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AP Physics Chapter 5

Force and Motion – I

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AP Physics

Take Quiz 5 Lecture Q&A

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Force and Acceleration

Acceleration causes

Force: an interaction that causes acceleration

Kinematics: Study of motion without consideration of force (Describe motion)

Mechanics: Study of force and motion (Explain motion)

Force What causes acceleration?

V

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Newton’s First Law of MotionLaw of Inertia

An object with no net force acting on it remains at rest or moves with constant velocity in a straight line.

No net force is needed to keep a constant velocity.

A net force is needed to change a velocity.

Net force = 0 Object at Equilibrium. A special case of Second Law of

Motion.

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Interpretation of Newton’s First Law

If no net force acts on a body, we can always find a reference frame in which that body has no acceleration.

Inertial Frame of Reference:

Newton’s First Law: Law of Inertia

– Fnet = 0 on a body, then a = 0 Inertial Frame– Fnet = 0 on the frame– The frame moves with constant velocity.

– Inertia: Resistance to force – Inertia Mass for now

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Newton’s Second Law of Motion

The acceleration of a body is directly proportional to the net force on it and inversely proportional to its mass.

Fa

m

F or Fnet: net force, total force, resultant force Force is the cause, and acceleration is the effect.

Unit of Force:

F = ma [F] =

F ma

[m] [a] 2

mkg

s Newton N

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Practice: An experimental rocket sled can be accelerated at a constant rate from rest to 1600 km/h in 1.8 s. What is the magnitude of the required average force if the sled has a mass of 500 kg.

0, 1600i f

kmv v

h

1000m

km

1h

444 ,3600

1.8 , 500 , ?

m

s s

t s m kg F

a

F

v

t

2

444. 0247.

1.8f i

mv v mst s s

ma 2 5500 247 / 1.24 10kg m s N

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Force

Force is a vector.

F ma

The acceleration component along a given axis is caused only by the sum of the force components along that same axis and not affected by force components along any other axis.

x x

y y

z z

F ma

F ma

F ma

= all forces in direction

all forces in direction

+xF x

x

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constantv F

3

3

3

x

y

z

F

F

F

3ˆˆ ˆ(3 ) (11 ) (4 )F N i N j N k

1 2ˆ ˆˆ ˆ ˆ ˆ(2 ) (3 ) ( 2 ) , ( 5 ) (8 ) ( 2 ) ,

ˆ ˆ(2 / ) (7 / )

F N i N j N k F N i N j N k

v m s i m s j

constant, 3 ?F

0 1 2 3F F F

0

1 2 2 ( 5 ) 3x xF F N N N

1 2 3 8 11y yF F N N N

1 2 2 ( 2 ) 4z zF F N N N

Example: 113-62Three forces act on a particle that moves with unchanging velocity of v = (2 m/s)i – (7 m/s)j. Two of the forces are F1 = (2N)i + (3N)j + (-2N)k and F2 = (-5N)i + (8N)j + (-2N)k. What is the third force?

3F

1 2F F

ˆˆ ˆ2 5 3 8 2 2N N i N N j N N k

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Two kinds of mass

Two methods to measure or use mass

0 0F ma m a

W mg

Inertial mass: measure mass by comparing acceleration of this object to acceleration of an object of known mass when the same force applied to both of them.

Gravitational mass: measure mass by comparing gravitational force on this object to gravitational force on an object of known mass at the same location

00

am m

a

Wg

m 0

0

WWg

m m 0

0

Wm m

W

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But what exactly is mass?

mass weight or size intrinsic characteristic how much a body resist to force a characteristic of a body that relates a force

on the body to the resulting acceleration of the body

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Free Body Diagram

Force Diagram Draw simple diagram Draw all forces acting on the object being

considered– Ignore all forces this object acting on other objects– Draw forces starting from center of object or at

points of action– Make sure each force giver can be identified

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Free-Body Diagram Example

W: Weight of object, gravity of Earth pulling on box

N: Normal force, force of incline supporting box

f: friction, force of incline surface opposing motion

v

Do not confuse velocity with forces.

T: tension, force of person pulling on box

f: friction, force of incline surface opposing motion

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Weight or gravity

Weight or gravity: gravitational force the Earth pulling on object around it

Always straight downward

– m: mass of object– g = 9.8 m/s2 near surface of Earth: acceleration due to gravity

When it is the only force acting on an object,

(acceleration due to gravity) free fall

Higher elevation, smaller g Lower latitude (closer to equator), smaller g

Fa

m

mg

mg

W mg

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Normal Force (N, but not Newton)

Given by the surface in contact to support the object.

NN

N

Fapp

• No contact No normal force• No tendency to move into surface No normal force

Always perpendicular to the surface in contact. (Normal = perpendicular)

Points from surface to the object

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Two kinds of frictions

• Friction: force opposing the motion or the tendency of motion between two rough surfaces that are in contact

sf

kf

– Static friction:

– Kinetic (or sliding) friction:

– Or simply: f

More next chapter.

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Tension

Tension is the force the person pulling on the box through the string. (Or string pulling on box.)

Tension is along the string and points away from the object of consideration.

Pulley can be used to change direction of tension. Reading of spring scale gives magnitude of tension in string.

T

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What is the tension/scale reading?

200N 200N

a) 0 N

b) 200 N

c) 400 N

scale

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What is the tension?

Wall

Answer:

Consider:

200N

Same force diagram as before.

reading =

tension =Wall?

b) 200 N

200N200N

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Weight and Apparent Weight

Apparent weight is the reading on the apparatus. Apparent weight is the normal force

the scale exerting on the object Apparent weight is the tension the

spring exerting on the object

T

Apparent weight does not have to be the same as the weight.

Weight is force of gravity of Earth pulling on the object.

W mgN

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Example: A weight-conscious penguin with a mass of 15.0 kg rests on a bathroom scale. What are a) the penguin’s weight W and b) the normal force N on the penguin? c) What is the reading on the scale, assuming it is calibrated in weight units?

kgm 0.15

) ?a W

) ?b N

) ?appc W

Let upward = +

W

F

F N W N

appW 147.N N

ma 0

N W0

215.0 9.8 147.m

mg kg Ns

147.W N

W

N

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Practice: An object is hung from a spring balance attached to the ceiling of an elevator. The balance reads 64 N when the elevator is standing still. What is the reading when the elevator is moving upward a) with a constant speed of 7.6 m/s and b) with a speed of 7.6 m/s while decelerating at a rate of 2.4 m/s2?

a) v = 7.6 m/s =

F

W=mg

T +W=64N

T = ?Define upward as the positive direction.

constant a 0

F

T

T W ma

m

64 6.53 0 64N kg N

T Wma

W ma

2

646.53

9.8 /

W Nkg

g m s

264 6.53 2.4 / 48.T W ma N kg m s N

This apparent weight is different from the weight.

b) a = -2.4 m/s2, T = ?

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Newton’s Third Law of Motion

When one object exerts a force on a second object, the second object also exerts a force on the first object that is equal in magnitude but opposite in direction. Action and reaction forces always exist together. Action and reaction forces are of the same kind of force.

(Both gravitational forces or both electromagnetic forces.)

Action and reaction forces act on two different objects and therefore do not cancel out each other when only one object is considered.

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Action and Reaction Forces

In general, the force acting on the object under consideration is the action force, and the other is the reaction.

Action

Reaction Force between box and ground:

– Action force: ground supporting the box, Fgb

– Reaction force: box pushing on the ground, Fbg

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How To Find Reaction Force

1. Identify the force giver and receiver of the force.

2. Reversing the order of force giver and receiver, you will get the reaction force.

Example: Action force: a b Reaction force: b a

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Example on Action-Reaction

table

box

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Example (2)

Wb: weight of box, force Earth pulling on box

N: normal force, force table supporting the box

Are N and W a pair of action and reaction force?

forces on box

No 2 different objects same fundamental force

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Example (3)

N: Force of box pushing on table

N

Ngt: normal force of ground supporting table

Wt: weight of table, gravitational force Earth pulling on table

forces on table

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Example (4)

Ntg: force of table pushing on ground (Earth), reaction force of Ngt.

Wt: reaction force to weight of table, gravitational force table pulling on Earth

Wt

Wb: reaction force to weight of box, force box pulling on Earth Wb

forces on earth

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Example (5)

Ntg

WtWb

Ngt

NWt

N

Wb

Action and reaction forces always exist in pairs and act on different objects.

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Example: 109-27A 40 kg girl and an 8.4 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible mass. The girl exerts a horizontal 5.2 N force on the rope. What are the acceleration magnitudes of (a) the sled and (b) the girl? c) How far from the girl’s initial position do they meet?

2) ?a a

1 240 , 8.4 , 15 , 5.2m kg m kg D m T N

T

W2

N2

m1m2

1) ?b a

22

2

5.20.62

8.4

F N m

m kg s

1

21

5.20.13

40

F N m

m kg s

2a

1a W1

N1

T

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Continues …

1) 15 , ?c D m x

1 2x x

21 1

1

2iv t a t

2 21 2

1 1

2 2a t a t D

1x

D

22 2

1

2iv t a t D

21 2

1

2a a t D 2

1 2

2Dt

a a

21

1

2a t

1

2 1

2a

1 2

D

a a

1

1 2

aD

a a

2

2 2

0.1315 2.6

0.13 0.62

m

s m mm m

s s

T

W2

N2

m1m2

W1

N1

T

1x2x

D

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Practice: An 80-kg person is parachuting and experiencing a downward acceleration of 2.5 m/s2. The mass of the parachute is 5.0 kg. a) What upward force is exerted on the open parachute by the air? b) What downward force is exerted by the person on the parachute?

MF +

Choose downward = + direction. m1 = 80 kg, a = 2.5 m/s2, m2 = 5.0 kg, M = m1 + m2 = 85 kg

W = Mg

f

a) f = ?

f

W2 = m2g

T

b) T = ?

2F

Chute & perso

n

Chute only

f

2 285 9.8 2.5 621

m mkg N

s s

2 2621 5.0 9.8 5.0 2.5 585

m mN kg kg N

s s

W f Ma

W Ma Mg Ma M g a

2T W f 2m a 2 2T f W m a 2 2f m g m a

What if we did force diagrams for individual masses?

2 2T f W m a 2 2f m g m a

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Another Approach: An 80-kg person is parachuting and experiencing a downward acceleration of 2.5 m/s2. The mass of the parachute is 5.0 kg. a) What upward force is exerted on the open parachute by the air? b) What downward force is exerted by the person on the parachute?

1F

+Choose downward = + direction. m1 = 80 kg, a = 2.5 m/s2, m2 = 5.0 kg, M = m1 + m2 = 85 kg

W1 = m1g

T

a) f = ?

f

W2 = m2g

T2F

Person only

Chute only

2 2584 5.0 9.8 2.5 621

m mN kg N

s s

1W T 1m a

2T W f 2m a

f 2 2T m g m a 2 2T W m a

T 1 1W m a 1 1m g m a 1m g a

2 280 9.8 2.5 584

m mkg N

s s

2T m g a

?

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Example:Two blocks, 3.0kg and 5.0kg, are connected by a light string and pulled with a force of 16.0N along a frictionless surface. Find (a) the acceleration of the blocks, and (b) the tension between the blocks.

3.0kg 5.0kg16.0N

W2

N2

T T F

N1

W1

+

Let right = +, m1 = 5.0 kg, m2 = 3.0 kg, F = 16.0 N

Consider only forces in the horizontal direction.

m1: 1F

m2: 2F 2 1F m a m a

2 1 1 2F m a m a m m a

21 2

16.02.0

3.0 5.0

F N ma

m m kg kg s

2 23.0 2.0 6.0

mT m a kg N

s

F T 1m a

T 2m a

Also applies

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Another Approach:Two blocks, 3.0kg and 5.0kg, are connected by a light string and pulled with a force of 16.0N along a frictionless surface. Find (a) the acceleration of the blocks, and (b) the tension between the blocks.

3.0kg 5.0kg16.0N

N2

T F

N

W

+

Let right = +, m1 = 5.0 kg, m2 = 3.0 kg, F = 16.0 N

Take the two masses as a system. Then

m1 and m2: F

m2: 2F

a

2 23.0 2.0 6.0

mT m a kg N

s

F 1 2m m a

T 2m a

21 2

16.02.0

3.0 5.0

F N m

m m kg kg s

Then we can apply 2nd on either m1 or m2 to find T.

W2

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Example: Two blocks, one of mass 5.0 kg and the other of mass 3.0 kg, are tied together with a massless rope as to the right. This rope is strung over a massless, resistance-free pulley. The blocks are released from rest. Find a) the tension in the rope, and b) the acceleration of the blocks. 3

5

+

W1

W2

T

T

+

Let downward = + for m1 = 5 kg, and upward = + for m2 = 3 kg.

Then two masses will have the same acceleration, a = ?. And the tensions will be the same, T = ?

m1:

m2:

F

F

1W T 1m a

2T W 2m a

1 1T W m a

1 1 2 2W m a W m a

1 2 1 2W W m a m a 1 2 1 2m g m g m m a 1 2 1 2m m g m m a

21 2

21 2

5 3 9.82.45

5 3

mkg kgm m g msa

m m kg kg s

1 1 1 1 1 2 25 9.8 2.45 36.8

m mT W m a m g m a m g a kg N

s s

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Another Approach: Two blocks, one of mass 5.0 kg and the other of mass 3.0 kg, are tied together with a massless rope as to the right. This rope is strung over a massless, resistance-free pulley. The blocks are released from rest. Find a) the tension in the rope, and b) the acceleration of the blocks. 3

5

W1

W2

+

Take the two masses as a single system. Let clockwise as the positive direction of the motions.

1 2 1 2 1 2F W W m g m g m m g

And this net force is acceleration a total mass of 1 2m m m So the acceleration of the system is

21 2

21 2

5 3 9.82.45

5 3

mkg kgm m gF msa

m m m kg kg s

Then we apply Newton’s second law on one of the mass to find the tension.

+

T

F

1 1 1 1 1 2 25 9.8 2.45 36.8

m mT W m a m g m a m g a kg N

s s

1W T 1m a

Let downward = + for 5kg. Then

Then the net force accelerating the system is:

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Example: 114-75In Fig. 5-66, a force F of magnitude 12 N is applied to a FedEx box of mass m2 = 1.0 kg. The force is directed up a plane tilted by = 37o. The box is connected by a cord to a UPS box of mass m1 = 3.0 kg on the floor. The floor, plane, and pulley are frictionless, and the masses of the pulley and cord are negligible. What is the tension in the cord?

N1

W1

Tx

y+

T

W2

FN2

1 23.0 , 1.0 , 37 , 12 ,om kg m kg F N

Set up the coordinates as in the diagram. Also decompose W2 into the x and y directions as W2x and W2y.

W2y

W2x

37o

3.0 kg

1.0 kgFv

?T

2xW 2 2 cosyW W 2 sin ,W

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Continues …

1yF 1

Ta

m

2 21

x

TF T W m

m

12

1 2

sinm

T F Wm m

Sub into the third eqn, we have

1xF

2xF

2 yF

1m

2m

T 1m a

2 21

x

TF W T m

m 1 2

21

sinm m

F W Tm

2

3.012 1.0 9.8 sin37 4.6

1.0 3.0om kg

N kg Ns kg kg

12

1 2

sinm

F m gm m

1 1N W 0

2xF T W 2m a

2 2 yN W 0

N1

W1

Tx

y+

T

W2

FN2

W2y

W2x

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Practice: 110-41A 5.00 kg block is pulled along a horizontal frictionless floor by a cord that exerts a force T = 12.0N at an angle = 25.0o above the horizontal. a) What is the acceleration of the block? b) The force F is slowly increased. What is its value just before the block is lifted (completely) off the floor? c) What is the acceleration of the block just before it is lifted (completely) off the floor?

oNFkgm 0.25,0.12,00.5 ) ?a a

yF

b) T = ?Right before the force is large enough to lift off the block

xF

a

xT cosT ma

2

cos 12.0 cos25.02.18

5.00

oT N m

m kg s

N =

T

yT N W yma sin 0T mg

ay =

0, 0

25.00 9.8

116sin sin 25.0o

mkgmg s N

x

y

T

Tx

Ty

W=mg

N

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Continues …

c) a = ?

2

cos 116 cos25.021.0

5.00

oT N ma

m kg s

Similar to part a),