Lecture 5

Newton’s Laws of Motion(sections 5.5-5.7)

Exam #1 - next Thursday!

- 21 multiple-choice problems- A calculator will be needed.- CHECK YOUR BATTERIES!- NO equations or information may be stored in your calculator. This is part of your pledge on the exam.- Scratch paper will be provided, to be turned in at the end of the exam.- A sign-in sheets will be used, photographs of the class will be taken, all test papers will be collected

A 71-kg parent and a 19-kg child meet at the center of an ice rink. They place their hands together and push.

(a) Is the force experienced by the child more than, less than, or the same as the force experienced by the parent?

(b) Is the acceleration of the child more than, less than, or the same as the acceleration of the parent? Explain.

(c) If the acceleration of the child is 2.6 m/s2 in magnitude, what is the magnitude of the parent’s acceleration?

3

4

On vacation, your 1300-kg car pulls a 540-kg trailer away from a stoplight with an acceleration of 1.9 m/s2

(a) What is the net force exerted by the car on the trailer? (b) What force does the trailer exert on the car? (c) What is the net force acting on the car?

5

6

Collision Course I

A small car collides

with a large truck.

Which experiences the

greater impact force?

a) the car

b) the truck

c) both the same

d) it depends on the velocity of each

e) it depends on the mass of each

Collision Course I

A small car collides

with a large truck.

Which experiences the

greater impact force?

a) the car

b) the truck

c) both the same

d) it depends on the velocity of each

e) it depends on the mass of each

According to Newton’s Third Law, both vehicles experience the

same magnitude of force.

A force F acts on mass m1 giving acceleration a1.

The same force acts on a different mass m2

giving acceleration a2 = 2a1. If m1 and m2 are

glued together and the same force F acts on this combination, what is the resulting acceleration?

Force and Two Massesa) ¾a1

b) 3/2a1

c) ½a1

d) 4/3a1

e) 2/3a1

F = m1 a1

F a1

m1

F m2

m1a3

a3 = ?

F

a2 = 2a1

m2

F = m2 a2

Mass m2 must be ( m1) because its

acceleration was 2a1 with the same

force. Adding the two masses

together gives ( )m1, leading to an

acceleration of ( )a1 for the same

applied force.

A force F acts on mass m1 giving acceleration a1.

The same force acts on a different mass m2

giving acceleration a2 = 2a1. If m1 and m2 are

glued together and the same force F acts on this combination, what is the resulting acceleration?

F = m1 a1

F a1

m1

F = (3/2)m1 a3 => a3 = (2/3) a1

F

a2 = 2a1m2

F = m2 a2 = (1/2 m1 )(2a1 )

Force and Two Massesa) ¾a1

b) 3/2a1

c) ½a1

d) 4/3a1

e) 2/3a1

F m2

m1a3

Newton’s Laws

In order to change the velocity of an object – magnitude or direction – a net force is required.

(I)

(II)

Newton’s third law: If object 1 exerts a force on object 2, then object 2 exerts a force – on object 1.

(III)

An archer shoots a 0.022-kg arrow at a target with a speed of 57 m/s. When it hits the target, it penetrates to a depth of 0.085 m.

(a) What was the average force exerted by the target on the arrow? (b) If the mass of the arrow is doubled, and the force exerted by the

target on the arrow remains the same, by what multiplicative factor does the penetration depth change? Explain.

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13

When you lift a bowling ball with a force of 82 N, the ball accelerates upward with an acceleration a. If you lift with a force of 92 N, the ball’s acceleration is 2a. Find (a) the weight of the bowling ball, and (b) the acceleration a.

When you lift a bowling ball with a force of 82 N, the ball accelerates upward with an acceleration a. If you lift with a force of 92 N, the ball’s acceleration is 2a. Find (a) the weight of the bowling ball, and (b) the acceleration a.

a)

b)

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Contact Force

Two blocks of masses 2m and m

are in contact on a horizontal

frictionless surface. If a force F0

is applied to mass 2m, what is

the force on mass m ?

a) 2F

b) F

c) ½ F

d) 1/3 F

e) ¼ F

2m m

F

17

The force F0 leads to a specific

acceleration of the entire system. In

order for mass m to accelerate at the

same rate, the force on it must be

smaller! For the two blocks together,

F0 = (3m)a. Since a is the same for

both blocks, Fm = ma

Contact Force

Two blocks of masses 2m and m

are in contact on a horizontal

frictionless surface. If a force F0

is applied to mass 2m, what is

the force on mass m ?

a) 2F

b) F

c) ½ F

d) 1/3 F

e) ¼ F

2m m

F

Two boxes sit side-by-side on a smooth horizontal surface. The lighter box has a mass of 5.2 kg, the heavier box has a mass of 7.4 kg. (a) Find the contact force between these boxes when a horizontal force of 5.0 N is applied to the light box. (b) If the 5.0-N force is applied to the heavy box instead, is the contact force between the boxes the same as, greater than, or less than the contact force in part (a)? Explain. (c) Verify your answer to part (b) by calculating the contact force in this case.

The boxes will remain in contact, so must have the same acceleration. Use Newton’s second law to determine the magnitude of the contact forces

The lighter box will require less net force than the heavier box did (for the same acceleration). If the external force is instead applied to the heavier box, the contact force must be less than it was before:

(b)

mmM

(a) Mm

Forces in Two Dimensions

The easiest way to handle forces in two dimensions is to treat each dimension separately, as we did for kinematics.

Normal Forces

The normal force is the force exerted by a surface on an object, to keep an object above the surface.

The normal force is always perpendicular to the surface.

The normal force may be equal to, greater than, or less than the weight.

No vertical motion, so

WeightApparent weight:

Your perception of your weight is based on the contact forces between your body and your surroundings.

If your surroundings are accelerating, your apparent weight may be more or less than your actual weight.

In this case the “apparent weight” is the normal force, and is equal to the sum of the gravitational attaction (actual weight) and the force required to accelerate the body, as specified

Below you see two cases: a physics student pulling or pushing a sled with a force F that is applied at an angle . In which case is the normal force greater?

a) case 1

b) case 2

c) it’s the same for both

d) depends on the magnitude of the force F

e) depends on the ice surface

Normal Force

Case 1

Case 2

In case 1, the force F is pushing down (in

addition to mg), so the normal force needs

to be larger. In case 2, the force F is

pulling up, against gravity, so the normal

force is lessened.

Normal Force

Case 1

Case 2

Below you see two cases: a physics student pulling or pushing a sled with a force F that is applied at an angle . In which case is the normal force greater?

a) case 1

b) case 2

c) it’s the same for both

d) depends on the magnitude of the force F

5) depends on the ice surface

On an Incline

a) case A

b) case B

c) both the same (N = mg)

d) both the same (0 < N < mg)

e) both the same (N = 0)

Consider two identical

blocks, one resting on a

flat surface and the other

resting on an incline. For

which case is the normal

force greater?

A B

a) case A

b) case B

c) both the same (N = mg)

d) both the same (0 < N < mg)

e) both the same (N = 0)

A B

N

WWy

x

y

f

On an Incline

Consider two identical

blocks, one resting on a

flat surface and the other

resting on an incline. For

which case is the normal

force greater?

In case A, we know that N = W.

In case B, due to the angle of the

incline, N < W. In fact, we can

see that N = W cos().

Translation Equilibrium

“translational equilibrium” = fancy term for not accelerating = the net force on an object is zero

example: book on a table

example: book on a table in an elevator at constant velocity

Ftop

FbotW

A weight on a string...if I pull the bottom string down, which string will break first?

a) top string

b) bottom string

c) there is not enough information to answer this question

How quickly is the string pulled? A sudden, strong tug is resisted by the inertia of the mass, protecting the top string. A gradual pull forces the top string to keep the system in equilibrium.

Before practicing his routine on the rings, a 67-kg gymnast stands motionless, with one hand grasping each ring and his feet touching the ground. Both arms slope upward at an angle of 24° above the horizontal.

(a) If the force exerted by the rings on each arm has a magnitude of 290 N, and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet?

(b) If the angle his arms make with the horizontal is greater that 24°, and everything else remains the same, is the force exerted by the floor on his feet greater than, less than, or the same as the value found in part (a)? Explain.

Before practicing his routine on the rings, a 67-kg gymnast stands motionless, with one hand grasping each ring and his feet touching the ground. Both arms slope upward at an angle of 24° above the horizontal.

(a) If the force exerted by the rings on each arm has a magnitude of 290 N, and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet?

(b) If the angle his arms make with the horizontal is greater that 24°, and everything else remains the same, is the force exerted by the floor on his feet greater than, less than, or the same as the value found in part (a)? Explain.

N

a)

b) if the angle is larger and everything else remains the same, the applied forces are more vertical. With more upward force from the arms, LESS normal force is required for zero acceleration

Lecture 6

Applications of Newton’s Laws

(Chapter 6)

33

a) N > mg

b) N = mg

c) N < mg (but not zero)

d) N = 0

e) depends on the size of the elevator

m a

A block of mass m rests on

the floor of an elevator that is

accelerating upward. What is

the relationship between the

force due to gravity and the

normal force on the block?

Going Up II

34

The block is accelerating upward, so

it must have a net upward force. The

forces on it are N (up) and mg

(down), so N must be greater than

mg in order to give the net upward

force!

a) N > mg

b) N = mg

c) N < mg (but not zero)

d) N = 0

e) depends on the size of the elevator

ΣF = N – mg = ma > 0∴ N > mg

m a > 0

mg

N

A block of mass m rests on

the floor of an elevator that is

accelerating upward. What is

the relationship between the

force due to gravity and the

normal force on the block?

Going Up II

Follow-up: What is the normal force if the elevator is in free fall downward?

35

You are holding your 2.0 kg

physics text book while standing

on an elevator. Strangely, the

book feels as if it weighs exactly

2.5 kg. From this, you conclude

that the elevator is:

Elevate Mea) in freefall

b) moving upwards with a constant velocity of 4.9 m/s

c) moving down with a constant velocity of 4.9 m/s

d) experiencing a constant acceleration of about 2.5 m/s2 upward

e) experiencing a constant acceleration of about 2.5 m/s2 downward

35

Use Newton’s 2nd law! the apparent weight:

You are holding your 2.0 kg

physics text book while standing

on an elevator. Strangely, the

book feels as if it weighs exactly

2.5 kg. From this, you conclude

that the elevator is:

Elevate Mea) in freefall

b) moving upwards with a constant velocity of 4.9 m/s

c) moving down with a constant velocity of 4.9 m/s

d) experiencing a constant acceleration of about 2.5 m/s2 upward

e) experiencing a constant acceleration of about 2.5 m/s2 downward

and the sum of forces:

give a positive acceleration ay

37

Frictional Forces

Friction has its basis in surfaces that are not completely smooth:

Kinetic friction

Kinetic friction: the friction experienced by surfaces sliding against one another

The frictional force is proportional to the contact force between the two surfaces (normal force):

The constant is called the coefficient of kinetic friction.

fk always points in the direction opposing motion of two surfaces

Frictional Forces

fk

fk

Naturally, for any frictional force on a body, there is an opposing reaction force on the other body

Frictional Forces

fk

fk

when moving, one bumps “skip” over each

other

fs

fs

when relative motion stops, surfaces settle

into one another

static friction

The static frictional force tries to keep an object from starting to move when other forces are

applied.

Static Friction

The static frictional force has a maximum value, but

may take on any value from zero to the maximum... depending on what is needed to

keep the sum of forces to zero.

The maximum static frictional force is also

proportional to the contact force

A block sits on a flat table. What is the force of static friction?

Static Friction

a) zero

b) infinite

c) you need to tell me stuff, like the mass of the block, μs, and what planet this is happening on

Characteristics of Frictional Forces• Frictional forces always oppose relative motion

•Static and kinetic frictional forces are independent of the area of contact between objects

• Kinetic frictional force is also independent of the relative speed of the surfaces.

(twice the mass = twice the weight = twice the normal force = twice the frictional force)

• Coefficients of friction are independent of the mass of objects, but in (most) cases forces are not:

Coefficients of Friction

Q: what units?

45

Measuring static coefficient of friction

N

W

fs

x

y

Wx

Wy

If the block doesn’t move, a=0.

at the critical point

Given the “critical angle” at which the block starts to slip, what is μs?

46

Acceleration of a block on an incline

N

W

fk

x

y

Wx

Wy

If the object is sliding down -

v

47

Acceleration of a block on an incline

N

W

fk

x

y

Wx

Wy

If the object is sliding up -

v

What will happen when it stops?