
It is vitally important to realise that F is the overall resultant and not Friction (FR).
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Example 2
If the object in Example 1 is slightly modified to take account of the fact that there is a pushing force 600N, calculate the acceleration.
The resultant of the two horizontal forces is 250N pushing the object to the right. So by setting up an equation of motion:
F = ma
250 = 3500 × a
a = 0.07ms-2
350N
3500g
a ms-2
600N
SS
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Questions A
1 Find the resultant force that will bring about an acceleration of 4 ms-2 for a particle of mass 3.75kg
2 A particle of mass 4.5kg is acted upon by forces (6i + 3j)N and (-2i + 7j)N. Calculate the acceleration of the particle in vector form.
3 A toy train of mass 1.25kg is pulled across a horizontal floor by a horizontal string. The tension in the string is 1.2N. Calculate the acceleration and the distance traveled by the train in the first 4 seconds.
4 A car of mass 750Kg experiences a resistive force of RN while being brought to rest in 9 seconds form a speed of 18ms-1. Calculate the magnitude of the force.
5 A car of mass 900Kg is under constant resistance to motion of 500N. Find the value of the engine force required to bring about an acceleration of 1.25ms-2.
6 A train of mass 4000kg produces a driving force of 3000N. If the train experiences constant resistance to motion of 1200N calculate the acceleration.
7 A stone of mass 1.25kg is dropped into a viscous liquid and fall vertically through it with an acceleration of 4.8ms-2. Find the resistive force acting against the stone.
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Vertical Motion
If a particle is falling in the earth’s atmosphere then it will accelerate at 9.8ms-2. By Newton’s first law the particle must be experiencing a force and the only force present is the weight, so by considering the equation of motion:
F = ma
Weight = mg
Example 3
A man of mass 95kg is traveling up in a lift. Given that the acceleration of the lift is 0.8ms-2, find the force exerted on the man by the floor of the lift.Calculate the same force when the lift is descending with the same acceleration.
95g
R
0.8ms-2
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Taking up to be positive and setting up an equation of motion:
F = ma
R – 95g = 95a
R – 931 = 95 × 0.8
R = 1007N
When the lift is traveling in the opposite direction let down be positive so the equation of motion becomes:
95g – R = 95a
931 –R = 95 × 0.8
R = 855N
It is worth pointing out that the reaction from the floor is less than the weight of the man when the lift is descending (and vice versa).
The next example introduces constant acceleration equations into the problem.
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Example 4
A ball of mass 0.9 kg falls from a height of 22m above horizontal ground. The ball reaches the ground after t seconds. The ball sinks into the ground a distance of 1.9cm before coming to rest. The ground is assumed to exert a constant resistive force of magnitude F newtons. Find:a) the value of t to 3 sig fig;b) the value of F to 3 sig fig.
a) Using constant acceleration equations to find t:
s = 22, a = 9.8, u = 0, t = ?
2
2
1s ut at2
22 0 0.5 9.8 t
t 2.12sec
= +
= + × ×
=
b) The ball is brought to rest in 1.9cm so by using constant acceleration equations we can find the deceleration and hence the force:
Firstly we need to find the speed with which the ball hits the surface
u = 0, v = ?, t = 2.12, a =9.8
1
v u at
v 9.8 2.12
v 20.776ms−
= +
= ×
=
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And now for the deceleration:
s = 0.019, a = ?, v = 0, u = 20.776
2 2
2
2
v u 2as
0 20.776 2 a 0.019
a 11359ms−
= +
= + × ×
= −
And finally using an equation of motion for the particle:
F = ma
F = 0.9 × 11359
F = 10.2KN
Note that we were asked for the magnitude hence the positive answer. How realistic is this answer? Ask a physics teacher for some other examples.
Example 5
A ball of mass 5kg falls from a height of 6m into a jar containing a viscous liquid. The upward force exerted by the liquid is of magnitude 75N. How far will the ball sink into the liquid? Calculate the total time that the ball is in motion.
Firstly we need to calculate the speed with which the ball hits the liquid.
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Assuming that down is positive:s = 6, a = 9.8, u = 0, v = ?
2 2
2
1
v u 2as
v 0 2 9.8 6
v 10.84ms−
= +
= + × ×
=
Secondly we can set up an equation of motion for the ball to work out the deceleration.
F = ma
5g – 75 = 5a
a = -5.2ms-2
Thirdly we need to calculate the distance that the ball travels through the liquid before coming to rest.u = 10.84, v = 0, a = -5.2 s = ?
2 2v u 2as
0 117.6 2 5.2 s
s 11.31m
= +
= − × ×
=
5g
75N
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Finally the total time taken by the ball in motion must be done in two parts seeing as before the ball hits the liquid it has an acceleration of 9.8ms-2 whereas, whilst falling through the liquid, it has an acceleration of -5.2ms-2.
Time to meet the surface of the liquid:
a = 9.8, u = 0, v = 10.84, t = ?
v = u + at
10.84 = 0 + 9.8t
t = 1.1 sec
Time to come to rest:
a = -5.2, u = 10.84, v = 0, t = ?
v = u + at
0 = 10.84 -5.2t
t = 2.1 sec
Therefore the total time in motion is 3.3 sec. The liquid is rather viscous, how realistic is the resistance force? What about a ball falling in to a jar of syrup?
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The next example introduces friction on a slope. The questions are beginning to get more challenging but a good diagram is always the best place to start. Examiners regularly report that the most successful candidates in mechanics M1 and M2 always draw diagrams.
Example 6
A ball of mass 2kg is projected up a line of greatest slope inclined at an angle of 30º to the horizontal. The coefficient of friction between the plane and the ball is 0.4. The initial speed of the ball is 5 ms-1. Find:a) the frictional force acting whilst the ball moves up the plane.b) the distance moved up the plane by the ball before it comes to instantaneous rest.
a) Since the ball is moving then Friction must be at its maximum value.
Resolving perpendicular to the plane gives:
R = 2g cos 30º
R = 16.97N
30º 2g
R
5ms-1
FR
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Using FR = μR:
FR = 0.4 × 16.97
FR = 6.79N
b) The frictional force and the weight component of the ball are trying to slow the ball down. By setting up an equation of motion for the ball we can calculate the deceleration.Assuming that uphill is positive:
F = ma
- FR – weight comp down the plane = ma
-6.79 - 2g cos 30º = 2a
a = -8.295ms-2
At the point the ball comes to instantaneous rest it will have zero velocity and we can use constant acceleration equations to calculate the distance traveled.
u = 5, v = 0, a = -8.295, s = ?
2 2v u 2as
0 25 2 8.295 s
s 1.51m
= +
= − × ×
=
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Motion of Two Connected Particles
Newton's Third Law
Before we can consider the motion of two connected particles we need to discuss Newton’s Third Law. This law states that action and reaction are equal and opposite.
If two bodies A and B are in contact and exert forces on each other, then the force exerted by A on B is equal in magnitude and opposite in direction to the force exerted by B on A.
This principle will be applied to tow truck problems and pulleys to name but two.
Consider the situation below where the cartoon car is towing a racing car.
The racing car is pulled forward by tension in the tow bar. The racing car will exert an equal but opposite force on the car. If the car is slowing down and there are no breaks on the racing car then some force must be acting in the opposite direction to the direction of motion of the two cars. In this case the tow bar will exert a thrust on both cars (arrows change directions).
T T
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Example 7
The AA man is towing a car along a straight horizontal road. The truck has a mass of 1500kg and the car has a mass of 850kg. The truck is connected to the car by a bar which is to be modelled as a light inextensible string. The truck’s engine produces a constant driving force of 2500N. The resistance to motion of the truck and the car are constant and of magnitude 750N and 400N respectively. Find:a) the acceleration of the truck and the car;b) the tension in the rope.When the truck and the car are traveling at 22ms-1 the tow bar breaks. If the magnitude of the resistance to motion of the truck remains at 750N calculate:c) the time difference in achieving a speed of 30ms-1 with and without the car in tow.
a) Setting up equations of motion for the car and truck separately gives:
Car T – 400 = 850a Truck 2500 – 750 – T = 1500a
Adding the two equations gives:
1350 = 2350a a = 0.574ms-2
2500N
400N
1500kg850kg
750N
T T
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b) Finding the tension:
Substituting the value into the car’s equation of motion gives:
T – 400 = 850 × 0.574
T = 888N
c) If we assume that the bar doesn’t break then the time required to reach 30ms-1 is calculated by using the constant acceleration equations.
u = 22, v = 30, a = 0.574, t = ?
v = u + at
30 = 22 + 0.574t
t = 13.9 sec
At the point that the tow bar breaks, the tension in the bar is no longer acting against the truck. Therefore the equation of motion of the truck becomes:
2500 – 750 = 1500a
a = 1.167ms-2
u = 22, v = 30, a = 1.167, t = ?
v = u + at
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30 = 22 + 1.167t
t = 6.9 sec
Therefore there is a time difference of 7 seconds.
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Pulleys
In all questions in M1 the pulley system will be smooth. This implies that the motion of the particles at the end of the string are unaffected by the string passing over the pulley. A further assumption is that the string is light and inextensible. These modelling assumptions make the problem simpler but we can still get pretty realistic answers. If two particles are connected by a string where the string passes over a smooth pulley, then we can assume that the particles will have equal accelerations but in opposite directions.
Example 8
Two particles P and Q are connected by a light inextensible string which passes over a smooth fixed pulley. The system is released from rest. Find:a) the magnitude of the acceleration;b) find the tension in the string.
QP
20g8g
TT
a ms-2
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To start the problem set up an equation of motion for particle P. The particle will accelerate upwards hence:
F = ma
T – 8g = 8a (1)
When considering particle Q, its weight will cause it to accelerate down hence the equation of motion is:
F = ma
20g – T = 20a (2)
By adding the two equations the tension will be eliminated:
12g = 28a
a = 4.2ms-2
Substituting the value of the acceleration into equation (1) will give the tension.
T – 8g = 8 × 4.2
T = 112N
The following example is more algebraic but this does not mean that it is more complicated.
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Example 9
Two particles P and Q of masses have masses 8m and Km, where K > 8. They are connected by a light inextensible string which passes over a smooth fixed pulley. The system is released from rest with the string taut and the hanging parts of the string vertical, as shown below. Initially P has an acceleration of
magnitude of 3 g4 .
a) Find, in terms of m and g, the tension, T, in the string.b) Find the value of K.
a) Adding forces to the system:
Q
P
Q
P
Kmg
8mg
TT
3 g4
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Setting up an equation of motion for particle P:
F = ma
T – 8mg = gm438 ×
T = 14mg (1)
b) Considering particle Q:
Kmg – T = gkm43×
Using (1):1 Kmg4 = 14mg
K = 56
The next problem involves a pulley system where one of the particles is being dragged across a horizontal table as the other particle is falling. The problem will be made more complex when the horizontal table is considered to be rough. The questions are increasing in complexity but the same basic principles apply.
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Example 10
A particle A, of mass 0.9kg, rests on smooth horizontal table and is attached to one end of a light inextensible string. The string passes over a smooth pulley P fixed at the edge of the table. The other end of the string is attached to a particle B of mass 1.8kg which hangs freely below the pulley. The system is released from rest with the string taut and B at a height of 3.2m above the ground. In the subsequent motion A does not reach the pulley before B reaches the ground. Find:a) the tension in the string before B reaches the ground.b) the time taken by B to reach the ground.Then, to make the model more realistic, assume that the coefficient of friction between the particle and the table is 0.3. Using this modification find the time taken by B to reach the ground.
a) Setting up equations of motion for the two particles gives:
A F = ma B F = ma
T = 0.9a 1.8g – T = 1.8a
AP
B
3.2m
1.8g
0.9g
R
T
T
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Adding the two equations gives:
1.8g = 2.7a
a = 6.53ms-2
Therefore:T = 0.9 × 6.53
T = 5.88N
b) The particle B is falling with acceleration 6.53ms-2. So by using constant acceleration equation we can find the time it takes to reach the floor.
u = 0, a = 6.53, s = 3.2, t = ?
2
2
1s ut at2
13.2 6.53 t2
t 1.01sec
= +
= × ×
=
Seeing as the particle is moving friction must be at its maximum value and hence FR = μR.Setting up equations of motion for the two particles with friction included gives:
A F = ma B F = ma
T - FR = 0.9a (1) 1.8g – T = 1.8a (2)
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Resolving vertically for A:
R = 0.9g
Using FR = μR FR = 0.27g
Adding equations (1) and (2) gives:
1.8g - 0.27g = 2.7a
a = 5.553ms-2
The new value of acceleration can now be used to calculate the new time:
u = 0, a = 5.553, s = 3.2, t = ?
2
2
1s ut at2
13.2 5.553 t2
t 1.07sec
= +
= × ×
=
In the next problem a particle is being pulled up a rough inclined plane by the motion of another particle falling towards a floor. This is very similar to an exam question and would be worth in excess of 10 marks on an M1 paper.
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Example 11
A particle, A of mass 6kg, rests on a rough plane inclined at an angle of 35º to the horizontal. The particle is attached to one end of a light inextensible string which lies in a line of greatest slope of the plane and passes over a light smooth pulley P fixed at the top of the plane. The other end of the string is attached to a particle B of mass 16kg. The particles are released from rest with the string taut. The particle B moves down with an
acceleration of 2 g5 .
Find:a) the tension, T, in the string.b) the coefficient of friction between the plane and A.
a) Setting up equations of motion for A and B gives:
A F = ma B F = ma
T - 6g sin 35º - FR = 6 × 2 g5 (1) 16g – T = 16× 2 g
5 (2)
Using (2) to find the tension:
T = 48 g 94.08N5
=
R P
B
x
T
T
F
6g35º
A
16g
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Resolving vertically for A:
R = 6gcos35º = 48.166
Using FR = μR FR = 48.166μ
Therefore equation (1) becomes:
1294.08 33.726 - 48.166μ g 5
− =
Rearranging for μ:
1260.35 - g5μ
48.166
μ 0.765
=
=
Extension
Assume that in the example above the particle is 5m above a level surface and that after 0.5sec the string breaks. Calculate the total time that the particle B is in flight and the distance that A moves up the plane before it comes to instantaneous rest (assume that it does not reach the pulley).
This is a rather large value for μ and this obviously accounts for the low value for the acceleration.What is the resultant force on the pulley?
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Momentum and Impulse
Momentum
The momentum of a body of mass m, having a velocity v is mv. The units of momentum are Newton seconds (Ns).
Momentum = mv
The momentum of a body is dependent upon its velocity therefore momentum is a vector quantity. This implies that direction is very important and great care must be taken with signs.
Example 12
Find the momentum of a hockey ball of mass 0.9kg hit at 18ms-1.
Momentum = mass × velocity
Momentum = 0.9 × 18 = 16.2Ns
Change in Momentum
If a particle experiences a change in velocity then, by definition, its momentum must change. Let the initial velocity be u and the final velocity be v then the change in momentum is given by:
Change in Momentum = mv – mu = m( v - u )
Example 13
If the hockey ball from example 12 hits a wall directly and returns with a velocity of 12ms-1. Find the change in momentum.
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Always draw a diagram.
Taking left to right as positive, therefore u = 18, v = -12
Change in Momentum = m( v - u )
= 0.9 × (-12 – 18 )
= -27Ns
The wall in the above example has experienced a force as the hockey ball hits it. This action takes place in a very short time period and is called the impulse.
Impulse
When a force F, is applied to a particle for a period of time t, then this quantity is defined as the impulse of the force. Obviously an impulse will bring about a change in velocity and therefore momentum will change.Therefore:
Impulse = F × t = m(v- u )
18ms-1
0.9kg
Before
12ms-1
0.9kg
After
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The derivation of the formula comes from the equation of motion and constant acceleration equations:
F = ma v = u + at
v uat−=
F = v um
t−
Therefore Ft = m(v – u)
Example 14
A particle of mass 7.5kg is acted on by a force for 6 seconds and in the process its velocity increases from 6ms-1 to 15ms-1. Find the magnitude of the force.
Impulse = change in momentum
F × t = m(v- u )
F × 6 = 7.5 (15 – 6)
F = 11.25N
Example 15
A ball of mass 1.2kg is moving vertically with a speed of 14 ms-1
when it hits a smooth horizontal floor. It rebounds with a speed 8 ms-1. Find the magnitude of the impulse exerted by the floor on the ball.
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Take care with signs
Assuming down to be positive
Impulse = Change in Momentum
= m( v – u )
= 1.2( -8 – 14 )
= 1.2( – 22 )
= -26.4 Ns
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The Principle of the Conservation of Momentum
When a collision occurs between two bodies, A and B, then the force exerted on A by B will be equal and opposite to the force exerted on B by A (by application of Newton’s third law). If no other forces are present then the change in momentum in one particle will equate to the loss of momentum in the other particle. Momentum is conserved and therefore the sum of the momentum of the particles before collision must equal the sum of momentum after the collision. This is referred to as the Principle of Conservation of Momentum.
If two particles of masses, m1 and m2, with initial velocities u1 and u2, collide then, given that their final velocities are v1 and v2 we can say that:
m1u1 + m2u2 = m1v1 + m2v2
The following examples illustrate the principle and once again it is always best to draw a diagram as this will help to avoid mistakes with signs and direction.
Example 16
Two particles A and B have masses of 1.2kg and 1kg respectively. Particle A is moving towards a stationary particle B with a velocity of 3 ms-1. Immediately after the collision the speed of A is 2.1ms-1 and its direction is unchanged. Find:a) the speed of B after the collision;b) the magnitude of the impulse exerted on A in the collision.
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a)
By conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
3 × 1.2 = 2.1 × 1.2 + v
3.6 – 2.52 = v
v = 1.08 ms-1.
b) Impulse = Change in Momentum
= m( v – u )
= 1.2( 2.1 – 3 )
= -1.08 Ns
3ms-1
1.2kg 1kg
Before
0ms-1
A B
1.2kg 1kg
After
2.1ms-1 v ms-1
A B
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Example 17
Two small balls A and B have masses 1.2kg and 2.5kg respectively. They are moving in opposite directions on a smooth horizontal surface when they collide directly. Immediately before the collision, the speed of A is 4.5ms-1 and the speed of B is 0.9ms-1. The speed of A immediately after the collision is 1.3ms-1. The direction of A remains unchanged after the collision. Find:a) the speed of B immediately after the collision;b) the magnitude of the impulse exerted on B in the collision.
a) By conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
1.2 × 4.5 – 2.5 × 0.9 = 1.2 × 1.3 + 2.5v
v = 0.636ms-1
b) Impulse = Change in Momentum
= m( v – u )
= 2.5 ( 0.636 - - 0.9 )
= 3.84Ns
4.5ms-1 0.9ms-1
1.2kg 2.5kg
Before
A B
1.2kg 2.5kg
After
1.3ms-1 v ms-1
A B
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Example 18
A locomotive A, of mass 1800kg is moving along a straight horizontal track with a speed of 9ms-1. It collides directly with a stationary coal truck, B, of mass 1000kg. In the collision, A and B are coupled and move off together.a) Find the speed of the combined train.b) After collision a constant breaking force of magnitude R Newtons is applied. The train comes to rest after 15 seconds. Find the value of R.
a) By conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
9 × 1800 = 2800 × v
v = 5.79ms-1
b) The combined train comes to rest in 15 seconds therefore we need to calculate the acceleration for use in an equation of motion.v = 0, u = 5.79, a = ?, t = 15
Using: v = u + at0 = 5.79 + 15aa = -0.386ms-2
0ms-1
1800kg 1000kg
9ms-1
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Assuming that the train acts as one body:
Equation of motion:
F = ma
R = -2800 × 0.386
R = 1080.8N = 1.10KN
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Jerk in a String
If two particles are connected by a light inextensible string and one of the particles is projected away from the other then at some point there will be a jerk in the string. At the instant before the jerk one of the particles will have momentum. As soon as the string becomes taut the particles will move onwards with the same velocity. The overall momentum must be conserved so therefore the velocity after the jerk must be lower than the initial velocity. This idea is best explained through an example.
Example 19
Two particles P and Q of masses 4kg and 7.5kg respectively are connected by a light inextensible string which is initially slack. Q is projected away from P with velocity 5ms-1. When the string becomes taught the two particles move on together with a common speed. Find the common speed and the impulse exerted on P by the string.
Using the conservation of momentum where v1 = v2 :
m1u1 + m2u2 = m1v1 + m2v1
4 × 0 + 7.5 × 5 = 4 × v1 + 7.5 × v1
37.5 = 11.5v1
v1 = 3.26ms-1
So the common speed is 3.26ms-1
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Impulse is the change in momentum so considering particle P:
Impulse = m(v – u)
= 4(3.26 – 0)
= 13.0Ns
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Questions B
1 A hockey ball of mass 0.125kg, is moving horizontally at 22ms-1 when it hits a vertical kick board at right angles. The ball rebounds horizontally at 12ms-1.a) Find, in Ns, the impulse of the force exerted by the ball on the kick board.
Given that the ball is in contact with the kick board for 0.15s:b) Find, in N, the force, assumed constant exerted by the ball on the kick board.
2 Two particles of mass 5kg and 10kg are connected by a light inextensible string which passes over a smooth fixed pulley. The system is released from rest with the string taut. Find the acceleration of the system and the tension in the string.
3 A particle of mass 0.8kg rests on rough horizontal table and is attached to one end of a light inextensible string. The string passes over a smooth pulley P fixed at the edge of the table. The other end of the string is attached to a particle P of mass 1.4kg which hangs freely below the pulley. The coefficient of friction between the particle and the table is 0.45. The system is released from rest with the string taut and B at a height of 1.2m above the ground. At the point of release A is 1.8m from P. Find:a) the acceleration of the particles;b) the time taken by B to reach the ground.c) the speed with which A hits P.
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4 Thomas the tank engine has a mass of 12000Kg and is moving along horizontal rails at 1.2ms-1, strikes buffers and is brought to rest in 0.45s. a) Calculate the impulse, in Ns, of the force exerted by the buffers on Thomas in bringing him to rest.b) Calculate the magnitude of this force assuming it to be constant.
5 Two particles A and B, of masses 4m and 2m respectively are moving towards each other on a smooth horizontal surface with speeds 9v and 3v respectively. The particles collide directly and after the collision A continues to move in the same direction but its speed is halved. Find:a) the speed of B after the impact.b) the magnitude of the impulse exerted by A on B.
6 A bullet is fired horizontally with a speed of 450ms-1 into a block of wood of mass 0.15kg that is placed on a smooth horizontal surface. Given that the block begins to move with a velocity of 9ms-1, find, in kg the mass of the bullet.
AP
B
1.2m
1.4g
0.8g
R
T
T
Page 41
7 Two particles P and Q of masses have masses 8kg and m, they are connected by a light inextensible string which passes over a smooth fixed pulley. The system is released from rest with the string taut and the hanging parts of the string vertical. After 1.5 seconds P has fallen 2.5m. Assuming that Q does not reach the pulley, calculate the tension in the string and the value of m.
8 Two particles A and B of masses 2kg and 3kg are connected by a light inextensible string which passes over a smooth fixed pulley, as outlined in the diagram below. The system is released from rest with the string taut. Find the acceleration of the system and the tension in the string if:a) both planes are smooth;b) both planes are rough and the coefficient of friction between the particles and the plane is 0.1.
A
RP
B
T
F
2g35º
3g
S
55º
T
Page 42
9 A particle, A of mass 4kg, rests on a rough plane inclined at an angle of 30º to the horizontal. The particle is attached to one end of a light inextensible string which lies in a line of greatest slope of the plane and passes over a light smooth pulley P fixed at the top of the plane. The other end of the string is attached to a particle B of mass 13kg. The particles are released from rest with the string taut. Given that the coefficient of friction between the particle A and the inclined plane is 0.35 calculate:a) the tension in the string and the acceleration of the system.b) the angle of inclination required to increase the acceleration by 50%.
10 A body of mass 6kg is moving with velocity (4i + 7j)ms-1 when an impulse is applied. The impulse causes its velocity to change to (-3i - 5j)ms-1. Find the impulse.
11 A body of mass 7.5kg is initially at rest on a smooth horizontal surface, experiences a force (8i - 13j)N for 3 seconds. Find the final velocity of the body and its speed.
R P
B
T
T
F
4g30º
A
13g
Page 43
12 A pile driver of mass 250 Kg strikes a pile of mass 450kg and drives it into the ground. The pile driver strikes the pile directly with a velocity of 8.5ms-1. The driver does not rebound and in the subsequent motion the pile and the drive move as one.a) Calculate the common speed of the pile and driver immediately after impact.b) Calculate the impulse exerted by the driver on the pile.c) The pile and the driver penetrate 0.55m into the ground before coming to rest. Assuming that the ground exerts a constant resistive force on the pile and driver, calculate the magnitude of the force.
Kinematics is the study of the motion of particles. In M1 all motion will have constant acceleration. This leads to the development of the constant acceleration equations.
Definitions
a = acceleration (ms-2)u = initial velocity (ms-1)v = final velocity (ms-1)t = time (second)s = displacement (metres)
If we know any of the three we can find the other two.
Page 2
From GCSE you should remember that:change in velocityacceleration
time
v-u a = t
so v = u + at (1)
=
∴
As mentioned above the acceleration is constant hence the average velocity is simply the average of u and v.
u vaverage velocity2+=
Another definition is that:displacement saverage velocity
time t= =
Therefore:u v s
2 t
u v s = t (2)2
+ =
+ ∴ By using equation (1) we can eliminate v.
2
u u at s = t2
1s = ut + at (3)2
+ + ∴
If equation (1) is rearranged to make t the subject:v-ut = a
Then by substituting into equation (2):
2 2
2 2
u v v-u s = 2 a
2as v - u
v = u + 2as (4)
+ ∴
=
Page 3
You must learn all four equations above and remember that they only apply to constant acceleration problems. Since velocity is a vector quantity, getting the direction right in these problems is vital.
Example 1
A particle is moving in a straight line from O to P with a constant acceleration of 4ms-2. Its velocity at P is 48ms-1 and it takes 12 seconds to travel from O to P. Find (a) the particle’s velocity at O and (b) the distance OP.With all problems of this nature, write down what you are given and the one required. It should then be obvious as to which equation you need to use.
a) a = 4, u = ?, v = 48, t = 12Obviously we need to use equation (1):
v = u + at 48 = u + 4 × 12
Therefore u = 0ms-1
b) a = 4, u = o, v = 48, t = 12, s = ?Using equation (3):
2
2
1s = ut + at2
1s 4 122
s = 288m
= × ×
Example 2
Percy starts from rest at a station S and moves with constant acceleration. He passes a signal box B 14 seconds later with a speed of 84kmh-1. Modelling the train as a particle, find the acceleration of the train in ms-2 and the distance in metres between the station and the signal box.
Page 4
u = 0, t = 14, v = 84kmh-1 a = ?, s = ?Using equation (1) :
v = u + at84000/60/60 = 0 + 14aa = 1.67ms-2
Once again using equation (3) 2
2
1s = ut + at2
1 5s = 142 3
s 163.33m = 163m
× ×
=
Example 3
ET is travelling up a hill on his BMX. He experiences a constant retardation of magnitude 3ms-2. Given that his speed at the bottom of the hill was 15ms-1 determine how far he will travel before he comes to rest.Retardation is simply deceleration.a = -3, u = 15, v = 0, s = ?There is no mention of time therefore we use equation (4):
2 2v = u + 2as
0 = 225 - 2 3 s
s = 37.5m
× ×
Page 5
Vertical Motion Under Gravity
A few assumptions need to be stated before continuing.1. Objects will be treated as particles.2. Motion will only be in a straight line.3. No evidence of spinning or turning of objects.4. Particles will have constant acceleration of g (9.8ms-2).
If an object is projected vertically upwards and it falls 3m below the point of release then the time taken can be calculated by setting s = -3m and a = -9.8 and substitute the values into equation (3). A lot of students work out the time the object takes to reach the top, then the time to return to point of release and finally the -3m part, but this much more complicated than what you need to do!
It is also worth noting at this point that time taken to reach maximum height and fall back down to the point of release are the same. This type of question regularly appears on exam papers and it is far quicker to use the first method.
Example 4
A Kinder Surprise falls off a shelf which is 0.9 m above the floor. Find:(a) the time it takes to reach the floor(b) the speed with which it will reach the floor.
a) s = 0.9, a = 9.8, u = 0, t = ?Using equation (3):
2
2
1s = ut + at2
10.9 = 9.8 t2
s 0.429sec
× ×
=
Page 6
b) s = 0.9, a = 9.8, u = 0, t = 0.429, v = ?Using equation (1):
v = u + atv = 0 + 9.8 × 0.429v = 4.20ms-1
Example 5
A bouncy ball B is projected vertically upwards from a point O with a speed of 42ms-1. Find:(a) the greatest height h above O reached by B(b) the total time before B returns to O(c) the total distance travelled by the particle.With questions of this nature one has to be careful as to which direction is positive.
a) At the greatest height, v = 0,u = 42, a = -9.8, s = ?Using equation (4):
2 2
2
2
v = u + 2as
0 = 42 - 2 9.8 s
42s = 19.6
s 90m
× ×
=
42ms-1
Max Height
Acceln = -9.8
Start/End
Page 7
b) When the ball returns to B its displacement will be zero!s = 0, a = -9.8, u = 42, t = ?Using equation (3):
2
2
1s = ut + at2
0 = 42t-4.9tFactorising gives:
7t( 0.7t - 6 ) = 0
Therefore t = 0,t = 8.57sec
c) In part (a) we found the distance to the top therefore we only need to double the answer.
Total distance = 180m
Example 6
A cricket ball is thrown vertically upwards with a velocity of15ms-1. Modelling the ball as a particle moving under gravityalone, find for how long its height exceeds 10 m.
We need to find the times at which the ball has a displacement of 10m.a = -9.8, u = 15, s = 10, t = ?Using equation (3):
2
2
2
1s = ut + at2
10 = 15t - 4.9t
4.9t - 15t + 10 = 0By using the quadratic formula we find the two values of t.
15 225 4 10 4.9t9.8
± − × ×=
t = 2.08 or 0.98.Therefore the ball is above 10m for 1.1 second.
Page 8
Questions A
1 A car moves with constant acceleration along a straight horizontal road. The car passes the point A with speed 7ms-1 and 5 seconds later it passes the point B, where AB = 53m.a) Find the acceleration of the car.When the car passes the point C, it has a speed of 26ms-1.b) Find the distance AC.
2 A competition diver makes a dive from a highboard into a pool. She leaves the board vertically with a speed of 4ms-
1upwards. When she leaves the board, she is 6m above the surface of the pool. The diver is modelled as a particle moving vertically under gravity alone and it is assumed that she does not hit the springboard as she descends.a) Find her speed as she reaches the surface of the pool.b) Find the time taken to reach the surface of the pool.c) State two physical characteristics that have been ignored in the model.
3 A car moves from rest at a point O and moves in a straight line. The car moves with constant acceleration 5ms-2 until it passes the point A when it is moving with speed 14 ms-1. It then moves with constant acceleration 2 ms-2 for 8 seconds until it reaches the point B.a) Find the speed of the car at B.b) Find the distance OB.
Page 9
4 An aircraft moves along a straight horizontal runway with constant acceleration. It passes a point A on the runway with speed 15ms-1. It then passes the point B on the runway with speed 39ms-1. The distance AB is 190m.a) Find the acceleration of the aircraft.b) Find the time taken by the aircraft in moving from A to B.c) Find the speed of the aircraft when it passes the mid point of A and B.
5 A racing car moves along a straight horizontal road with constant acceleration. It passes the point O with speed 11ms-1. It passes the point A, 4 seconds later with speed 55ms-1.a) Find the acceleration of the car.b) Find the distance OA.c) Find the speed of the car as it passes the midpoint of OA.
6 A body is projected vertically upwards from ground level at a speed of 49ms-1. Find the length of time for which the body is at least 78.4m above the ground.
7 A body is projected vertically upwards from ground level at a speed of 14ms-1. Find the height of the body above the level of projection after:a) 1 second of motionb) 2 seconds of motionFind the distance travelled by the body in the 2nd second of motion.
Page 10
8 A balloon is moving vertically upwards with a steady speed 3ms-1. When it reaches a height of 36m above the ground an object is released from the balloon. The balloon then accelerates upwards at a rate of 2ms-2. Finda) the greatest height of the object above the ground.b) the speed of the object as it hits the ground.c) the time taken by the object from leaving the balloon to striking the ground.d) the speed of the balloon as the object hits the ground.
9 A cyclist travels on a straight road with a constant acceleration of 0.6 ms-2. P and Q are 120m apart. Given that the cyclist increases his speed by 6ms-1 as he travels from P to Q. Find:a) the speed of the cyclist at Pb) the time taken to travel from P to Q
10 Oblivion at Alton Towers has a vertical drop of 60m. Assuming that the ride starts from rest, calculate the speed of the ride at the bottom of the pit.
Page 11
Speed Time Graphs
Definitions
For constant acceleration problems the speed time graph will be a straight line.The gradient of the graph is the acceleration.The area under the graph represents the distance traveled.
Example 1
A body starts from rest, accelerates uniformly to a velocity of 8ms-1 in 2 seconds, maintains that velocity for a further 5 seconds, and then retards uniformly to rest. The entire journey takes 11 seconds.Find:
a) the acceleration of the body during the initial part of the motion.
b) The retardation of the body in the final part of the motion.c) The total distance traveled by the body
In these type of questions it is vital that you make a sketch of the motion.
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9 10 11 12
tim e (s)
velo
city
(m/s
)
Page 12
a) As the definitions suggested earlier the acceleration is simply the gradient of the velocity time graph. This is:
grad = 8/2 = 4ms-2.b) Similarly for the retardation:
grad = 8/4 = 2ms-2.c) The total distance travelled by the body is equal to the area of under the graph.
Area = 0.5 × 8 × ( 11 + 5 ) = 64m
Example 2
A car accelerates uniformly from rest to a speed of 10ms-1 in T seconds. The car then travels for 4T seconds and finally decelerates uniformly to rest in a further 70s. The total distance traveled by the car is 1250m. Find:a) the value of T.b) the initial acceleration of the car.Once again, sketch the journey.
a) the area under the graph must equal 1250.1250 = 10T + 40T + 700/2900 = 50TT = 18sec.
b) The car accelerates to 10ms-1 in 18 seconds, therefore the acceleration is:
Acceln = 10/18 = 0.556ms-2
T 5T 5T+70
10ms-1
Page 13
Example 3
A train is traveling along a straight track between the points P and Q. It starts from rest at P and accelerates at 3ms-2 until it reaches a speed of 54ms-1. It continues at a constant speed of 54ms-1 for a further 180seconds and then decelerates at a constant deceleration of 6ms-2.a) sketch a speed time graph for the train’s journey.b) calculate the total time for the journey from P to Q.c) calculate the distance between P and Q.
a) The train will take 54/3 seconds to reach the constant speed (18seconds). It will take 54/6 seconds to decelerate to rest (9 seconds). Hence the graph will have the following shape.
b) the total time for the journey is 207 seconds.c) the total distance is once again equal to the area under the graph.Area = (54 × 18)/2 + 54 × 180 + (54 × 9)/2Distance = 10449m = 10.4Km
18 198 207
54ms-1
Page 14
Problems involving two vehicles
Example 4
Two particles A and B are traveling along a straight path PQ of length 20m. A leaves P, heading for Q, from rest with acceleration of 2ms-2 and at the same time B leaves Q, in the direction of P, from rest with a constant acceleration of 5ms-2. Find how far from A the two particles collide?Let the point of collision be x metres from P and hence (20 – x)m from Q.So for particle A: u = 0 a = 2 t = T s = xUsing:
2
2
1s = ut + at2
x = t (i)So for particle B: u = 0 a = 5 t = T s = 20 - xUsing
2
2
1s = ut + at2
20 - x = 2.5tSubstituting (i) gives:
2 220 - t = 2.5t
t 2.39sec=Using (i) again:
x = 2.392
x = 5.71m
Page 15
Questions B
1 The diagram shows the sketch of a velocity time graph for a particle moving in a straight line. Find the value of T given that:
a) the distance covered is 100m.b) the constant speed attained by the particle is 10ms-1
c) the rate of acceleration is twice the rate of retardation.d) The time spent at constant speed is equal to the total
time spent accelerating and decelerating.
2 Two trains, P and Q, run on parallel straight tracks. Initially both trains are at rest. At time t = 0, P moves off with constant acceleration for 15 seconds until it reaches a speed of 30ms-1 . P then continues at a constant speed. At time t = 35seconds Q moves off with the same acceleration until it reaches a speed of 55ms-1 . Q then continues at this constant speed. Train Q overtakes P whilst both trains are travelling at constant speeds after time T. Sketch a speed time graph for the journeys and find the value of T.
3 A particle Q starts from rest at a point O and accelerates at 3ms-2 until it reaches a speed V ms-2. Q then continues at the constant speed for 90 seconds before decelerating at 1.5 ms-2 to come to rest at the point R. If the entire journey takes 150 seconds find the distance QR. Make sure that you draw a speed time graph.
T
10ms-1
Page 16
4 A racing car moves in a straight line. The car accelerates at 8ms-2 for 5 seconds, maintains a steady speed for 15 seconds and then decelerates to rest in 9 seconds. Sketch a velocity time graph for the car and calculate the total distance travelled.
Equilibrium of a lamina under parallel forces.................................. 5Example 5............................................................................................. 5
Non uniform beams.................................................................................8Example 7............................................................................................. 8
The moment of a force is the measure of its capacity to turn the body on which it is acting.
Moment = Force × Perpendicular Distance
Example 1
A door of width 1.2m is being pushed by a force of 25N. Find the moment about the hinge.
Moment = Force × Perpendicular Distance
= 25 × 1.2 = 30Nm
Page 2
Example 2
Forces of magnitude 6N, 8N and 5N are applied to a light rod AB, of length 8m, as outlined in the diagram below.
Calculate the sum of the moments about the midpoint of the rod.
Moment of 6N force is given by: 6 × 4 = 24Nm clockwiseMoment of 8N force is given by: 8 × 2 = 16Nm clockwiseMoment of 5N force is given by: 5 × 4 = 20Nm anti-clockwise
Therefore the sum of the moments is 20Nm clockwise.
It is more likely however that the forces are not being applied at right angles to the object.
The diagram above shows a force F, acting on an object P at a given angle θ and given distance d. The force brings about a turning effect at P. The dotted line is the perpendicular distance. x = d × Sinθ.
A B8m
2m
6N 8N 5N
θ
F
Px
d
Page 3
Therefore the moment about P is given by:
Moment = Force × Perpendicular Distance from line of action of force to pivot point
Moment = FdSinθ
Example 3
Two forces are applied to a light rod as outlined in the diagram below. Find the sum of the moment about P.
The 6N force gives a 30Nm clockwise moment.
The dotted line has been added to show the perpendicular distance. The 8N force gives an anticlockwise moment of magnitude:
8 × 3 × Sin30 = 12Nm.
P
5m
3m30º
6NN
8N
P
5m
3m30º .
6NN
8N
Page 4
Therefore the system is experiencing a clockwise moment of magnitude 18Nm.
Example 4
A point Q(4,6) is acted upon by a force (2i + 5j)N. Calculate the sum of the moment about the origin.
The 2i force gives a clockwise moment of: 2 × 6 = 12NmThe 5j force gives a anticlockwise moment of: 5 × 4 = 20NmTherefore the sum of the moments is 8Nm anticlockwise.
Find the sum of the moments about the point R(7,3).The 2i force gives a clockwise moment of: 2 × 3 = 6NmThe 5j force gives a clockwise moment of: 5 × 3 = 15NmTherefore the sum of the moments is 21Nm clockwise.
4
6 Q 2i
5j
7
3 R
Page 5
Equilibrium of a lamina under parallel forces
If a system is in equilibrium then the following must hold:
1. The component of the resultant force in any direction must be zero.
2. The algebraic sum of the moments about any point must be zero.
For uniform non light systems the weight acts through the centre of mass. A rod is said to be uniform if it has even weight distribution and therefore the centre of mass acts at the centre of the rod.
Example 5
A uniform rod PQ of length 7m and mass 12kg is pivoted at the point R where PR is 2.5m. Calculate the mass of the particle that must be attached at P to maintain the rod in a horizontal position.
The moment at P about R must equate to the moment at G about R.The mass of the rod gives a clockwise moment of:
12g × 1 = 12gNm
The mass added at P gives an anticlockwise moment of: xg × 2.5 = 2.5xNm.
Therefore: 12gNm = 2.5xNm
P Q7m
2.5m
xgN 12gN
RG
Page 6
x = 4.8Kg
Uniform Beams
Example 6
A uniform beam AB of mass 40kg and length 12m is supported in a horizontal position at C and D, where AC = 1.5m and DB = 1m. A man of mass 80Kg stands on the beam at the point E where EB = 2.5m. Find the reactions at C and D.
Examiners always suggest that a diagram is VITAL.
Resolving vertically gives:
X + Y = 120g (1)
Taking Moments about C gives:
The mass of the beam gives a clockwise moment of: 40g × 4.5 = 180gNmThe mass of the man gives a clockwise moment of: 80g × 8 = 640gNmThe reaction at D gives an anticlockwise moment of: 9.5YNm
Therefore:9.5Y = 820g
A
40g
DC
E
80g
X Y
B1.5m 1m12m
Page 7
= 1640Y g19 = 845.9N
Using equation (1) from above
X + Y = 120g
X + 845.9 = 120g
X= 330N
Page 8
Non uniform beams
Example 7
A non uniform beam AB of mass 20kg and length 12m has an object of mass 40kg placed at a point 8m from A. The beam is in equilibrium in a horizontal position resting on a support C at the midpoint of AB. Find the position of the centre of mass.
Taking moments about C gives:
40g × 2 = 20g × y
Y = 4m
A
C
E
40g
X
B
12m
20g
y
Page 9
Questions
(some diagrams have forces missing, you need to figure out which ones!)
1 A uniform rod AB of weight 70N and length 5m. It rests in a horizontal position supported at point C and D, where AC = 0.4m. the reaction on the rod at C has magnitude 9N. Find
a) the magnitude of the reaction on the rod at Db) the distance AD.
2 A uniform rod AB of length 6m and mass 40Kg. It is supported by two smooth pivots in a horizontal position at A and C where AC = 3m. A woman of mass 75Kg stands on the rod which remains in equilibrium. The magnitudes of the reactions at the two pivots are equal to R Newtons.
Finda) the value of R.b) the distance of the woman from A.
A
DC
9 Y
B
A
C
B
Page 10
3 A non uniform plank of wood AB of length 10m and mass 100Kg is smoothly supported in a horizontal position at A and B. An object of mass 90Kg is put on the plank at C, where AC = 6m. The plank is in equilibrium and the magnitudes of the reactions at A and B are equal. Find:a) the magnitude of the reaction R, on the plank at B.b) the distance, x, of the centre of mass of the plank from A.
4 A uniform plank AB has weight 80N and length xm. The plank rests in equilibrium on two supports at A and C, where AC = 3m. A rock of weight 20N is placed at B and the plank remains in equilibrium. The reaction on the plank at C has magnitude 70N.a) find the value of x
The support at A is now moved to a point D on the plank and the plank remains in equilibrium with the rock at B. The reaction on the plank at C is now two times the reaction at D.b) find the distance AD. 5 A non uniform rod, AB, of length 7m and mass 10Kg is suspended in equilibrium in a horizontal position by ropes attached to the points E and F of the rod, where AE = 2m and AF = 6m. The tensions in the ropes are equal. Find the distance of the centre of mass from A.
A
C
B6m
A
C
Bxm
Page 11
Extension
1 A large log AB is 10m long. It rests in a horizontal on two supports C and D, where AC = 1m and BD = 1m. An estimate of the weight of the log is required, but the log is too heavy to lift off the supports. When a force of magnitude 1100N is applied vertically to the log at A, the log is about to tilt about D. a) state the value of the reaction on the log at C for this case.b) by modeling the log as a uniform rod, estimate the weight of the log.
The force at A is removed and a force vertically upwards is applied at B. The log is about to tilt about C when the force has a magnitude of 1600N. By modeling the log as a non uniform rod, with the distance of the centre of mass of the log being x metres from A, find:c) a new estimate for the weight of the log.d) the value of x.
Example 1.............................................................................................. 3Example 2 ........................................................................................... 5
Resolving Forces into Components...................................................... 6Resolving Several Forces into Components.................................. 6Example 3............................................................................................. 7Equilibrium of Coplanar Forces........................................................8Example 4............................................................................................. 8
Rough and Smooth surfaces........................................................... 13Limiting Equilibrium.......................................................................... 13Coefficient of Friction (μ).............................................................. 14Example 5............................................................................................ 15Non Horizontal Forces.....................................................................16Example 6............................................................................................ 17
Objects on Inclined Planes................................................................. 18Example 7............................................................................................ 18Varying Values of Friction.............................................................. 19Other external forces on inclined planes...................................20Example 8........................................................................................... 20Example 9........................................................................................... 22
Questions B........................................................................................... 24Solutions to Exam type questions.................................................... 27
Statics is the study of stationary objects. We will consider a variety of situations where bodies are acted upon by a number of forces. A few of the concepts introduced in our work on vectors will be built upon in this unit.
Page 2
Resolving ForcesIn the vectors unit we were made aware of the fact that the resultant of two vectors is the diagonal of a parallelogram, as highlighted in the diagram below.
This idea can be applied to forces:
In a real world sense the path R is the direction that a particle would take if it were to be acted upon by the forces F1 and F2. This principle can be applied to more than two forces.
Suppose that a particle is acted upon by the forces F1, F2 and F3.
The diagrams below should explain the path that the particle will follow.
F1
F2 R
F1
F2
F3
Page 3
Firstly find the resultant of the forces F1, and F2 to give R.
Second find the resultant of R and F3 .
The path R2 shows the direction of motion of the particle as it is the resultant of the three forces F1, F2 and F3.
Example 1
Two forces act on a particle as outlined in the diagram below. Find the resultant force acting on the object and the angle it makes with the 10N force.
We need to find the magnitude and direction of the force R.
F1
F2
RF3
F1
F2
RF3
R2
10N
6N50º
Page 4
Using Cosine Rule:
2 2 2R 6 10 2 6 10cos130
R 213.1
= + − × ×
=
Using Sine rule to find the angle that the resultant R makes with the 10N force:
Sin 130 Sin R 6
6Sin 130Sin =213.1
18.4°
θ=
θ
θ =
Later work will involve more than two forces but the method used is the one we introduced in the vectors unit (resolving into i and j components).
Questions involving forces will also be given in i, j notation.
10N
6N50º 130º
6NR
Page 5
Example 2
Forces E, F and G are applied to a particle. Find the resultant of the three forces in terms of i and j components. Find the magnitude and the direction of the resulting force.
E = (7i – 3j)N, F = (-3i + 8j)N G = (-2i – 2j)N
Since the forces are in Cartesian components, R is found by adding the forces.
Therefore R = (2i + 3j)N
Using Pythagoras to find the magnitude:
2 2R 2 3
R 13
= +
=
And finally the angle θ:
3Tan 2
= 56.3°
θ =
θ
3j
2i
θ
Page 6
Resolving Forces into ComponentsThis concept was introduced in earlier work on vectors as it is far easier to deal with a number of forces if we can split them into their horizontal and vertical components.
The diagram below shows a 9N force acting at an angle of 30◦ to the horizontal. Find the components of the force in the horizontal and vertical direction.
x component = 9 × Cos 30º = 239 N
y component = 9 × Sin 30º = N29
Resolving Several Forces into Components
All we have to do is find all of the x components and add them together to give the x component of the resultant force. This idea is applied to the y component and from this we can find the direction of the resultant.
30º
9N
x
y
Page 7
Example 3
The diagram below shows a number of forces. Find the resultant of the forces and its direction.
The best way to attack this problem is to use a table:Force x
componenty component
3N - 3 04N 4 × Cos 30º 4 × Sin 30º6N 6 × Cos 40º -6 × Sin 40ºTotal 5.06 -1.86
The resultant can be represented pictorially as:
The resultant has a magnitude of:
N39.5
86.106.5 22
=
+=
In the direction of:
°−=
−− 2.20
06.586.11Tan
30º
4N
x
y
40º
6N
3N
5.06N
-1.86Nθ
Page 8
Equilibrium of Coplanar Forces
If a system is being acted upon by a series of forces that all lie in the same plane then it will be in equilibrium if their resultant vector is zero.
Example 4
A particle is in equilibrium under the forces (8i + 10j)N, (-6i -5j)N and (ai + bj)N. Find the values of a and b.
Equating coefficients gives:
8 + a – 6 = 0 10 – 5 + b = 0a = -2 b = -5
Page 9
Questions A1 In each of the following diagrams find the magnitude of the resultant and the angle it makes with the x axis.a) b)
c)
2 Find the angle between a force of 6N and a force of 5N given that their resultant has magnitude 9N.
3 The angle between a force of QN and a force of 3N is 150º. If the resultant of the two forces has a magnitude 8N find the value of Q.
x
y
6N
3N
40º
6N
x
y
35º
8N
2N
x
y
3N
Page 10
4 Each of the following diagrams shows a number of forces. Find the magnitude of their resultant and the angle it makes with the x axis.
5 For each of the sets of axis below find the sum of the components in the direction of:a) the x axis b) the y axis
(i) (ii)
10º
8N
x
y
6N
3N
80º
6N
x
y
30º
5N
4N
10º
7N
x
y
45º
6N
5N
55º
15º
9N
x
y7N
45º 65º
9N
x
y
10N
80º
Page 11
(iii) (iv)
6 Find the magnitude of the force (-3i + 8j)N and the angle it makes with the direction of i.
7 Find the resultant of the following forces:(2i + 4j)N, (3i - 5j)N, (6i + 2j)N and (-5i + 8j)N
8 Find the magnitude of the resultant in question 7 and find the angle it makes with the direction of i.
9 The resultant of the forces (5i - 2j)N, (7i + 4j)N, (ai + bj)N and (-3i + 2j)N is a force (5i + 5j)N. Find a and b.
10 Find the resultant of forces 5N, 7N, 8N and 5N in the directions, north, north east, south west and south east respectively, giving your answer in the form ai + bj
11 Two forces, P and Q, are such that the sum of their magnitudes is 45N. The resultant of P and Q is perpendicular to P and has a magnitude of 15N. Calculate:a) the magnitude of P and Q;b) the angle between P and Q.
30º
7N
x
y
45º
6N15º
8N
x
y
35º
6N
6N
45º
Page 12
12 Forces of magnitude 3N, 6N and 2N act at a point as shown in the diagram below. Show that the component of the resultant force in the i direction is zero. Calculate the magnitude of the resultant force and state its direction.
3Nx
y
2N
6N
60º
Page 13
Friction
Rough and Smooth surfaces
A block of mass M Kg on a horizontal table is acted upon by a force P Newtons. From Newton’s Third Law, it is known that equal and opposite forces act on the block and on the plane at right angles to the surfaces in contact.
The force F acts to oppose the motion and this is called the frictional force. If the surface were to be perfectly smooth then the block would accelerate across the surface. In general the surface is unlikely to be smooth and the block would move if the force P was greater than the frictional force.
Limiting Equilibrium
The frictional force in the above situation is not constant, but increases as the force P increases until if reaches a value Fmax. The block is then on the point of moving and the system is said to be in a state of limiting equilibrium.
PF
R
Mg
Page 14
Coefficient of Friction (μ)
Friction is proportional to the normal reaction and in limiting equilibrium it is given by:
Fmax = μR
Where μ is the coefficient of friction for the two contact surfaces.
Friction can also be expressed as:
F ≤ μR
When friction is less than μR motion will not take place.
Consider the following points:
1. At the point where friction can’t increase any further, motion is about to take place.
2. Note that friction is only dependent upon the nature of the surfaces in contact and not upon the contact area.
3. For perfectly smooth surfaces μ = 0. 4. Friction will never be larger than that necessary to prevent
motion. 5. It can be assumed that friction will have a maximum value
μR when motion occurs.6. Friction always acts to oppose the motion of an object and
great care must be taken with objects on slopes as friction could be acting either up or down the plane (see examples 7 and 8).
Page 15
Example 5
A block of mass 7.5Kg rests on a rough horizontal plane, the coefficient of friction between the block and the plane is 0.55. Calculate the frictional force acting on the block when a horizontal force P is applied to the block and the magnitude of any acceleration that may occur.
There is no motion perpendicular to the plane.
a) 15N b)65Na) Resolving perpendicular to the plane gives:
R = 7.5g
Friction will act to oppose motion and at its maximum valueFmax = μR, therefore:
μR = 0.55 × 7.5g
μR = 40.4N
In this example it is possible for friction to increase until it reaches a value of 40.4N. There is only a pushing force of 15N therefore friction will prevent motion.
PFR
R
7.5g
Page 16
b) Seeing as the pushing force is greater than 40.4N the block will accelerate across the surface. Using the value of μR from above (since we have reached the value of Fmax) and by setting up an equation of motion for the block we get:
F = ma
65 - 40.4 = 7.5 × a
a = 3.28ms-2
Note the difference between F and Fmax : The F is the sum total of all the forces acting in the horizontal direction.
Non Horizontal Forces
When the force P acting on the block of mass M is inclined at an angle to the horizontal, two effects must be considered.
(1) The vertical component of P alters the size of the normal reaction R. One needs to consider the direction of the applied force and its effect on the value of R.
(2) Only the horizontal component of the force P will bring about motion in the block.
It is worth noting at this point that since R has altered there will also be a change in the frictional force (F = μR).
P
F
R
Mg
θ
Page 17
Example 6
A box of mass 2Kg lies on a rough horizontal floor with the coefficient of friction between the floor and the box being 0.5. A light string is attached to the box in order to pull the box across the floor. If the tension in the string is TN, find the value that T must exceed for motion to occur if the string is 30º above the horizontal.
If motion is to take place then FR = μR.
Resolving forces perpendicular to the plane gives:
2g = R + T × Sin 30º
R = 19.6 – 0.5TUsing FR = μR
FR = 9.8 – 0.25T (1)
Resolving parallel to the plane gives:
FR = T × Cos 30º (2)
Equating (1) and (2) gives:
9.8 – 0.25T = 3T2
T = 8.78N
T
FR
R
2g
30º
Page 18
Objects on Inclined PlanesWhen objects are on inclined planes it is easier to resolve the forces parallel to the plane and perpendicular to the plane. This concept is best shown through an example.
Example 7
A particle of mass MKg rests in equilibrium on a rough plane inclined at an angle 20º to the horizontal. Find the normal reaction R and the frictional force in terms of M and g.
The normal reaction by definition only reacts to the component of the weight force that acts perpendicular to the plane (the orange line in the diagram below). Equally the frictional force is only acting against the component of the weight that is acting parallel to the plane (the pink line).
20º
R
FR
Mg
20º
R FR
Mg
20º
Page 19
This leads to the following two statements that will be used over and over again in M1.
R = Mg Cos 20º
FR = Mg Sin 20º
This is only valid if there are no other external forces.
Varying Values of Friction
In the introduction of the coefficient of friction we suggested that friction can vary. Now that we know how to resolve forces parallel and perpendicular to the plane we can use this new skill to explain the point.
Consider the case below with a 2kg mass on a rough surface inclined at an angle of 20º, where the coefficient of friction between the object and the surface is 0.4.
Resolving forces perpendicular to the plane gives:
R = 2g Cos 20º = 18.4N
Therefore:μR = 7.37N
Resolving parallel to the plane gives:
20º
R
FR
2g
Page 20
FR = 2g Sin 20º = 6.70N
The condition FR ≤ μR is upheld and as a result there is no motion down the plane.
If the slope was raised to 30º, resolving forces perpendicular to the plane gives:
R = 2g Cos 30º = 17.0NTherefore:
μR = 6.79N
Resolving parallel to the plane gives:
FR = 2g Sin 30º = 9.8N
Friction is working against the parallel component of the weight (mg Sin θ). By definition Sin θ increases as the angle increases therefore friction must increase to prevent motion, but it can only increase to the point where F = μR. In the second part of the example the condition F ≤ μR is no longer upheld and therefore the object would slide down the slope. The object in the example above would be in a state of limiting equilibrium for an angle between 20º and 30º (calculate the exact value).
Other external forces on inclined planes
Example 8
A mass of 2.5Kg rests on an inclined plane at 50º to the horizontal, and the coefficient of friction between the mass and the plane is 0.3. Find the force P, acting parallel to the plane, which must be applied to the mass in order to just prevent motion down the plane.
Page 21
Seeing as the mass is about to slide down the plane, friction must act up the plane.
Resolving the forces parallel to the plane gives:
Fmax + P = 2.5g Sin 50º
Resolving perpendicular to the plane:
R = 2.5g Cos 50º
Using Fmax = μR:0.3 × ( 2.5g Cos 50º ) + P = 2.5g Sin 50º
P = 14.0N
The next problem introduces the idea of forces acting on a particle on an inclined plane where the force acts at an angle to the plane.
R
Fmax
P
50º2.5g
Page 22
Example 9
A building block of mass 1.25kg is placed on an incline plane at an angle of 30º to the horizontal. The coefficient of friction between the box and the plane is 0.2. The box is kept in equilibrium by a light inextensible string which lies in a vertical plane. The string makes an angle of 15º with the plane. The box is in limiting equilibrium and is about to move up the plane. The tension in the string is T Newtons. Modelling the box as a particle, find the value of T.
Note that the frictional force is acting down the slope as the box is at the point of moving up the plane.Once again we need to resolve the forces into their components, but this time we must resolve them parallel and perpendicular to the plane.
Resolving parallel to the plane:(don’t forget to resolve the tension force)
1.25g × Sin 30 + FR = T × Cos 15
6.125 + FR = 0.966T
FR = 0.966T - 6.125 (1)
15º
T
30º
R
FR
1.25g
Page 23
Resolving perpendicular to the plane:
1.25g × Cos 30 = R + T × Sin 15
10.61 = R + 0.259T
R = 10.61 – 0.259TUsing F = μR
0.966T - 6.125 = 0.2 × ( 10.61 – 0.259T )
1.0178T = 8.247
T = 8.10N
Page 24
Questions B1 A block of mass 3Kg is initially at rest on a rough horizontal plane. The coefficient of friction between the block and the plane is 0.6. Calculate the pushing force required to cause it to accelerate along the surface at:a) 8ms-2 b) 0.1ms-2
2 A block of mass 12kg rests on a rough horizontal surface. The coefficient of friction between the block and the plane is 0.4. Calculate the frictional force when a horizontal force of 65N acts on the block. If the block moves, find the acceleration.
3 A block of mass 7kg rests on a rough horizontal surface. The coefficient of friction between the block and the plane is 0.15. A light string is attached to the box in order to pull the box across the floor. If the string makes an angle of 45º with the horizontal find the value that the tension in the string must exceed in order to bring about motion.
4 Referring to question 3 what value of tension would be required to give the block an acceleration of 3ms-2.
5 A block of mass 4.5kg rests on a rough horizontal surface. The coefficient of friction between the block and the plane is 0.25. State whether or not the block will slide when the surface is moved horizontally with an acceleration of of:a) 0.9ms-2. b) 2ms-2. c) 3.5ms-2.
6 For each of the following diagrams below the value of μ = 0.3. In each case calculate the magnitude of the force X if the body is just on the point of moving.
Page 25
a) b)
c) d)
e)
7 A body of mass 6.5kg is placed on an incline plane at an angle of 35º to the horizontal. The coefficient of friction between the box and the plane is 0.2. A horizontal force of 25N is applied to the body. Find the frictional force and the value of the acceleration. State the direction of motion.
8 A body of mass 4kg is released from rest on a rough surface that it inclined at an angle of 40º to the horizontal. If
X
30º
R
F
2g
X
10º
R
F
5g
X
FR
3g45º
X
FR
2.5g45º
15º
XF
R
3g40º
Page 26
after 2 seconds the body has a velocity of 4.9ms-1 down the surface, find the coefficient of friction between the body and the surface.
9 A horizontal force of 0.9N is just sufficient to prevent a brick of mass 750g from sliding down a rough plane inclined at an angle of 20º. Calculate the coefficient of friction between the plane and the brick.
10 A box of mass 5kg is placed on an incline plane at an angle of 35º to the horizontal. The coefficient of friction between the box and the plane is 0.45. Find the horizontal force that must be applied so that:a) the box is just prevented from sliding down the planeb) the box is just at the point of moving up the planec) the box moves up the plane with an acceleration of 1.25ms-2.
Page 27
Solutions to Exam type questions1
A box of mass 7.5kg is held in limiting equilibrium on a rough plane by the action of a horizontal force of magnitude 55N acting in a vertical plane through a line of greatest slope. The plane is inclined at an angle of 18º to the horizontal, as shown in the diagram above. By modeling the box as a particle and noting that it is on the point of moving up the slope:a) find the normal reaction R.b) find the coefficient of friction between the box and the plane.
a) Resolving perpendicular to the plane gives:
R = 7.5g × Cos 18 + 55 × Sin 18
R = 90.5N
(b) Resolving parallel to the plane gives:
F + 7.5g × Sin 18 = 55 × Cos 15F = 30.4N
Using F = μR:30.4 = μ × 90.5μ = 0.336
55N
FR
7.5g18º
Page 28
2 A swing ball is attached to one end of a light inextensible string; the other end of the string is attached to the top of a fixed vertical pole. A child applies a horizontal force, Q, of magnitude 45N to the particle P. P is in equilibrium under gravity with the string making an angle 25º with the vertical pole. By modeling the ball as a particle find:
a) the tension, T, in the string.b) the weight of the ball.
a) All problems of this type can be solved by resolving forces into their components.
Resolving horizontally:
Q = T cos 65º
45T = cos 65º
T 106N=
b) Resolving vertically:
Mg = T sin 65º
Mg = 96.5N
25º
PQ
T
Mg
Page 29
3 A ring of mass 0.45kg is threaded onto a fixed, rough horizontal pole. A light inextensible string is attached to the ring. The string and the pole lie in the same vertical plane. The ring is pulled downwards by the string which makes an angle θ with the
horizontal, where tan θ = 34 . The tension in the string is 2.2N.
Find the coefficient of friction between the ring and the pole.
Adding the weight, the reaction and frictional forces to the diagram gives:
Resolving vertically gives:
0.45g + T Sin θ = R
Given that Tan θ = 34 hence Sin θ = 3
5
Therefore: R = 0.45g + 2.2 × 0.6
R = 5.73N
Resolving horizontally gives:T Cos θ = F
T
θ
T
θ
R
F
0.45g
Page 30
Given that Tan θ = 34 hence Cos θ = 4
5
F = 2.2 × 0.8
F = 1.76
Using F = μR1.76 = μ × 5.73
μ = 0.307
4 A particle of weight 98 is attached at S to the ends of two light inextensible strings PS and QS. The other ends of the string are attached to two fixed points P and Q on a horizontal ceiling. The particle hangs in equilibrium with PS and QS inclined to the horizontal at angles of 25 and 60 degrees respectively.
Calculate the tension in both strings.
Adding the two tension forces to the diagram gives:
P
S
Q
10g
P
S
Q
10g
T1
T2
Page 31
Seeing as the object is in equilibrium then the forces must resolve to zero
Resolving horizontally gives:
T2 Cos 25 º = T1 Cos 60 º
Therefore: T2 = 0.552T1 (1)
Resolving horizontally gives:
T2 Sin 25 º + T1 Sin 60 º = 10g
Using (1) from above:
Sin 25 º × 0.552T1 + T1 Sin 60 º = 10g
T1 = 89.2N
Therefore: T2 = 49.2N
Page 1
VectorsRecap of GCSE content......................................................................... 2
Cartesian unit vectors and components.............................................7i, j notation........................................................................................... 7Example 2............................................................................................. 7Adding vectors in i, j notation......................................................... 7Example 3............................................................................................. 8Subtracting vectors in i, j notation................................................8Magnitude of a vector in i, j notation............................................ 8Example 4............................................................................................. 9Equal Vectors..................................................................................... 10Components of a vector................................................................... 10Example 5............................................................................................ 11
Vectors and Mechanics........................................................................ 12Position Vectors................................................................................ 12Relative position vectors................................................................. 12Velocity as a Vector......................................................................... 13Relative Velocity............................................................................... 13Acceleration as a Vector.................................................................14Example 6............................................................................................ 14Example 7............................................................................................ 15Example 8............................................................................................ 16
Vectors are first introduced at GCSE and students regularly ask how they apply to the real world. One simple example would be to consider a swimmer attempting to cross a river. If the swimmer can maintain a velocity of 1ms-1 and the river is flowing at 3ms-1
then the swimmer will obviously move on a diagonal path. The path that the swimmer tackles is said to be the resultant of the two forces (swimmer and current). This idea is explained in the diagram below.
The swimmer will be moving with a speed of 110ms−
Recap of GCSE content
Vector quantities require direction and magnitude to be truly defined. Scalar quantities are completely specified by their magnitude. Examples of each would be a car moving with a velocity of 25ms-1 on a bearing of 045º, and a second car traveling along a road with a speed of 15ms-1.
All vectors are represented by a directional line segment
Swimmer1ms-1
River flow3ms-1
Swimmer’s Path
Page 3
PQuuur
Vectors can also be represented in bold type and the triangle law is written as:
The magnitude (length) of a vector is given as PQuuur
.
Vectors can only be equal if they have the same magnitude and direction.e.g.
Vectors parallel and same magnitude AB MN∴ =uuur uuuur
Remember that direction is important
BS SB≠uuur uuur
a
bc
a + b = c
B
S
S
B
Page 4
Vectors may have different magnitudes but still be parallel.e.g.
KR xBQ=uuur uuur
(where x is a scalar quantity).
Adding Vectors
The earlier real world example introduced the triangle law by default. In more formal terms it can be said that for two vectors uuurOA and
uuurOB :
OA AB OB+ =uuur uuur uuur
This can be shown quite easily by using column vectors.
The triangle law can be applied to prove the parallelogram law.
The vector OBuuur
is said to be the resultant and is the diagonal of the parallelogram OABC
This idea is used in the following example:
Q
R
B
K
A
B
O
AO
C B
Page 5
Example 1
A particle P is acted upon by two forces P and Q of magnitude 6N and 10N respectively. The angle between the two vectors is 140˚. Find the magnitude of the resultant and the angle it makes with the force P.
This problem is solved by applying sine and cosine rule since the diagonal of the parallelogram is the resultant force.
c2 = a2 + b2 – 2abCosC
c2 = 36 + 100 – 2 × 6 × 10 × Cos 140
c = 15.87N
The resultant has a magnitude of 15.9N.
6N
10N
140˚
6N
10N
140˚CB
A
b
a
c
Page 6
The resultant makes an angle with force P of:
°=
=
9.23
87.15140
10
B
SinSinB
Page 7
Cartesian unit vectors and componentsA unit vector is a vector with magnitude of 1 unit.
i , j notation
The diagram above shows the two unit vectors i and j. By definition they are vectors of magnitude one unit along the x and y coordinate axis respectively.
Example 2
A vector R = (6i+4j)represents a displacement of:
6 units in the direction of the unit vector i, 4 units in the direction of the unit vector j.
And is displayed diagramatically as:
Adding vectors in i , j notation
When vectors are given in terms of unit vectors i and j, you can add them together by adding their terms involving i and j separately.
6i
4jR
Page 8
Example 3
Consider the two vectors a and b where: a = 2i + 3j b = 4i – 2j
a + b = (2i + 3j) + (4i – 2j)
=6i + j
Subtracting vectors in i , j notation
Using the same vectors a and b as in the example above
a = 2i + 3j b = 4i – 2j
a – b = (2i + 3j) – (4i – 2j)
= (-2i +5j)
Please take care with signs as these questions are not mathematically difficult but students are sometimes prone to make silly mistakes.
Magnitude of a vector in i , j notation
When a vector R is given in terms of the unit vectors i and j you can find its magnitude by using Pythagoras’ Theorem.
Page 9
Using the vector R from the earlier example:
2 2 2R 6 4
R 10
= +
=
The following example combines unit vectors and i,j notation.
Example 4
Find the unit vector in the direction of the vector 4i 7j= +a
The magnitude of a is:
2 24 7
65
a
a
= +
=
So the unit vector in the same direction as a is
( ))1 4 765
a i ja
= +
In more simple terms the vector a is 65 in length and the unit
vector in the same direction of a must be 165 times the vector a.
6i
4jR
Page 10
Equal Vectors
Two vectors are equal if and only if the i components are equal and the j components are equal.
Components of a vector
Any two vectors can be written as a single vector (triangle law). However, sometimes it is useful to reverse this process. This process is called resolving a vector into Cartesian components. This idea is used extensively in mechanics when dealing with a number of forces (see Statics chapter).
The diagram shows the vector R, which makes an angle θ with the x axis. The X and Y components can be given as:
X = R Cos θ
Y = R Sin θ
i
j
R
θx
y
X
Y
Page 11
Example 5
A force of magnitude 8N acts on a particle at an angle of 30º to the horizontal. Find the x and y components of the force.
X = 8 Cos 30 = 6.93N
Y = 8 Sin 30 = 4N
8N
30ºx
y
Page 12
Vectors and MechanicsSo far very little in the way of mechanics has been discussed. The positions of particles and their motion can be described by the use of vectors.
Position Vectors
If a particle is moving in a plane, where O is a fixed point, then the position of P is defined by OP r=
uuur
The vector r is known as the position vector of P relative to O.
Relative position vectors
As the name suggest we aim to find the vector of one particle relative to another.
Imagine two particles P and Q with position vectors OPuuur
and OQuuuur
respectively.
The vector PQuuur
gives the position vector of Q relative to P. It’s called the relative position vector (ie how do you get from P to Q).
P
O
r
P
O
rp
Qrq
Page 13
Velocity as a Vector
We discussed earlier that velocity is a vector quantity but defined more formally:
The velocity of a particle is a vector in the direction of motion whose magnitude is equal to the speed of the particle (usually denoted by v)
Relative Velocity
We have discussed relative position vectors, and seeing as velocity is a vector we should be able to consider relative velocity vectors. But what do they mean?
Imagine that two trains are traveling at equal velocities in the same direction at 90kmh-1. As far as passengers on the trains are concerned they will appear to be stationary. In this case we can say that their relative velocity is zero. If on the other hand the two trains were traveling in the same direction but at velocities of 90kmh-1 and 50kmh-1 then the relative velocity would be 40kmh-1. This principle can be applied to vectors:
Consider two particles C and D. If their velocities are vc and vd
respectively then the velocity of D relative to C is vd – vc.
In real terms the vector vd – vc is the velocity vector required to get from C to D. Unfortunately as time passes the particles C
vd – v
cvc
vd
Page 14
and D will get further apart and the vector between them will be a multiple of vd – vc.
The only other property of the vector vd – vc is that if we imagine that C and D are aeroplanes that set off at the same time from the same point, then it is the direction that a passenger on a plane C would look to see plane D.
Acceleration as a Vector
Considering that velocity can be a vector, and that acceleration is the rate of change velocity. It follows that acceleration can be a vector. Using the constant acceleration equations find final velocities.
Example 6
A particle has is moving with velocity (8i – 12j)ms-1 and it experiences an acceleration of (3i + 6j)ms-2 for 4 seconds. Find the final velocity.
The constant acceleration equation needed is v = u + at, however I find that it is best to write the equation in words:
Final velocity = initial velocity + (acceleration × time)
The following example uses the equation of motion. A number of vector exam questions will involve ideas from other areas of the M1 course and therefore you need to have a decent grasp of the whole course.
Page 15
Example 7
A particle Q, of mass 7.5Kg is moving under the action of a constant force F. Initially the velocity of Q is (12i – 16j)ms-1 and 8 seconds later it is (32i + 16j)ms-1.a) find, in vector form, the acceleration of Q.b) calculate the magnitude of F.
a) Given that acceleration is change in velocity over time:
The next example uses more concepts but shouldn’t cause any problems!
Page 16
Example 8
A particle A, of mass 3.5Kg is acted upon by two constant forces (6i – 3j)N and (8i + 10j)N.a) Find, in vector form, the resultant force F acting on A.b) Find, in degrees to 3 sig fig, the angle between F and i.c) Find the magnitude of the acceleration of A.d) Given that the initial velocity is (-6i + 7j)ms-1, find the speed of A after 6 seconds.
a) The resultant force is simply the sum of the forces (as they are in i,j form):
F = (6i – 3j) + (8i + 10j)
F = (14i + 7j)N
b) The diagram below shows the force F and angle θ:
7Tanθ14
θ 26.6°
=
=
F7j
14iθ
Page 17
c) The acceleration can be found by using the equation of motion for the particle.
F = ma
(14i + 7j) = 3.5 × a
a = (4i + 2j)ms-2
d) Seeing as the acceleration is constant we can use constant acceleration equations:
u = (-6i + 7j), a = (4i + 2j), t = 6
v = u + at
v = (-6i + 7j) + 6 × (4i + 2j)
v = (18i + 19j)ms-1
Speed is the magnitude of the velocity:
2 2
1
v (18 19 )
Speed 26.2ms−
= +
=
Page 18
AS Exam QuestionsA few basic principles are used over and again in M1 exam questions. Hopefully the solutions given below should give a few ideas as to what might appear. Examiner Reports usually state that vectors is the worst answered question on the paper, but with a little thought half marks is easily attainable.
Example 9
A particle B moves with constant acceleration (3i + 7j)ms-2. At time t its velocity is v ms-1. When t = 0, v = (12i – 14j)ms-1.a) find the time when B is moving parallel to the vector i.b) find the speed of B when t = 8.c) find the angle between the direction of motion of B and the vector i when t = 8.
a) The particle will be moving parallel to the vector i when it has no j component.
The velocity at any time is given by:
Velocity = initial velocity + ( acceleration × time )
= (12i – 14j) + t(3i + 7j) (1)
We are only interested in the j component and particularly when it is zero:
-14j + 7tj = 0
Therefore: t = 2
Page 19
b) By substituting t = 8 into equation (1) we can find the velocity of the particle. Speed is the magnitude of the velocity.
Velocity = (12i – 14j) + t(3i + 7j)
When t = 8 Velocity = (12i – 14j) + 8 × (3i + 7j)
= (36i + 42j) ms-1
Speed = 2 236 42+ = 55.3ms-1
c) The velocity at any time gives the direction of motion.
Therefore the direction is given by:
1 42θ Tan36
θ 49.4
−
°
=
=
36i
42j
θ
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Example 10
A particle has position vector (3i + 7j) and is moving with speed 39ms-1 in the direction (12i – 5j). Find the position vector at time t = 5 and the distance from the point (3,4) at this time.
The hint in the question is that the velocity of the particle must be a multiple of (12i – 5j)ms-1. The magnitude of this vector is 13, but the speed of the particle is 39ms-1, therefore the velocity of the particle must be (36i – 15j)ms-1. This one little trick is often used but if you don’t spot it the question is almost impossible.
The position vector of the particle at any time is given by:
Position = initial position + (velocity × time)
r = (3i + 7j) + t(36i – 15j)When t = 5
r = (183i – 68j)
The distance between the point and the particle can be found by Pythagoras theorem.
Distance = 2 272 180+ = 193.9m
(3,4)
(183,-68)
72
180
X
Y
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The final two examples are of the more challenging type. Setting the equations for the position vectors is pretty straight forward as long as you follow the rule:
r = initial position + (velocity × time)
In most cases the ships or aeroplanes are on a collision course and therefore you must set the equations equal to each other and equate coefficients. These questions may carry in excess of 10 marks but you should be able to make a start.
Example 11
A command post O monitors the movement of two of its ships in the Gulf. At 1200 hrs a battleship (B) has position (-2i + 10j) km relative to O and has constant velocity of (3i + 2j) kmh-1. A frigate (F) is at the point with position vector (4i + 5j) km and has constant velocity (-3i + 7j) kmh-1, where i and j are unit vectors directed due east and due north respectively. a) The captain of one ship has been taken ill, show that the two ships will collide.The command post contacts the battleship and orders it to reduce its speed to move with velocity (2i + 2j) kmh-1.b) Find an expression for the vector BF
uuurat time t hours after
noon.c) Find the distance between B and F at 1400 hrs.d) Find the time at which F will be due north of B.
a) The position of each ship is given by it’s position vector:
position = initial position + (velocity × time)
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So for the battleship:
rb = (-2i + 10j) + t(3i + 2j)
And for the frigate:
rf = (4i + 5j) + t(-3i + 7j)
If the two ships are to collide then for some value of t their respective i and j components must be equal.
Therefore by equating i’s:
-2 + 3t = 4 -3t
t = 1
Substituting the value of t = 1 into rb and rb gives the same position vector of (i + 12j). Therefore the two ships will collide after one hour at the point with position vector (i + 12j).
b) The position vector for the battleship must change to take account of the new velocity:
gives the distance between the two ships, at 1400 hrs, t = 2
= (6i - 5j) + 2(-5i + 5j)
= (-4i + 5j)
Magnitude = Km41
B
F
O E
N
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d) If F is due north of B, then BFuuur
will have no i component.
BFuuur
= (6i - 5j) + t(-5i + 5j)
6i -5ti = 0
t = 1hr 12mins
Example 12
Two cars P and Q are moving on straight horizontal roads with constant velocities. The velocity of P is 25ms-1 due east and the velocity of Q is (10i + 8j)ms-1. Initially P is at rest at the origin and Q has the position vector 230im relative to the origin. At time t seconds the position vectors of P and Q are r metres and s metres respectively.a) Find expressions for r and s in terms of t. b) Write an expression for PQ
uuur.
c) Find the time when the bearing of Q from P is 045ºd) Find the time when the cars are 200m apart.
a) The position vector for car P is given by:
r = initial position + (velocity × time)
r = 0 + 25ti
Therefore for Q:
s = 230i + t(10i + 8j)
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b) Using the triangle law:
0P PQ OQ
PQ OQ 0P
+ =
= −
uur uuur uuuur
uuur uuuur uur
Where 0Puur
= r and OQuuuur
= s
Therefore: PQuuur
= s – r
= 230i + t(10i + 8j) – 25ti
= 230i + t(8j – 15i)
c) If the bearing of Q from P is 045º, then the vector PQuuur
must be parallel to the vector (i + j).
Therefore230i + t(8j – 15i) = m(i + j)
Equating coefficients:
230 - 15t = m (1) 8t = m (2)
P
Q
O x
y
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Substituting gives:
230 - 15t = 8t
t = 10
The bearing of Q from P is 045º after 10 seconds.
d) The magnitude of PQuuur
will give the distance that the two cars are apart. We need the value of t where PQ 200=
uuur
Rewriting PQuuur
gives:
PQuuur
= (230 – 15t)i + 8tj
2 2
2 2
2 2
2
PQ (230 15t) 64t
PQ 52900 6900t 225t 64t
200 52900 6900t 161t
161t 6900t 12900 0
t 1.79
= − +
= − − +
= − −
+ − =
=
uuur
uuur
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Example 13
At 1200 hrs the position vectors of two helicopters A and B are rA and rB as outlined below. Use the velocity vectors vA and vB to give the position vectors of A and B at a time t hours after noon.
rA = (3i + 5j) rB = (5i + 2j) vA = (i + 2j) vB = (2i + 3j)
Find an expression at time t hours after noon for the position vector of B relative to A.
If d is the distance in Km between the two helicopters find the value of d2 in terms of t.
Find the time at which the helicopters are closest together. Give the value of the minimum distance.
a) For Helicopters A and B the position vectors at time t hours after noon are:
b) The position vector of B relative to A is given by:
rB – rA = 2i – 3j + t(i + j)
= (2 + t)i + (t - 3)j
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c) The magnitude of the vector in part (b) gives the distance between the two particles.Therefore:
d2 = (2 + t)2 + (t – 3) 2
d2 = 2t2 - 2t + 13
By completing the square we can find the minimum value:
d2 = 2[t2 - t + 6.5]
Remembering to halve the t coefficient:(-0.5)2 = 0.25 we need 6.5
d2 = 2[(t – 0.5)2 + 6.25]
The minimum value occurs when t = 0.5. Therefore the minimum distance between the two helicopters is √12.5 Km
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Questions1 For each of the vectors below find the magnitude and the angle the vector makes with the positive x axis.a) (6i + 3j) b) (i + 4j) c) (-4i + 3j) d) (-5i - 3j)e) (3i - 2j)
2 The table below gives the magnitude and direction of a number of vectors. Express each in the form ai + bj, where i and j are unit vectors in the direction Ox and Oy respectively.
Vector Magnitude Direction(as measured from positive x axis)
P 6 30ºQ 9 60ºR 4.5 150ºS 12 240º 3 Find the speed of a body moving with velocity (6i - 8j)ms-1
4 Find the speed of a body moving with velocity (-3i + 10j)ms-1
5 A particle is moving with velocity (2i + aj)ms-1 has speed 5.2ms-1. Find the two possible values of a.
6 A particle has an initial position vector of (6i - 8j)m. If the particle moves with constant velocity of (5i + 3j)ms-1, find its position vector after:a) 3 seconds b) 7 seconds.
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7 How far is the particle in question 6 from the origin after 12 seconds?
8 A particle has initial position vector (7i + 5j)m. The particle moves with constant velocity of (2pi + qj)ms-1 and after three seconds it has position vector (-17i - j)ms-1.
9 Given that a = 3i + 6j and b = -2i + 5j, find the magnitude of each of the following and angle that they make with the positive x axisa) a + b b) a – b c) 2a – b d) 4a + 2b
10 Two joggers Chris (C) and Gwyneth (G) are moving with constant velocities across a level plane. At a certain instant Chris and Gwyneth have position vectors (-60i +170j) and (90i – 100j) respectively. Thirty seconds later they meet at the point with position vector (300i + 20j).a) find in vector form the velocity of C and G.b) Calculate the magnitude of the velocity of C relative to G.
11 The velocities of two particles P and Q are (ai – 7j)ms-1 and (5i + bj)ms-1 respectively. The velocity of Q relative to P is (2i – 3j)ms-1. Find a and b.
12 The velocities of two particles P and Q are (14i + 7j)ms-1 and (5i + 12j)ms-1 respectively. Find:a) the speed of Q.b) the velocity of Q relative to P.c) the angle between the velocity in part (b) and the vector j.
