02_physics_lecture_04 Mar 2013

MS101: PhysicsMS101: PhysicsDr. Ahmed Amin HusseinDr. Ahmed Amin Hussein

[email protected]@gmail.com20132013

May 3, 2023 Prepared By: Dr. Ahmed Amin 1

Chapter 2Force

May 3, 2023 2Prepared By: Dr. Ahmed Amin

Forces Vector addition Newton’s First and Third Laws

Acceleration Gravity Contact Forces Tension Fundamental Forces

§2.1 Forces§2.1 Forces


Simply, a force is a “push” or “pull” on an object.

Force is a vector --- it has magnitude and direction.

Forces can be represented by arrows.

The size of the arrow represents the magnitude and the point shows its direction.

Addition of ForcesAddition of Forces


= 0

Balanced and Unbalanced Balanced and Unbalanced ForcesForces


• Forces occur in pairs and they can be either balanced or unbalanced

Balance ForcesBalance Forces


EqualEqual forces acting on one object in forces acting on one object in oppositeopposite directions directions

•acceleration is zero •speed is constant •Net Forces = 0 (NO CHANGE IN MOTION)

Unbalanced ForcesUnbalanced Forces


An unbalanced force always causes a change in motion

When unbalanced forces act in opposite directions you can find the

net force

– Net force

• Magnitude– The difference between the two forces

• Direction– Direction of the largest force

Calculate net forceCalculate net force


Four people are pulling on the same 200 kg box with the forces shown.

Calculate the acceleration of the box.

How can a force be How can a force be measured?measured?


By hanging masses on a spring we find that the spring stretch applied force.

The unit of force is the newton (N).

Example [A]Example [A]


QQ Two forces act on an object.Two forces act on an object.

In which situation is it impossible for the object to be in In which situation is it impossible for the object to be in equilibrium? equilibrium?

AA The two forces act in the same direction.The two forces act in the same direction.

BB The two forces act through the same point.The two forces act through the same point.CC The two forces are of the same type.The two forces are of the same type.

DD The two forces are the same size.The two forces are the same size.

Calculate force using Calculate force using equilibriumequilibrium


Two chains are used to lift a small boat. One of the chains has a force of 600

Newtons.

Find the force in the other chain if the mass of the boat is 150 kilograms.

Vectors versus scalarsVectors versus scalars


A vector is a quantity that has both a magnitude and a direction. A force is an example of a vector quantity.

A scalar is just a number (no direction). The mass of an object is an example of a scalar quantity.

NotationNotation


Vector: F

or F

The magnitude of a vector: .For or

FF

Scalar: m (not bold face; no arrow)

The direction of vector might be “35 south of east”; “20 above the +x-axis”; or….

§§2.2 Graphical Vector 2.2 Graphical Vector AdditionAddition


To graphically represent a vector, draw a directed line segment.

The length of the line can be used to represent the vector’s length or magnitude.

Example (1):Example (1):


To add vectors graphically they must be placed “tip to tail”. The result (F1 + F2) points from the tail of the first vector to the tip of

the second vector.

For collinear vectors:

F1

Fnet

F2

F1

Fnet

F2

§§2.3 Vector Addition Using Components2.3 Vector Addition Using Components


Vector Addition: Place the vectors tip to tail as before. A vector may be moved any way you please provided that you do not change its length nor

rotate it. The resultant points from the tail of the first vector to the tip of the second (A+B).

•The trigonometric ratios can be used to calculate sides or angles in right-angled triangles.

• Each trig ratio can be written in different ways.

AB: Opposite AC: Adjacent BC: Hypotenuse

Opp = Hyp x sin Adj = Hyp x cos Opp = Adj x tan

Example (1)Example (1)


Three coplanar forces act at a point. The magnitudes of the forces are 5 N,

6N and 7 N, and the directions in which the forces act are shown in the

diagram. Find the magnitude and direction of the resultant of the three

forces?

Answer: Magnitude is 5.09 N and direction is 9.40 anti-clockwise from force of magnitude 7 N



Vector A has a length of 5.00 meters and points along the

x-axis. Vector B has a length of 3.00 meters and

points 120 from the +x-axis. Compute A+B (=C).

A x

y

B

120

C

Example continuedExample continued


A x

y

B

120By

Bxadjopp

cossintan

hypadjcos

hypoppsin

m 50.160cosm00.360cos 60cos

m 60.260sinm00.360sin60sin

BBBB

BBBB

xx

yy

and Ax = 5.00 m and Ay = 0.00 m



The components of C:

m 2.60m 2.60 m 00.0m 3.50m 501 m 00.5

yyy

xxx

BAC.BAC

x

y

C

Cx = 3.50 m

Cy = 2.60 m

The length of C is:

m 36.4

m 60.2m 50.3 22

22

yx CCC C

The direction of C is:

6.367429.0tan

7429.0m 3.50m 60.2tan

1

x

y

CC

From the +x-axis

§2.4 Newton’s First Law§2.4 Newton’s First Law


Newton’s 1st Law (The Law of Inertia):

If no net force acts on an object, then its speed and direction of motion do not change.

Inertia is a measure of an object’s resistance to changes in its motion.

The net force is the vector sum of all the forces acting on a body.

321net FFFFFi

i


If the object is at rest, it remains at rest (speed = 0).

If the object is in motion, it continues to move in a straight line with the same speed.

No force is required to keep a body in straight line motion when effects such as friction are negligible.

An object is in translational equilibrium if the net force on it is zero.

Free Body DiagramsFree Body Diagrams


Must be drawn for problems when forces are involved.

Must be large so that they are readable.

Draw an idealization of the body in question (a dot, a box,…). You will need one free body diagram for each body in the problem that will provide useful

information for you to solve the given problem.

Indicate only the forces acting on the body. Label the forces appropriately. Do not include the forces that this body exerts on any other body.

Do not include fictitious forces. Remember that ma is itself not a force!

You may indicate the direction of the body’s acceleration or direction of motion if you wish, but it must be done well off to the side of the free body diagram.


§§2.5 Newton’s Third Law2.5 Newton’s Third Law


Newton’s 3rd Law:

When 2 bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction. Or, forces come in pairs.

Mathematically:1221 FF

EquilibriumEquilibrium


The restoring force from a

wall is always exactly equal

and opposite to the force you

apply, because it is caused

by the deformation resulting

from the force you apply.

FBDFBD


Consider a box resting on a table

(a) If F1 is the force of the Earth on the box, what is the interaction partner of this force?

The force of the box on the Earth.

(b) If F2 is the force of the box on the table, what is the interaction

partner of this force?

The force of the table on the box.

F1

F2

External forcesExternal forces


Any force on a system from a body outside of the system.

Pulling a box across the floorF

Internal forcesInternal forces


Fext

Force between bodies of a system

Pulling 2 boxes across the floor where the two boxes are attached to each other by a rope.

§2.6 Gravity§2.6 Gravity


Gravity is the force between two masses. Gravity is a long-range or field force. No contact is needed between the bodies. The force of gravity is always attractive!

r is the distance between the two masses M1 and M2 and G = 6.6710-11 Nm2/kg2.

221

rMGMF

M2

r

M1F21 F12

1221 FF


Let M1 = mass of the Earth. 22 MrGMF E

Here F = the force the Earth exerts on mass M2. This is the force known as weight, w.

.222 gMMrGMwE

E

N/kg 8.9 where 2 E

E

rGMg Near the surface of the Earth

km 6370kg 1097.5

E

24E

rM


What is the direction of g?

Note thatmFg is the gravitational force per unit mass. This is

called the gravitational field strength. It is often referred to as the acceleration due to gravity.

What is the direction of w?

ExampleExample


What is the weight of a 100 kg astronaut on the surface of the Earth (force of the Earth on the astronaut)? How about in low Earth orbit? This is an

orbit about 300 km above the surface of the Earth.

On Earth: N 980mgw

In low Earth orbit:

N 890)( 2

oE

Eo rR

GMmrmgw

Their weight is reduced by about 10%. The astronaut is NOT weightless!

§2.7 §2.7 Contact ForcesContact Forces


Contact forces: these forces arise because of an interaction between the atoms in the surfaces in contact, and are responsible for most

visible interactions between macroscopic collections of matter..

Normal force: this force acts in the direction perpendicular to the contact surface.

Normal force of the ground on the box

Normal force of the ramp on the box

N

w

N

w

Non-contact forcesNon-contact forces


In the Standard Model of modern physics, the four fundamental forces of

nature are known to be non-contact forces. The strong and weak interaction

primarily deal with forces within atoms, while gravitational effects are only

obvious on a macroscopic scale. Molecular and quantum physics show that

the electromagnetic force is the fundamental interaction responsible for

contact forces. The interaction between macroscopic objects can be roughly

described as resulting from the electromagnetic interactions between protons

and electrons of the atomic constituents of these objects.

ExampleExample


FBD for box

mgwN

wNFy

that So

0Apply Newton’s 2nd law

This just says the magnitude of the normal force equals the magnitude of the weight; they

are not Newton’s third law interaction partners.

Consider a box on a table

N

w

x

y

FrictionFriction


a contact force parallel to the contact surfaces.

Static friction acts to prevent objects from sliding

The force of static friction is modeled as

where s is the coefficient of static friction and N is the normal force.

Kinetic friction acts to make sliding objects slow down.

The force of kinetic friction is modeled as

where k is the coefficient of kinetic friction and N is the normal force.

.ss Nf

.kk Nf

ExampleExample


A box full of books rests on a wooden floor. The normal force the floor exerts on the box is 250 N.

(a) You push horizontally on the box with a force of 120 N, but it refuses to budge. What can you say about the coefficient of friction between

the box and the floor?

FBD for box

0)2(

0)1(

sx

y

fFF

wNFApply Newton’s 2nd Law

N

w

x

y

F

fs


From (2): 48.0 NFNfF sss

This is the minimum value of s, so s > 0.48.

(b) If you must push horizontally on the box with 150 N force to start it sliding, what is the coefficient of static friction?

Again from (2): 60.0 NFNfF sss


(c) Once the box is sliding, you only have to push with a force of 120 N to keep it sliding. What is the coefficient of kinetic friction?

FBD for box

N

w

x

y

F

fk

0)2(

0)1(

kx

y

fFF

wNFApply Newton’s 2nd Law

From 2:

48.0N 250N 120

k

NF

NfF kk


Consider a box of mass m that is at rest on an incline. Its FBD is:

There is one long-range force acting on the box: gravity.

There is one contact force acting on the box from the ramp.

If the net force acting on the box is zero, then the contact force from the ramp must have the same magnitude as the weight force, but be in the opposite

direction.

FRB

w x

y


The force FRB can be resolved into components that are perpendicular and parallel to the ramp.

The perpendicular component is what we call the normal force.

The parallel component is the static friction force.

FRB

w

N

fs

x

y



Let the box on the ramp have a mass 2.5 kg. If the angle between the incline and the horizontal is 25, what are the magnitudes of the weight force, normal

force, and static friction force acting on the box?

FRB

w

N

fs

x

y

0sin

0cos

sx

y

fwF

wNF

Apply Newton’s

2nd Law

N 31.925sinN 5.24sin

N 2.2225cosN 5.24cosN 5.24N/kg 8.9kg 5.2

wfwNmgw

s

The forces are:

§2.8 §2.8 TensionTension


This is the force transmitted through a “rope” from one end to the other.

An ideal cord has zero mass, does not stretch, and the tension is the same throughout the cord.

Example Example


A pulley is hung from the ceiling by a rope. A block of mass M is suspended by another rope that passes over the pulley and is

attached to the wall. The rope fastened to the wall makes a right angle with the wall. Neglect the masses of the rope and the pulley. Find the tension in the rope from which the pulley

hangs and the angle .

FDB for the mass M

w

T

y

MgwT

wTFy

0Apply Newton’s 2nd Law to the mass M.



FBD for the pulley:

x

y

T

TF

0sin

0cos

TFF

TFF

y

x

Apply Newton’s 2nd Law:

sincos FFT This statement is true only when = 45 and

MgTF 22

§2.9 Fundamental Forces§2.9 Fundamental Forces


The four fundamental forces of nature are:

• Gravity which is the force between two masses; it is the weakest of the four.

• Strong Force which helps to bind atomic nuclei together; it is the strongest of the four.

• Weak Force plays a role in some nuclear reactions.

• Electromagnetic is the force that acts between charged particles.

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