chapter 4 dynamics: newton’s laws of...

Chapter 4

Dynamics: Newton’s Laws of

Motion

• Force

• Newton’s First Law of Motion

• Mass

• Newton’s Second Law of Motion

• Newton’s Third Law of Motion

• Weight – the Force of Gravity; and the Normal Force

• Applications Involving Friction, Inclines

Recalling Recalling LastLast LectureLectureRecalling Recalling LastLast LectureLecture

When a cord, string or rope pulls

on an object, it is said to be under

tension, and the force it exerts is

called a tension force.

The figure shows you pulling a SYSTEM consisting of two boxes connected by a

cord.

The force you apply on the system is .

Tension in a Flexible CordTension in a Flexible Cord

The force you apply on the system is .

Note: If you consider each individual component of the system, the force

is actually only applied on box A and NOT on box B.

The force will be transmitted to box B via the tension in the cord .


In the figures:

Fig. (b) is a diagram of all forces applied on box A.

Note that box A exerts a force on the cord connecting

the two boxes. According to Newton’s third law,

the cord responds with a force (tension force), , the cord responds with a force (tension force), ,

of same magnitude but opposite direction.

Fig. (c) is a diagram of forces applied on B. Now the cord

pulls box B.

� If we neglect the mass of the cord, this pull will

correspond to a force equal to the tension

found on the other end of the cord (connected to A).

Note that this is NOT true if a mass is assigned to the

cord.

Problem 4.78 (textbook: (a)What minimum force F is needed to lift the piano

(mass M) using the pulley apparatus shown in Fig. 4–60? (b) Determine the tension

in each section of rope: and


, , , T3T2T1 FFF ,T4F

Problem 4.78 (textbook):

(a)

To find the minimum force, assume that the piano is moving with

a constant velocity. Since the piano is not accelerating,

For the lower pulley, since the tension in a rope is the same

throughout, and since the pulley is not accelerating, it is seen that

T 4F Mg=

T1 T 2 T1 T1 T 22 2F F F Mg F F Mg+ = = → = =

Also, since , then

(b)

Draw a free-body diagram for the upper pulley.

From that diagram, we see that

T 2F F=

2F Mg=

T 3 T1 T 2

3

2

MgF F F F= + + =

T1 T 2 T 3 T 42 3 2 F F Mg F Mg F Mg= = = =

Lower

Pulley

Upper

Pulley

Fr

T1Fr

T1Fr

T2Fr

T2Fr

T3Fr

T4Fr

On a microscopic scale, most surfaces are rough.

When we try to SLIDE an object across a table, the

roughness of both surfaces (object’s and table’s

surfaces) tends to oppose the object‘s motion. This

effect is called sliding friction, or kinetic friction.

FrictionFriction and and InclinedInclined PlanesPlanes

effect is called sliding friction, or kinetic friction.

The opposition to the body’s motion is due to a force

called friction force.


The friction force acts in the opposite direction to the object’s motion.

In the figure, the forces applied on the box are

: the applied force;

: the normal force;

: the gravitational force;

: the friction force.: the friction force.

The reason for the friction force is not very well

understood, though it has been very well modeled.

Experiments have shown that the magnitude of the friction force is approximately

proportional to the magnitude of the normal force between two surfaces, and does

not depend on the total surface area of contact between the two surfaces.

(4.5)


The term is called coefficient of kinetic friction, and its value depends on the

nature of the two contacting surfaces.

Sliding friction (or kinetic friction) is not present when the object is not SLIDING.

(4.5)

Sliding friction (or kinetic friction) is not present when the object is not SLIDING.

However, you might have experienced the fact that it takes different (magnitude)

forces to slide a same object across different surfaces.

This is due to the so called static friction. The force of static friction is parallel to the

two contacting surfaces.

Note: Let me stress again � there is no sliding force if the object moves (or is at

rest) without sliding.


The force of static friction varies with the applied force on the object � The more

force you apply, the bigger is the force of static friction.

There will be a critical applied force that will ultimately set the object in motion. This

point will correspond to the maximum force of static friction:

(4.6)

The tem is called the coefficient of static friction.

Note that, at any point, it is true that force of static

friction will be such that:

(4.6)


The magnitude of the friction forces depends on the nature of the two surfaces in

contact.

Problem 4-36 (textbook: If the coefficient of kinetic friction between a 35-kg crate

and the floor is 0.30, what horizontal force is required to move the crate at a steady

speed across the floor? What horizontal force is required if is zero?



A free-body diagram for the crate is shown.

The crate does not accelerate vertically, and so

Also, the crate does not accelerate horizontally

(steady speed), and so

NF m g=

P frF F=

mgr

NFr

frFr

PFr

Putting this together, we have

If the coefficient of kinetic friction is zero, then the horizontal force required is 0 N,

since there is no friction to counteract. Of course, it would take a force to START the

crate moving, but once it was moving, no further horizontal force would be necessary

to maintain the motion.

P frF F=

( )( )( )2 2

P fr N0.30 35 kg 9.8m s 103 1.0 10 N

k kF F F mgµ µ= = = = = = ×

Problem 4-38 (textbook Suppose that you are standing on a train accelerating at

0.20 g. What minimum coefficient of static friction must exist between your feet and

the floor if you are not to slide?



A free-body diagram for you as you stand on the train is shown.

You do not accelerate vertically, and so

NF mg=

mgr

NFr

frFr

The maximum static frictional force is , and that must be greater than or equal

to the force needed to accelerate you.

The static coefficient of friction must be at least 0.20 for you to not slide.

s NFµ

fr N 0.20 0.20

s s sF ma F ma mg ma a g g gµ µ µ≥ → ≥ → ≥ → ≥ = =


The problem with inclined plans are no different from the problems we have

approached so far.

However, remember:

1- The normal force is ALWAYS perpendicular to the plane where the object lies on.

� So, in an inclined plane its is not going to be in the vertical direction.� So, in an inclined plane its is not going to be in the vertical direction.

(See next slide)


Three forces can ALWAYS be identified acting on an object moving on an inclined

place:

• Gravity (vertical);

• Friction (along the surface);

• Normal (perpendicular to the surface).friend 1

friend 2

So, it is convenient to define your

coordinate system such that the x

direction coincides with the inclined

line, and y in the direction perpendicular

to the plane.

You can surely have other applied forces:

� for instance, you (or you and a group

of friends) can push or pull this object.

you

Problem 4-41 (textbook A 15.0-kg box is released on a 32º incline and accelerates

down the incline at 0.30 m/s2. Find the friction force impeding its motion. What is the

coefficient of kinetic friction?


NFr

frFr

y

x

θ

θ

mgr

frF


A free-body diagram for the box is shown.

Write Newton’s 2nd law for each direction:

Notice that the sum in the y direction is 0, since

y

x

θ

θ

mgr

NFr

frFr

fr

N

sin

cos 0

x x

y y

F mg F ma

F F mg ma

θ

θ

= − =

= − = =

∑∑

Notice that the sum in the y direction is 0, since

there is no motion (and hence no acceleration) in

the y direction. Solve for the force of friction.

Now solve for the coefficient of kinetic friction. Note that the expression for the

normal force comes from the y direction force equation above.

mg

( ) ( )( )fr

2 o 2

fr

sin

sin 15.0 kg 9.80m s sin 32 0.30m s 73.40 N 73 N

x

x

mg F ma

F mg ma

θ

θ

− = →

= − = − = ≈

( )( )( )fr

fr N 2 o

73.40 Ncos 0.59

cos 15.0 kg 9.80m s cos 32k k k

FF F mg

mgµ µ θ µ

θ= = → = = =

Problem 4-65 (textbook A bicyclist of mass 65 kg (including the bicycle) can coast

down a 6.0º hill at a steady speed of 6.0 Km/h because of air resistance. How much

force must be applied to climb the hill at the same speed and same air resistance?



Consider a free-body diagram for the cyclist coasting downhill at a constant speed.

Here we call Ffr the friction due to air resistance (and not sliding or static friction).

Since there is no acceleration, the net force in each direction must be zero. Write

Newton’s 2nd law for the x direction.

This establishes the size of the air friction force at

6.0 km/h, and so can be used in the next part.

fr frsin 0 sin

xF mg F F mgθ θ= − = → =∑ y

θ

NFr

frFr

θ6.0 km/h, and so can be used in the next part.

Now consider a free-body diagram for the cyclist

climbing the hill. Fp is the force pushing the cyclist

uphill. Again, write Newton’s 2nd law for the x direction,

with a net force of 0.

xθ

mgr

θ

y

x

θθ

mgr

NFr

frFr

PFr

fr Psin 0

xF F mg Fθ= + − = →∑

( )( )( )P fr

2 o 2

sin 2 sin

2 65 kg 9.8 m s sin 6.0 1.3 10 N

F F mg mgθ θ= + =

= = ×

AssignmentAssignment 55

Textbook (Giancoli, 6th edition), Chapters 4 and 6:

Due on Thursday, October 23, 2008

- Problem 52 - page 102 of the textbook

- Problem 87 - page 105 of the textbook

- Problems 6 and 8 - page 162 of the textbook

Tutorial next week, Oct. 15, CL-127 �� solving an old midterm.

MidtermMidterm

Date: October 16, 1pm – 2:15pmRemind:

- There are four problems � Full mark will be considered only when all

problems have been completed

- You ARE allowed to use a calculator

- A equation sheet will be provided with some useful equations and identities.

- You are NOT allowed to:

- Use laptops;

- Have cell phones (please, turn it off if you have one);

- Use or consult any material other than those provided by me;

-Make sure you bring your photo ID with you

- Do not forget to write your name and UofR ID on each of the notebook used

to solve your problems..

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