08/04/23 13:26EPF0014 Chapter 4 1

Dynamics

1. Definition of Force and Mass

2. Newton 1st, 2nd & 3rd Laws of motion

4. Free body Diagrams

3. Types of Forces: Normal, Frictional, Tension and Gravitation Forces. etc

3.

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Lesson ObjectivesAt the end of the lesson, students should be able to:

1. define force, mass and inertia.

2. state Newton’s laws of motion.

3. explain everyday phenomena in terms of Newton’s laws of motion.

4. use free-body diagrams to solve problems involving forces and accelerations.

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Dynamics

• Displacement, velocity and acceleration are basic kinematics quantities. However, when we begin to think about why objects move, this is the science of Dynamics.

• The influence that changes the basic kinematics quantities of a particle is called a force.

• A force, simply put, is a push or a pull.

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Force • The concept of force gives us a quantitative description of the

interaction between two bodies or between a body and its environment.

• When a force involves direct contact between two bodies, we call it a contact force. An example is frictional force.

• There are also non-contact forces or action-at-a-distance forces, including gravitational and electrical forces, which act even when the bodies are separated by empty space.

• The force of gravitational attraction that the earth exerts on a body is called the weight of the body.

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The idea of a force is made explicit in Newton’s laws of motion.

Sir Isaac Newton(British Physicist)

1642-1727

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

Inertia is the natural tendency of an object to remain at rest or in motion at constant speed along a straight line.

Every body continues in its state of rest, or of uniform motion (constant velocity) in a straight line, unless a net external force acts upon it.

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Lighter Side

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Newton’s 2nd Law of Motion

The acceleration of an object is directly proportional to the

net force acting on it in the same direction.

amF net

The proportionality constant is a quantity known as the mass of the body denoted by m. Mass is a measure of an object’s inertia.

netFa

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• Net force is the vector sum of all the forces acting on it.

• SI Unit of Force: kg.m/s2 = newton (N)

amF net

For a given net force, the magnitude of acceleration is inversely proportional to the mass.

ma

1

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When Fnet = 0,

If a = 0, then F = 0

then a = 0

1st Law

(Newton’s 1st Law is a special case of 2nd Law)

amF net

Newton’s 2nd Law of Motion

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Newton’s 3rd Law of Motion

To every action there is always an equal and opposite reaction.

Action and reaction of forces are equal and opposite.

The two forces acts on different bodies, so resultant force 0.

BAAB FF

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The astronaut pushes on the spacecraft with a force +P and it pushes back on him with a force –P.

Example:

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Conceptual Question

• Why do you forward when your car suddenly comes to a halt? Why are you pressed backward against the seat when your car rapidly accelerates? In your explanation, refer to the appropriate one of Newton’s three laws of motion.

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Reasoning and Solution• When the car comes to a sudden halt, the upper part of the body continues

forward (as predicted by Newton's first law) if the force exerted by the lower back muscles is not great enough to give the upper body the same deceleration as the car. The lower portion of the body is held in place by the force of friction exerted by the car seat and the floor.

• When the car rapidly accelerates, the upper part of the body tries to remain at a constant velocity (again as predicted by Newton's first law). If the force provided by the lower back muscles is not great enough to give the upper body the same acceleration as the car, the upper body appears to be pressed backward against the seat as the car moves forward.

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Conceptual Question

• The net external force acting on an object is zero. Is it possible for the object to be traveling with a velocity that is not zero? If your answer is yes, state whether any conditions must be placed on the magnitude and direction of the velocity. If your answer is no, provide a reason for your answer.

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Reasoning and Solution

• If the net external force acting on an object is zero, it is possible for the object to be traveling with a nonzero velocity. According to Newton’s second law, F = ma, if the net external force F is zero, the acceleration a is also zero. If the acceleration is zero, the velocity must be constant, both in magnitude and in direction. Thus, an object can move with a constant nonzero velocity when the net external force is zero.

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When you are solving a dynamics problem, it is very helpful to draw a free body diagram.

In such a diagram, a sketch of each object in the problem is drawn, showing all the forces acting on it.

Free body diagram

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The weight of an object is the force due to the attraction between it and the Earth.

Some common forces

• Weight

W = mg

W

The weight always acts downward, towards the center of the earth.

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• Normal ForceWhen an object is pressed against a surface, there is a force in the direction of the normal to the surface.

W = mg

N = mg

W = mg

N = mg cos

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When a rope is attached to an object and pulled taut, the rope is under tension.

• Tension

The direction of tension is always away from the object.

W = mg

W = mg

W = mg

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When an object slides over a surface there is a force resisting the motion. The frictional force is parallel to the surface and in the direction opposite to that of the motion.

• Friction

f

motion

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(a) Weight W and the normal reaction force FN on a stationary body.

(b) The resultant frictional force FsF due to the applied force FT but

with no resultant motion, (c) the resultant frictional force FkF due to the applied force FT that

causes the body to move with velocity v.

w

FN

ww

FN FN

FTFTFsF FkF

v

(a) (b) (c)

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Frictional Force• Normal force, FN is one of the force that a surface exerts on an object

with which it is in contact-namely the component that perpendicular to the surface.

• The component parallel to the surface is the frictional forces, which are

• static friction force, FsF when the body at rest (stationary).

• kinetic friction, FkF when the body is moving

• By definition FN and FsF or FkF are always perpendicular to each other.

• The direction of the friction force is always such as to oppose relative motion of the two surfaces.

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Static Friction FSF • Frictional forces can act even

when there is no relative motion of the two surfaces in contact. This is called static friction force.

• When the force FT is gradually

increased, the frictional force FsF

also increases staying equal in magnitude to FT as long as the

body remains at rest.

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Static Friction • For some given pair of surfaces the maximum value of FsF depends on

the normal force FN and can have any value between zero and a

maximum value just before the body moves.

• In such cases it is found that FsF is approximately proportional to FN , i.e

NSFSFNSF FFFF Thus

where the proportionality constant SF is called the

coefficient of static friction.

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Kinetic Friction FkF

• At some point, FT becomes greater than the maximum frictional

force FsF that the surface can exert and the body starts to slide along

the surface with velocity v.

• In many instances, the magnitude of the sliding friction FkF is found

to be approximately proportional to FN, that is

NkFkFNkF FFFF Thus

where the proportionality constant kF is called

the coefficient of kinetic friction.

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Typical Coefficients of Friction

Rubber on concrete(dry) 0.80 0.90

Steel on steel 0.57 0.74

Glass on glass 0.40 0.94

Wood on leather 0.40 0.50

Copper on steel 0.36 0.53

Rubber on concrete (wet) 0.25 0.30

Steel on ice 0.06 0.10

Waxed ski on snow 0.05 0.10

Teflon on Teflon 0.04 0.04

Materials Kinetic, µk Static, µ s

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Measurement of Friction

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Measurement of Friction

• If is small enough that the block does not slide, then we must have

0

0

SFx

Ny

FsinmgF

cosmgFF

c

NSFSFcNSF FFFF

tan Thus

have Wetan

sF

cosNF

mg

sinmgFSF

tanNSF FF •By eliminating mg from (1) and (2) gives

•Increase until the block just begins to slide. At this critical angle c

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• In general, if an object is on the verge of slipping when the surface on which it rests is tilted at an angle c, the coefficient of static friction between the object and the surface is sF = tan c..

cNsFcNSF tanFtanFF sFor

•Its independent of the mass of the object.

•Fsf has a maximum value.

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Apparent Weight

• Apparent weight is the force that the object exerts on the scale with which it is in contact, in an acceleration system.

• For example, the sensation of feeling heavier or lighter in an accelerating elevator.

• The apparent weight of a passenger with a mass m riding in an elevator with acceleration ay can be understood better by looking at the free body diagram.

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The free-body diagram showing the forces acting on the person riding in the elevator accelerating upwards at ay.

yN maWFF

(Apparent weight is equal to the normal force acting on the body)

y

yN

agm

maWF

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)ag(mF yN

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Weightlessness• When ay is positive, the elevator accelerated upward, and Fa is

greater than the passenger’s weight, w=mg.

• When the elevator is accelerated downward, ay is negative, and Fa is less than the weight, w=mg.

• When the chain broken the elevator accelerated with ay= -g, when its in free fall. In this case Fa =0 and the passenger seems to be weightless. 0 ggmFa

• Similarly, an astronaut orbiting the earth in a space capsule experiences apparent a weightlessness.

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Example• A 5.0 kg salmon is weighed by hanging it from a a fish scale attached to

the ceiling of an elevator. What is the apparent weight of the salmon, Wa, if the elevator (a) is at rest, (b) moves with an upward acceleration 2.5 m/s2, or (c) moves with a downward acceleration of 3.2 m/s2

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Solution: The weight W = mg,

(a) F = Wa- W = may = 0, (ay=0)

So The apparent weight, Wa= mg = 5.0 kg (9.81m/s2) = 49N

(b) It moves up so F = Wa-W =may = ma where

ay = 2.5m/s2.

Thus Wa= mg + ma = 49N + (5kg)(2.5m/s2) = 62 N

(c) It moves down, F = Wa-W = may = -ma, where

ay= -3.2m/s2 Thus Wa= mg – ma = 49N - (5kg)(3.2m/s2)

= 33N

It is observed that when it moves (accelerating) up its apparent weight is greater than the original weight and when it moves down is less.

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Example:• A block of mass m1 slides on a frictionless tabletop. It is connected

to a string that passes over a pulley and suspends a mass m2. Find the acceleration of the masses and the tension in the string.

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To solve:• There is no friction in the pulley and we consider the string is

massless.

• The tension T is the same throughout and it applies a force of magnitude T to each body.

• The weights are m1g and m2g.

• If the string does not stretch, the two masses must move equal distances in equal time. Their speeds are equal at any instance.

• When the speeds change, they changed with equal amounts and gives equal acceleration at the same time.

• Draw separate free body diagram for each body and apply Newton’s 2nd Law

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N

For m2:

Fx = 0

Fy = W2 T

= m2 g T = m2 a (2)Combine (1) and (2):m2 g m1 a = m2 a

For m1:

Fy = N – W1 = 0

Fx = T = m1 a (1)

21

2

mm

gma

21

211 mm

gmmamT

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ExerciseA 4-kg block is connected to a 2-kg block by means of a massless rope through a frictionless pulley as shown in the figure. If the coefficient of static friction between the 4-kg block and the surface is 0.3, what is the acceleration of the blocks?

mg

N

Ff T

mg

T

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N

For m2:

Fx = 0

Fy = W2 T

= m2 g T = m2 a (2)Combine (1) and (2):m2 g m1 a = m2 a

For m1:

Fy = N – W1 = 0

Fx = T = m1 a (1)

21

2

mm

gma

21

211 mm

gmmamT

m/s 273

24

8192.

.

= (4)(3.27) = 13.1 N

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mg

NFf T

mg

Ty

x

For the 2-kg block :

Fx = 0

Fy = W2 T

= m2 g T = m2 a (2)

For the 4-kg block: Fy = N – W1 = 0

Fx = T – Ff = m1 a

T – m1 g = m1 a (1)

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Add (1) and (2):m2 g m1 g = m1 a + m2 a

21

12

mm

gmgma

gmamT 11

2m/s 31

24

8194308192

.

...

819430314 ...

T – m1 g = m1 a (1) m2 g T = m2 a (2)

From (1):

= 17.0 N

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ExampleA jet plane is flying with a constant speed along a straight line at an angle of 30 above the horizontal. The plane has a weight, W = 86500 N, an its engine provide a forward thrust, T = 103000 N. What are the lift force, L, perpendicular to the wings and the air resistance force, R, opposite the motion act on the plane?

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Solutionyx

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Example• A flatbed truck slowly tilts its bed upward to dispose of

a 95.0-kg crate. For small angles of tilt the crate stays put, but when the tilt angle exceeds 23.2 the crate begins to slide. What is the coefficient of static friction between the bed of the truck and the crate.

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Choose the positive x axis along the surface of the bed. They are three forces acting on the crate, NF, Wg and Fs. .

When the crate begins to slide it is in its verge of slipping, The static force Fs=-sN.

We resolve the force vectors acting on the crate into x and y-components.

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0

sinmgW

NFF

N

x

smax,sx,s

x

Thus

00

cosmgN

macosmgNWFNF yyy,syy

Since the crate is at rest,

cosmgW

F

NN

y

y,s

y

0

00

00

sinmgcosmg

masinmgNWFNF

s

xsxx,sxx

4290223

or Thus

..tantancosmg

sinmg

sinmgcosmg

s

s

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Conceptual Checkpoint:

A car drives with its tires rolling freely. Is the friction between the tires

and the road

a) Kinetic or b) static?

Reasoning and Discussion

A reasonable-sounding answer is that because the car is moving, the friction between its tires and the road must be kinetic friction – but this is not the case.

Actually, the friction is static because the bottom of the tire is in static contact with the road. To understand this, watch your feet as you walk. Even though you are moving, each foot is in static contact with the ground once you step down on it. Your foot doesn’t move again until you lift it up and move it forward for the next step. A tire can be thought of as a succession of feet arranged in a circle, each of which is momentarily in static contact with the ground.

Answer:

(b) The friction between the tires and the road is static friction.

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A tire can be thought of as a succession of feet arranged in a circle, each of which is momentarily in

static contact with the ground.

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ExerciseA crate rests on the flatbed of a truck that is initially traveling at 15 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 38 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding?

Direction of

motion of truck

mg

NFf

0220

2 asvv

222

0 m/s 3m382

m/s15

2

s

va

30819

3.

.g

a

maF mamgN

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In moving a 35.0-kg desk from one side of a classroom to the other, a professor finds that a horizontal force of 275 N is necessary to set the desk in motion, and a force of 195 N is necessary to keep it in motion at a constant speed. What are the coefficients of (a) static and (b) kinetic friction between the desk and the floor?  

(a) 0.81 (b) 0.57

mg

NFf F

(a) Ff = µs N = 275 N

µs = 275 N/343.4 = 0.81

Fy = N – mg = 0

N = mg = 35.0 × 9.81

= 343.4 N

(b) Ff = µk N = 275 N

µk = 195 N/343.4 = 0.57

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The coefficients of static and kinetic friction between a 50.0-kg box and a horizontal surface are 0.500 and 0.400 respectively. (a) What is the acceleration of the object if a 250-N horizontal force is applied to the box? (b) What is the acceleration if the applied force is 235 N?  

(a) (b) 021.1 m s

mg

NFFf

Maximum static friction = µs N = 0.500 × 490.5 N = 245.25 N

Fy = N – mg = 0

N = mg = 50.0 × 9.81 = 490.5 N

(a) Fx = ma

250 N – µk N = 250 N – (0.400 × 490.5 N) = 53.8 N

a = 53.8 N / 50.0 kg = 1.1 m/s2

(b) Applied force = 235 N < maximum static friction

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A hockey player hits a puck with his stick, giving the puck an initial speed of 5 m/s. If the puck slows uniformly and comes to rest in a distance of 20 m, what is the coefficient of kinetic friction between the ice and the puck? 

 0.064

v2 = v02 + 2as

v2 - v02

2s= 0.63 m/s2

F = µN = ma

µ mg = ma

µ = ma / mg = a/g = 0.63/9.81 = 0.064

0 - 52

2(20)a = =

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One 5-kg bucket is hanging by a massless cord from a 4-kg paint bucket, also hanging by a massless cord, as shown in the figure.

(a) Determine the tension in each cord if the buckets are at rest.

(b) If the two buckets are pulled upward with an acceleration of 1.60 m/s2 by the upper cord, calculate the tension in each cord.

4 kg

5 kg

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In the figure shown, m1 = 2.0 kg and m2 = 5.0 kg and the coefficient of kinetic

friction between m1 and the inclined plane is 0.20. What is the acceleration of

the blocks and in which direction are they are moving?

TN

m1 g

m2 g

T

Ff

y x

Fx = T – m1g sin - Ff = m1a

Fy = N – m1g cos = 0

Fy = m2g - T = m2a

= T – m1g sin 37° - N = m1a

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30°

Two forces F1 and F2 of equal magnitude are applied to a brick lying on the floor as shown in the figure above. If the coefficient of static friction between the brick and the floor is 0.4, what is the minimum value of F required to start the brick to move?