MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics

Physics 8.01 Fall 2012

Problem Set 2: Applications of Newton’s Second Law Solutions

Problem 1 (a) The static friction force

!fs can have a magnitude

!fs ! fs " µsN . Suppose you have a

block with a rope attached on opposite sides (right and left sides). (i) Describe the direction and magnitude of the static friction force as you increase your pull on the right side until the block just slips. (ii) Describe the direction and magnitude of the static friction force as you increase your pull on the left side until the block just slips. (iii). Describe the direction and magnitude of the static friction force when you pull both sides with the same magnitude of force. Answer. (i) As you increase your pull on the right, the static friction force increases in magnitude and is directed oppose your pull (to the left). When the block just slips the magnitude of the static friction force reaches its maximum value ( fs )max = µsN . (ii) When you pull on the left, the direction of the static friction force is now pointing to the right. The magnitude of the static friction force increases until the block just slips when the static friction force reaches its maximum value ( fs )max = µsN . (iii) When you pull both sides equally the static friction force is zero. (b) A basketball player is jumping vertically upward in order to shoot the ball. Her legs are flexed and pushing on the floor so that her body is accelerated upward. Draw free-body force diagrams for the player and the Earth. Identify all action-reaction pairs of forces. Which force acting on the player has the largest magnitude? Explain why. Solution

The external forces on the player are gravity and the contact force. Denote these as

!Fe, p

G

and !Ce, p . Gravity is directed down and

!Fe, p

G = mp!g . The contact force is directed up.

Internal forces in the player’s legs are not considered in a free-body diagram. The forces on the Earth are the gravitational force that the player exerts on the Earth and the contact force. Denote these forces as

!C p,e and

!Fp,e

G . !C p,e is directed down and

!Fp,e

G is directed

up. !Fe, p

G and !Fp,e

G form an action-reaction pair, as do !Ce, p and

!C p,e . We can express this

as

!Fe, p

G = !!Fp,e

G

!Ce, p = !

!C p,e

(1)

The player is accelerating upward, so the net force must be upward; the sum must be directed upward, so the upward force

!Ce, p must have the larger magnitude.

(c) The position vector ( )tr! of an object moving in a circular orbit of radius r is shown in the figure below with constant angular speed ! . At time t , the object is located at the point P with coordinates (r,!(t)) and position vector given by

!r(t) = r r(t) . Derive a vector expression for (i) the velocity and (ii) the acceleration. You must show all your work, in particular how you take derivatives.

Solution: In figure below,

we see that a vector decomposition expression for r(t) and !(t) in terms of i and j is given by r(t) = cos!(t) i + sin!(t) j (1) !(t) = " sin!(t) i + cos!(t) j (2) So we can write the position vector as

!r(t) = r r(t) = r(cos!(t) i + sin!(t) j) (3) The velocity is then

!v(t) = d!r(t)dt

= r ddt(cos!(t) i + sin!(t) j) = r(" sin!(t) d!(t)

dti + cos!(t) d!(t)

dtj) (4)

where we used the chain rule to calculate that

ddt(cos!(t) = " sin!(t) d!(t)

dt (5)

ddt(sin!(t) = cos!(t) d!(t)

dt (6)

We now rewrite Eq. (4) as

!v(t) = r d!(t)dt

(" sin!(t)i + cos!(t) j) (7)

Finally we substitute Eq. (2) into Eq. (7) and obtain an expression for the velocity of a particle in a circular orbit

!v(t) = r d!(t)dt

!(t) (8)

We denote the rate of change of angle with respect to time by the Greek letter ! ,

! "

d#dt

(9)

which can be positive (counterclockwise rotation in the figure below left), zero (no rotation), or negative (clockwise rotation in figure below right).

This is often called the angular speed but it is actually the z -component of a vector called the angular velocity vector.

!! =

d"dt

k =! k . (10)

The SI units of angular velocity are [rad ! s"1] .

Angular velocity, ! " d# / dt > 0

Thus the velocity vector for circular motion is given by

!v(t) = r! "(t) # v" "(t) , (11) where the ! -component of the velocity is given by

v! = rd!dt

. (12)

We shall call v! the tangential component of the velocity. We have assumed that the object is moving at a constant speed, d! / dt = 0 . The acceleration is then

!a(t) = d!v(t)dt

= r! d"(t)dt

(13)

Recall from Eq. (2) that !(t) = " sin!(t)i + cos!(t) j . So we can rewrite Eq. (13) as

!a(t) = r! ddt(" sin#(t)i + cos#(t) j) (14)

We again use the chain rule (Eqs. (5) and (6)) and find that

!a(t) = r! " cos#(t) d#(t)dt

i " sin#(t) d#(t)dt

j$%&

'() (15)

Recall that ! " d# / dt , and from Eq. (1) r(t) = cos!(t) i + sin!(t) j . Therefore the acceleration becomes

!a(t) = !r" 2r(t) . (16) The radial component of the acceleration is given by ar = !r" 2 < 0 . (17) Because ar < 0 , that radial vector component

!a r (t) = !r" 2 r(t) is always directed towards the center of the circular orbit.

Problem 2 A person is standing on top of a hill that slopes downwards uniformly at an angle ! with respect to the horizontal. The person throws a stone at an initial fixed angle ! from the horizontal with an initial speed of v0 . You may neglect air resistance. (a) What is the horizontal range of the stone when the stone strikes the ground? (b) Challenge: What angle ! should the person throw the stone such that the stone travels furthest down the hill?

Solution: a) In the absence of air friction or other forces other than gravity, the stone’s horizontal component of velocity will be constant, vx = v0 cos! . Taking the top of the hill as the origin, the stone’s x -coordinate as a function of time is x = (v0 cos!) t . The stone’s vertical component of velocity is

vy = v0 y ! g t = v0 sin" ! g t (1) and its vertical position is

y = (v0 sin!)t "12

g t2

=xv0 sin!xv0 cos!

"12

gx

v0 cos!#

$%&

'(

2

= x tan! " x2 g2v0

2 cos2!,

(2)

where t = x / (v0 cos!) has been used to express y in terms of x . The position of any point on the downward slope is y = !x tan" (3) and the stone will land at the point where the expressions in Equations (2) and (3) are equal,

x tan! " x2 g

2v02 cos2!

= "x tan# . (4)

The x = 0 root of Equation (4) represents the stone being at the top of the hill, and may be neglected. Solving Equation (4) for x gives

x = (tan! + tan")cos2!2v0

2

g

= (sin! cos! + cos2! tan")2v0

2

g=

12

sin2! + cos2! tan"#$%

&'(

2v02

g

(5)

where we have used the identity 2sin! cos! = sin2! (b) From Eq. (5), we can see that the horizontal distance is a function of the angle ! . If we maximize the horizontal distance then the stone will also go furthest down the hill . Thus we set

0 =

dxd!

=dd!

sin! cos! + cos2! tan"( ) 2v02

g. (6)

This becomes

0 = cos2! " sin!2cos! tan#( ) 2v0

2

g= cos2! " sin 2! tan#( ) 2v0

2

g,

which simplifies to the condition that

0 = cos2! " sin2! tan# .

We can now solve for the angle ! maximize the horizontal distance that the stone will travel

tan2! = co tan" , or

! =

12

tan"1(co tan#) . (7)

Problem 3 A person clings to a rope (assumed massless) that passes over a pulley. The person of mass

mp is balanced by a block of mass mb hanging at the other end of the rope. Initially both the person and block are motionless. The person then starts climbing the rope by pulling on it with a constant force in order to reach the block. The person moves a distance L relative to the rope. Does the block move as a result of the person’s climbing? If so, in which direction, and by how much?

Solution: The coordinate system is shown above and force diagrams on the block and person are shown in the figures below.

We are told that the person and block are initially balanced so

m ! mp = mb .

As the person pulls up on the rope, there is a force down on the rope, creating a tension T in the rope. This tension is transmitted through the rope, and so is also the force on the object. Both the person and block satisfy Newton’s Second Law,

m g ! T = may . (1) Because T > mg , the person and the block accelerate upwards with the same acceleration,

ay < 0 . The length of the rope between the person and the block is

l = yp + !R + yb , (2) where R is the radius of the pulley. As the person climbs,

yp and yb change by the same (negative) amount. So, if a length of rope L passes through the person’s hands, both the person and the object rise a distance / 2L .

Problem 4 A cart rolls on a track inclined at an angle ! to the horizontal. The mass of the cart is M . There is a rolling friction force given by F Nµ= , where N is the magnitude of the normal (i.e., perpendicular to the track) force the track exerts on the cart. Assume the cart is rolling down the track in the figure at the left, and is rolling up the track in the figure to the right. Find a vector expression for the acceleration of the cart when it is rolling (a) down the track (b) up the track.

Solution For the cart rolling down, the friction force is directed upward along the track, and for the cart rolling up, the friction force is directed down along the track. The direction of i is chosen pointing down the incline in both cases.

For both cases, rolling up and rolling down, the magnitude N of the normal force must be equal in magnitude to the component cosM g ! of the gravitational force normal to the plane because the acceleration in the j -direction is zero. With the model for the friction force given in the problem, cosF N M gµ µ != = . (1) The component sinM g ! of the gravitational force directed down the incline is the same for both cases. Thus we have: (a) Rolling down: Newton’s Second Law in the i -direction is:

M g sin! " µM g cos! = Mad (2) Therefore

!ad = (g sin! " µ g cos!)i (3) (b) Rolling up: Newton’s Second Law in the i -direction is: M g sin! + µM g cos! = M au (4)

!au = (g sin! + µg cos!)i (5) So the acceleration in both cases is downward and au >ad . (6)

Problem 5 An object with mass m is released from rest and slides a distance d down a roof that is inclined at an angle ! with respect to the horizontal. The contact surface between the roof and object has a coefficient of kinetic friction µk . The edge of the bottom of the roof is at a height h above the ground. What is the horizontal distance from the edge of the roof to the point where the object hits the ground?

Solution: It is clear that the distance the object travels will depend on how fast it is moving when it leaves the roof, and that this speed will be larger for larger values of d and ! . Consider then the two stages of the motion, the first when the object is sliding down the roof and the second where it is in free fall. We shall use Newton’s second Law to find the acceleration for the sliding down the roof and then use the kinematic equations for position and velocity along the roof to find the speed of the object just when it reaches the end of the roof. We can then use that speed as an initial condition for the projectile motion trajectory. We can then solve for the time that it takes to reach the ground and hence solve for the horizontal distance the object traveled from edge of the roof. For the first stage, choose a coordinate system with the positive i -direction down the roof (the steepest downward direction) and the positive j -direction to be perpendicular to the roof (with positive upward component).

The forces on the object are gravity

m!g = mg icos! " jsin!( ) (1)

and the contact force

!C =!N +!fk = N j! fk i . (2)

The components of the vectors in Newton’s Second Law, !F = m!a , are

mg sin! " fk = max (3)

N ! mg cos" = may . (4)

The condition that the object remains on the roof is expressed as

ay = 0 and the model for kinetic friction is fk = µk N . (5) The condition

ay = 0 in Eq. (4) gives N = mg cos! , (6) and so Eq. (5) becomes fk = µkmg cos! . (7) Using this in Eq. (3) allows solution for the x -component of acceleration ax = g(sin! " µk cos!) . (8) Let the time that the object takes to slide down the roof be t1 and the speed of the object just as it leaves the roof be v0 . Then, we have that

d =

12

ax t12 =

12

(ax t1)2

ax

(9)

We note that v1 = ax t1 . (10) So Eq. (9) becomes

d =

v12

2ax

(11)

We can now solve Eq. (11) for the speed of the object the instant it leaves the roof v1 = 2 ax d = 2 g d(sin! " µk cos!) . (12) For the second stage, the object is in free fall. For this stage, it should be clear that the coordinate system used for the first stage will cause great difficulty. So, take the positive i -direction to be horizontal and the positive j -direction to be vertically upward, so that the i - j plane contains the object’s motion.

We then have ax = 0 , (13)

ay = !g . (14) Additionally, if we now “reset our clock” so that t = 0 when the object leaves the roof,

vx ,0 = v1 cos! , (15)

vy ,0 = !v1 sin" . (16)

where v1 is the speed of the object when it just leaves the edge of the roof. If we further take the origin to be the point on the ground directly below the point where the object leaves the roof, x0 = 0 , y0 = h . The equations describing the object’s motion as a function of time t are then

x(t) = x0 + vx ,0 t +

12

ax t2 = v1 cos! t , (17)

y(t) = y0 + vy ,0 t + 1

2ay t2 = h ! v1 sin" t ! 1

2gt2 . (18)

When the object hits the ground, y = 0 , the x -coordinate may be denoted

x f and the

time when the object hits may be denoted t f . Eq. (18) then becomes

0 = h ! v1 sin" t f !

12

gt f2 (19)

This is a quadratic equation in

t f , and the quadratic formula gives

t f =

1g

!v1 sin" ± v12 sin2" + 2gh( ) (20)

and substitution of the positive solutions into Eq. (17) gives

x f =v1 cos!

g"v1 sin! + v1

2 sin2! + 2gh( )=

v12

gcos! " sin! + sin2! +

2ghv1

2

#

$%

&

'(

. (21)

We can rewrite Eq. (11) as v1

2 / g = d(sin! " µk cos!) (22) Then substitute Eq. (22) into Eq. (21) which yields

x f = d(sin! " µk cos!)cos! " sin! + sin2! +

2hd(sin! " µk cos!)

#

$%

&

'( (23)

Note that the gravitational constant g does not appear in this expression. Further simplification is of questionable value.

Problem 6 A truck is traveling in a straight line on level ground, and is accelerating uniformly with an acceleration of magnitude a . A rope (which of course is massless and inextensible) is tied to the back of the truck. The other end of the rope is tied to a bucket of mass M . The bucket tosses wildly when the truck starts to accelerate, but due to friction it soon settles into a position at a fixed distance behind the truck, with the rope hanging straight at a fixed angle, as shown in the diagram. Although friction is needed to cause the rope to settle to an equilibrium position, we will assume that it can otherwise be neglected. Express your answers in terms of the given variables M , g , and a . You may not need them all.

a) Find the angle ! at which the rope will settle. b) What will be the tension T of the rope once it settles into this angle? c) Now suppose that the truck comes to a downhill section of road, at an angle !

relative to the horizontal, as shown in the diagram. Suppose that the truck continues to accelerate with an acceleration of magnitude a . Once the rope again settles to a fixed angle !" relative to the truck, what will that angle be? (Express your answers in terms of the given variables M , g , ! , and a . You may not need them all.)

d) What will be the new tension !T in the rope? (Express your answers in terms of

the given variables M , g , ! , and a . You may not need them all. Solution: a) The free body diagram is shown in the figure below.

Applying Newton’s Second Law, we get that

T sin! = Ma . (1) T cos! " Mg = 0 . (2) Thus T cos! = Mg . (3) Dividing Eq. (1) by Eq. (3) yields tan! = a / g . (4) Thus ! = tan"1(a / g) . (5) b) What will be the tension T of the rope once it settles into this angle? Square Eqs. (1) and (2), add, and then take the square root T = M g 2 + a2 (6) c) Now suppose that the truck comes to a downhill section of road, at an angle ! relative to the horizontal, as shown in the diagram. Suppose that the truck continues to accelerate with an acceleration of magnitude a . Once the rope again settles to a fixed angle !" relative to the truck, what will that angle be? (Express your answers in terms of the given variables M , g , ! , and a . You may not need them all.)

The force diagram is now shown in the figure below.

Apply Newton’s Second Law down the hill: T sin !" + Mg sin# = Ma . (7) Therefore !T sin !" = Ma # Mg sin$ . (8) Apply Newton’s Second Law normal to the hill: !T cos !" # Mg cos$ = 0 . (9) Therefore !T cos !" = Mg cos# . (10) Dividing Eq.(8) by Eq.(10) yields

!T tan !" =

a # g sin$g cos$

. (11)

Thus

!" = tan#1 a # g sin$

g cos$%&'

()*

(12)

d) What will be the new tension !T in the rope? (Express your answers in terms of the given variables M , g , ! , and a . You may not need them all.) Square Eqs. (8) and (10), add, and then take the square root !T = ( Ma " Mg sin# )2 + ( Mg cos# )2 (13) !T = M (a2 + g 2 " 2ag sin# ) . (14) Alternatively: choose horizontal (positive i to the right) and vertical (positive j down) unit vectors. The acceleration then decomposes as

!a = acos! i + asin! j . (15) Newton’s Second Law becomes !T sin( !" +# ) = Macos# , (16) Mg ! "T cos( "# +$ ) = Masin$ . (17) Eq.(17) can be rewritten as !T cos( !" +# ) = Mg $ Masin# . (18)

Divide Eq.(16) by Eq.(18) yielding

tan( !" +# ) =

acos#g $ asin#

. (19)

Thus

!" = tan#1 acos$

g # asin$%&'

()*#$ . (20)

Problem 7 The system shown on the left above is made up of two massive blocks, three massless, frictionless pulleys and 3 ropes of fixed length. Find the acceleration of block 2 after the system is released from rest.

Solution: The figure below shows a coordinate system and the tensions in each of the three ropes that are useful in solving the problem.

Newton’s Second Law F = ma on block 1 yields

m1g ! T1 = m1

d 2 y1

dt2 . (1)

and F = ma on block 2 yields

m2g ! T2 = m2

d 2 y2

dt2 . (2)

The middle pulley will accelerate at some finite rate. However since it has no mass, unless the sum of the forces on it is zero, it would accelerate at an infinite rate. Thus

T1 = 2T3 . (3) Similarly, the sum of the forces on the lower pulley must be zero. T3 = 2T2 . (4) Together these last two equations imply that T1 = 4T2 . (5) The length of the rope with tension T1 is given by l1 = ( y1 ! s1) + y3 . (6) The constraint that the length of that rope is constant requires that

0 =

d 2l1dt2 =

d 2 y1

dt2 +d 2 y3

dt2 . (7)

Therefore

d 2 y1

dt2 = !d 2 y3

dt2 . (8)

The length of the rope with tension T3 is given by l3 = (d ! y3) + y4 ! y3 . (9) The constraint that the length of that rope is constant requires that

0 =

d 2l3

dt2 =d 2 y4

dt2 ! 2d 2 y3

dt2 . (10)

Hence

d 2 y4

dt2 = 2d 2 y3

dt2 . (11)

The length of the rope with tension T2 is given by l2 = (d ! y4 ) + y2 ! y4 . (12) The constraint that the length of that rope is constant requires that

0 =

d 2l2

dt2 =d 2 y2

dt2 ! 2d 2 y4

dt2 . (13)

Hence

d 2 y2

dt2 = 2d 2 y4

dt2 . (14)

These three constraint conditions for the accelerations imply (after a little algebra) that

d 2 y2

dt2 = !4d 2 y1

dt2 . (15)

Collecting our results we have that

m1g ! T1 = m1

d 2 y1

dt2 (16)

m2g ! T2 = m2

d 2 y2

dt2 (17)

T1 = 4T2 (18)

d 2 y2

dt2 = !4d 2 y1

dt2 . (19)

We can now solve for the accelerations of the two objects. We can rewrite Eq. (17) using Eqs. (18) and (19) as

m2g ! (1 / 4)T1 = !4m2

d 2 y1

dt2 . (20)

Then multiply Eq. (16) by !(1 / 4) yielding

!(1 / 4)m1g + (1 / 4)T1 = !(1 / 4)m1

d 2 y1

dt2 . (21)

Now add Eqs. (20) and (21) yielding

(m2 ! (1 / 4)m1)g = !(4m2 + (1 / 4)m1)

d 2 y1

dt2 . (22)

Therefore we can solve for the acceleration of block 1,

d 2 y1

dt2 =((1 / 4)m1 ! m2 )g4m2 + (1 / 4)m1

. (23)

Finally we use Eq. (19) and find that

d 2 y2

dt2 =(4m2 ! m1)g

4m2 + (1 / 4)m1

. (24)

The units are correct. We are looking for an acceleration and we have a dimensionless fraction times the acceleration of gravity. If m1 = 0 , body 2 is simply in free fall with acceleration g . In the limit m1 >> m2 , body 2 accelerates upward at 4 times the rate at which body 1 falls. These are the results we would expect on simple physical grounds.

Problem 8 In the system shown above m1 > m2 . The pulleys are massless and frictionless, and the rope joining the blocks has no mass. The coefficient of static friction between the blocks and the tables is greater than the coefficient of kinetic friction:

µs > µk . The downward acceleration of gravity is g .

a) Imagine that when the system is released from rest body 3 accelerates downward at a constant rate of magnitude a, but only one of the other blocks moves. Which block does not move, and what is the magnitude and direction of the friction force holding it back.

b) Now consider the case where, when released from rest, all three blocks begin to

move. Find the accelerations of all three blocks and the tension in the rope. Solution a) In this part

ay3 is a given quantity. Let T be the tension in the rope. The

free body diagram is shown below.

Choose positive j pointing down. Then Newton’s Second Law becomes

Fy = m3ay3 . (1)

m3g ! 2T = m3ay3 . (2) which we can solve for the tension

T =

12

m3(g ! ay3) . (3)

T exceeds the static friction force limit on one block but not the other which is given for the ith block by the expression

fs,max,i = µs Ni = µsmig . (4)

Therefore the lightest block moves first., hence block 1 with mass m1 does not move. The free body diagram for block 1 is

Choose positive i to the left. Therefore Newton’s Second Law becomes f1 ! T = m1ax1 = 0 . (5) Therefore

f1 = T =

12

m3(g ! ay3) . (6)

Solution b): Now consider the case where, when released from rest, all three blocks begin to move. The obvious dynamical variables are the accelerations ax1 , ax2 , and

ay3 . The

other variable that influences the motion of each of the masses is the tension T . These are the four variables we will try to find. Hence we need four equations that must be solved to find the dynamics of this system. The free body diagrams for block 1 is shown below. We choose positive i to the left and positive j pointing down in each diagram. Newton’s Second Law for the Block 1 are

f1 _T = m1ax1 . (7)

m1g ! N1 = m1ay1 = 0 . (8) Therefore N1 = m1g . (9) Because block 1 is moving

f1 = µk N1 = µk m1g . (10) So Eq. (7) becomes µk m1g ! T = m1ax1 . (11) The free body diagrams for block 2 is shown below. We choose positive i to the right and positive j pointing down in each diagram.

Newton’s Second Law for the Block 2 are f2 ! T = m2ax2 . (12)

m2g ! N2 = m2ay2 = 0 . (13) Therefore N2 = m2g . (14) Because block 2 is moving f2 = µk N2 = µk m2g . (15) So Eq. (12) becomes µk m2g ! T = m2ax2 . (16) The free body force diagram for block 3 is shown below. We choose positive j pointing down.

The analysis is the same has in the previous part so we collect our three equations Eqs. (2), (11), and (16),

m3g ! 2T = m3ay3 . (17)

µk m1g ! T = m1ax1 . (18) µk m2g ! T = m2ax2 . (19)

Let L be the length of the rope and R the radius of the pulleys.

Then these quantities are related by the geometrical constraint

L = x1 + x2 + 2y3 +

!2

R +!2

R + !R . (20)

Because we assumed that the string does not stretch, the quantity L is a constant and hence if we take two derivatives of Eq. (20) we have that

0 = ax1 + ax2 + 2ay3 . (21)

We now have four equations in the four unknowns the accelerations ax1 , ax2 , and

ay3

and the tension T . We solve Eq. (18) for ax1 ,

ax1 = µk g !

Tm1

. (22)

solve Eq. (19) for ax2 ,

ax2 = µk g !

Tm2

. (23)

and solve Eq. (17)

ay3 = g !

2Tm3

. (24)

Now substitute these results into Eq. (21) yielding

0 = µk g !

Tm1

+ µk g !Tm2

+ 2 g !2Tm3

"

#$%

&'. (25)

We now solve Eq. (25) for the tension

2g(µk +1) = T 1

m1

+1

m2

+4m3

!

"#$

%&. (26)

or

T =2g(µk +1)

1m1

+ 1m2

+ 4m3

!

"#$

%&

. (27)

We now solve for the accelerations. Eq. (22) becomes

ax1 = µk g !2g(µk +1)

1+m1

m2

+4m1

m3

"

#$%

&'

. (28)

solve Eq. (23) for ax2 ,

ax2 = µk g ! T = µk g !2g(µk +1)

m2

m1

+ 1m2

+4m2

m3

"

#$%

&'

. (29)

and solve Eq. (17) for ay3 ,

ay3 = g !4g(µk +1)m3

m1

+m3

m2

+ 4"

#$%

&'

. (30)