Lecture 6

Newton’s Laws of Motion

Exam #1 - next Thursday!- 20 multiple-choice problems- No notes allowed; equation sheet provided- 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.

Cart on Track I

a) slowly come to a stop

b) continue with constant acceleration

c) continue with decreasing acceleration

d) continue with constant velocity

e) immediately come to a stop

Consider a cart on a

horizontal frictionless

table. Once the cart

has been given a push

and released, what will

happen to the cart?

Cart on Track I

a) slowly come to a stop

b) continue with constant acceleration

c) continue with decreasing acceleration

d) continue with constant velocity

e) immediately come to a stop

Consider a cart on a

horizontal frictionless

table. Once the cart

has been given a push

and released, what will

happen to the cart?

After the cart is released, there is no longer a force in

the x-direction. This does not mean that the cart

stops moving!! It simply means that the cart will

continue moving with the same velocity it had at the

moment of release. The initial push got the cart

moving, but that force is not needed to keep the cart

in motion.

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 m1 a3

a3 = ?

F

a2 = 2a1m2

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 m2 m1 a3

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

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 sum of the gravitational attaction (actual weight) and the force required to accelerate the body, as specified

Newton’s Third Law of Motion

Forces always come in pairs, acting on different objects:

If object 1 exerts a force on object 2, then object 2 exerts a force – on object 1.

These forces are called action-reaction pairs.

Some action-reaction pairs

Action-reaction pair?

a) Yes

b) No

Newton’s 3rd: F12 = - F21

action-reaction pairs are equal and opposite, but they act on different bodies

Newton’s Third Law of Motion

Although the forces are the same, the accelerations will not be unless the objects have the same mass.

Q: When skydiving, do you exert a force on the earth? Does the earth accelerate towards you?

Is the magnitude of the acceleration of the earth the same as the magnitude of your acceleration?

Contact Force

2m m

F

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) ½F

d) 1/3F

e) ¼F

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

Newton’s Third Law of Motion

Contact forces:

The force exerted by one box on the other is different depending on which one you push.

Assume the mass of the two objects scales with size, and the forces pictured are the same. In which case is the magnitude of the force of box 1 on box 2 larger?

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

(a)

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)

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.

Forces in Two Dimensions

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

x

y

Fmom,x

35oFmom,y

Σ Fy = 0 ay = 0 ax = Fx/m = 33N / 19 kg = 1.7 m/s2

mass of child + sled = 19 kg

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

Below you see two cases: a physics student pulling or pushing a sled with a force F that is applied at an angle θ. 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 θ. 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

θ

θ

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(θ).

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)

Ftop

FbotW

A weight on a string with identical string hanging from bottom …

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

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)

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?

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?

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

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.