General Physics I

Spring 2011

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Applying Newton’s Laws

Friction• When you push horizontally on a

heavy box at rest on a horizontal

floor with a steadily increasing

force, the box will remain at rest

initially, i.e., remain in

equilibrium. Since the box is in

equilibrium, the net force in the

horizontal direction is zero, i.e.,

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horizontal direction is zero, i.e.,

there must be a force opposing

your push. This force is the force

of static friction. Static friction is

friction that acts when surfaces

in contact are at rest relative to

each other.

Static Friction

• As you continue to increase the applied force, the

magnitude of the static friction (fs) rises to match the

magnitude of the applied force and the box remains at

rest. Eventually, the box will begin to move relative to

the floor when the applied force exceeds the

magnitude of the maximum static friction force (fs,max)

for the two surfaces. Experimentally, it has been found

that f is approximately proportional to the

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that fs,max is approximately proportional to the

magnitude of the normal force acting on each surface:

fs,max = µsn,

where µs is the coefficient of static friction for the two

surfaces.

Kinetic Friction

• When there is relative motion between the surfaces, the

friction force becomes kinetic friction (fk). Kinetic friction is in

the opposite direction to the motion of one surface relative

to the other.

• It has also been found experimentally that the magnitude of

the kinetic friction force is approximately proportional to the

magnitude of the normal force:

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magnitude of the normal force:

fk = µkn,

where µk is the coefficient of kinetic friction.

• For the same two surfaces, usually µk < µs.

• Note that both surfaces experience the friction force,

according to Newton’s third law.

Kinetic Friction

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Kinetic friction is independent of speed and area of contact.

Static and Kinetic Friction

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Graph of friction force vs. time as the box is pushed.

Friction and Driving

• Under normal circumstances, the force that propels your car forward is static friction between the tires are the road. The bottom surface of a tire does not slip relative to the road. Note that friction is in the same direction as the motion of the car. That is why it is important to note that friction opposes relative motion between surfaces. (The tire pushes back on the road; friction opposes this attempted backward relative motion and so acts in the forward direction.)

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motion and so acts in the forward direction.)

• When you are braking, it is also better that static friction slows you down. If the brakes lock, the tires slip and the friction will be kinetic friction, which is usually less than the maximum static friction force. Anti-lock brakes prevent this from happening.

• When the roads are icy, the coefficients of static and kinetic friction are much smaller. The tires slip much more easily and the friction forces are much reduced, making driving more difficult.

Workbook: Chapter 5, Question 21

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Interacting Objects• As we saw before, when

objects interact, they experience forces that come in action/reaction pairs due to Newton’s third law. Each force that constitutes a pair acts on a different object. These forces have to be treated carefully when solving

An�

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carefully when solving problems involving interacting objects. Note that the forces of a pair have the same magnitude and this is a useful piece of information in solving problems involving interacting objects. The figure and free-body diagrams illustrate these ideas.

A

A

Aw�

handF�

BonAF�

B

Bn�

Bw�

AonBF�

AonB BonAF F=� �

Interacting Objects: Accelerations

• In many cases, the accelerations of

two interacting bodies will be

related. For example, for two

blocks being pushed while they are

in contact, their accelerations are

equal. Otherwise, the blocks would

separate. Applying Newton’s 2nd

law to each block yields:A

An�

handF�

BonAF�

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law to each block yields:A

Aw�

B

Bn�

Bw�

AonBF�

,

Block A. ---- (1)

Block B. ---- (2)

Since we can substitute for

in Eq. (1) to get or,

.hand

hand

x BonA Ahand

x BAonB

BonA AonB

BBonA A

BA

F

F

F F F m a

F F m a

F F

F m a m a

m m a

−

= − =

= =

=

=

= +

∑

∑

x

x

Interacting Objects: Accelerations• For two objects connected by a

rope or string that doesn’t

stretch or shrink, the

magnitudes of the accelerations

are the same. In the example to

the right, though the directions

of the accelerations are

different, their magnitudes are

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different, their magnitudes are

the same. Thus,

• In terms of components,

(Note that the acceleration of B

is in the negative y direction and

so is actually negative.)

.BA

a a=� �

.ByAx

a a=−

Bya

Ropes and Pulleys• In this course, we will

assume that ropes or strings

are massless and pulleys are

massless and frictionless.

• The tension in a massless

rope is the same everywhere

in the rope, including at both

ends (which are connected to

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ends (which are connected to

other objects). The rope

exerts a force on a connected

object equal to the tension.

(The object of course exerts

a reaction force on the rope

having the same magnitude

as the tension.)

Ropes and Pulleys• When a rope or string passes

over a pulley, the magnitude of

the tension in the rope is the

same on both sides of the

pulley.

T�

Frictionless surface

y

y

13

A

An�

Aw�

AT�

B

BT�

Bw�

BAT T=� �

x

Workbook: Chapter 5, Question 26, 29

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Textbook: Chapter 5, Problem 74

B

BT�

Bw�

mB = 100 kgA

AT�

Aw�

y

15

212

First, we have to find the acceleration of the 100-kg block.

Use: ( ) ( ) .

Knowns: 1 m, 0 m, 6.0 s, ( ) 0. Unknown: .

2 ( )Solving for gives:

(

yi i yBf

yi i yBf

yi if

yB yB

y y v t a t

y y t v a

y y v ta a

= + ∆ + ∆

=− = ∆ = =

− − ∆=

∆2

2 2

2( 1 m 0 0) 0.0556 m/s .) (6.0 s)t

− − −= =−

Problem 74 (continued)

2 2

Apply Newton's 2nd law to the 100-kg block.

.

So, ( ).

100 kg[9.8 m/s ( 0.0556 m/s )] 974 N.

Apply Newton's 2nd law to block of unknown mass.

y B B B yB

B B B yB B B yB B yB

B

y A A A y

F T w m a

T w m a m g m a m g a

T

F T w m a

= − =

= + = + = +

= + − =

= − =

∑

∑ . So,

( ). Thus,

A

= + = + = +

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2

2 2

( ). Thus,

.( )

Now, 974 N. Further, 0.0556 m/s .

974 NHence, 99 kg.(9.8 m/s 0.0556 m/s )

A A A yA A A yA A yA

AA

yA

B yBA yA

A

T w m a m g m a m g a

Tm

g a

T T a a

m

= + = + = +

=+

= = =− =

= =+