AP Physics Chapter 5
Force and Motion – I
1
AP Physics



2
Take Quiz 5
Lecture
Q&A
Force and Acceleration

Acceleration causes V

What causes acceleration?  Force
Force: an interaction that causes acceleration



3
Kinematics: Study of motion without
consideration of force (Describe motion)
Mechanics: Study of force and motion (Explain
motion)
Newton’s First Law of Motion
Law of Inertia

An object with no net force acting on
it remains at rest or moves with
constant velocity in a straight line.




4
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.
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
–
–
5
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
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.
a
F
 F  ma
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] = [m] [a]  kg 
6
m
s
2
 N ew to n  N
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.
km  1000 m   1 h 
m
v i  0, v f  1600
  444 ,


s
h  km   3600 s 
 t  1.8 s , m  500 kg , F  ?
a 
v
t

v f  vi
t
444.

m
s
1.8 s
0
 247.
m
s
2
2
5
F  ma  500 kg  247 m / s  1.24  10 N
7
 F x = + all forces in  x direction
Force


8
 all forces in  x direction
Force is a vector.
  Fx  m a x

F  m a    Fy  m a y

  Fz  m a z
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.
Example: 113-62
Three 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?
F1  (2 N ) iˆ  (3 N ) ˆj  (  2 N ) kˆ , F2  (  5 N ) iˆ  (8 N ) ˆj  (  2 N ) kˆ ,
v  (2 m / s ) iˆ  (7 m / s ) ˆj  c on sta n t, F3  ?
v  constant   F  0  F1  F2  F3  0
 F3    F1  F2 
 F3 x    F1 x  F2 x     2 N  (  5 N )   3 N

  F3 y    F1 y  F2 y     3 N  8 N    1 1 N

 F3 z    F1 z  F2 z      2 N  (  2 N )   4 N
    2 N  5 N  iˆ   3 N  8 N  ˆj    2 N  2 N  kˆ 


F3  (3 N ) iˆ  (11 N ) ˆj  (4 N ) kˆ
9
Two kinds of mass
Two methods to measure or use mass

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.
F  ma  m0a0  m 

10
a0
m0
a
Gravitational mass: measure mass by comparing gravitational
force on this object to gravitational force on an object of
known mass at the same location
W
W0
W
W
m0

 m 
 g 
W  mg  g 
W0
m
m0
m
But what exactly is mass?




11
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
Free Body Diagram



Force Diagram
Draw simple diagram
Draw all forces acting on the object being
considered
–
–
–
12
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
Free-Body Diagram Example
v
N: Normal force, force of
incline supporting box
f: friction, force of
incline surface
opposing motion
13
Do not confuse velocity
with forces.
T: tension, force of
person pulling on box
W: Weight of object, gravity of
Earth pulling on box
Weight or gravity


Weight or gravity: gravitational force the Earth pulling
on object around it
Always straight downward
W  mg
–
–

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,
a 
F
m

14


mg
m
 g
(acceleration due to gravity) free fall
Higher elevation, smaller g
Lower latitude (closer to equator), smaller g
Normal Force (N, but not Newton)

Given by the surface in contact to support the object.
•
•


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
N
Fapp
N
15
N
Two kinds of frictions
•
Friction: force opposing the motion or the
tendency of motion between two rough
surfaces that are in contact
Static friction:
–
Kinetic (or sliding) friction:
–
Or simply:
More next chapter.
16
fs
–
f
fk
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
17
What is the tension/scale reading?
200N
18
a)
0N
b)
200 N
c)
400 N
scale
200N
What is the tension?
Answer: b) 200 N
Consider:
reading =
Wall?
200N
Wall
tension =
200N
200N
Same force diagram as before.
19
Weight and Apparent Weight

Weight is force of gravity of Earth
pulling on the object.
N
W  mg

Apparent weight is the
reading on the apparatus.



20
Apparent weight is the normal force
the scale exerting on the object
Apparent weight is the tension the
spring exerting on the object
Apparent weight does not have to be the
same as the weight.
T
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?
m  15 . 0 kg
Let upward = +
a )W  ?
W  m g  15.0 kg  9.8
m
s
2
 147. N
N
b)N  ?
 F  N W 
  N  W  0  N  W  147. N
 F  ma  0 
c )W app  ?
W app  N  147. N
21
W
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?
W=64N  m 
a)
W
g

64 N
9.8 m / s
2
 6.53 kg
T
+
v = 7.6 m/s = constant  a  0
T=?
Define upward as the positive direction.
 F  T W
 F  ma

  T  W  ma

W=mg
 T  W  m a  64 N   6.53 kg    0   64 N
b) a = -2.4 m/s2, T = ?
T  W  m a  64 N   6.53 kg     2.4 m / s
2

48. N
This apparent weight is different from the weight.
22
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.



23
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.
Action and Reaction Forces


In general, the force
acting on the object under
consideration is the action
force, and the other is the
reaction.
Force between box and
ground:
–
–
24
Action
Reaction
Action force: ground supporting the box, Fgb
Reaction force: box pushing on the ground, Fbg
How To Find Reaction Force
1.
2.
Identify the force giver and receiver of the
force.
Reversing the order of force giver and
receiver, you will get the reaction force.
Example:


25
Action force: a  b
Reaction force: b  a
Example on Action-Reaction
box
table
26
Example (2)
N: normal force, force table supporting the box
forces on box
Wb: weight of box, force Earth pulling on box
Are N and W a pair of action and reaction force? No
27
 2 different objects
 same fundamental force
Example (3)
Ngt: normal force of ground supporting table
forces on table
N
N: Force of box pushing on Wt: weight of table, gravitational force
Earth pulling on table
table
28
Example (4)
Wb: reaction force to
weight of box, force box
pulling on Earth
Wb
Wt
Wt: reaction force to weight of
table, gravitational force table
pulling on Earth
forces on earth
Ntg: force of table pushing on ground (Earth),
reaction force of Ngt.
29
Example (5)
N
Ngt
Wb
Wb
Wt
N
Wt
30
Action and reaction forces
always exist in pairs and
act on different objects.
Ntg
Example: 109-27
A 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?
m1  40 kg , m 2  8.4 kg , D  15 m , T  5.2 N
a )a2  ?
a2 
 F2
N2

m2
5.2 N
 0.62
8.4 kg
2
m2
W2
m1
31
m
s
b ) a1  ?
F
a1   1
N1

5.2 N
40 kg
 0.13
m
s
2
T
T
m1
W1
N1
N2
m2
Continues …
m1
T
T
W2
W1
 x2
c ) D  15 m ,  x1  ?
 x1   x 2  D
 x1
D
1
1


2 
2 
v
t

a
t

v
t

a
t
1
 1i

2
 2i
 D
2
2




1
2
a1t 
2
 x1 
32
1
2
1
2
a 2t  D 
a1t
2
2
1
2
 a1  a 2  t  D  t 
2
2D
2
a1  a 2
0.13
m
2
 2D 
a1
s
a1 
 15 m  2.6 m
D 


m
m
2
a1  a 2
 a1  a 2 
0.13 2  0.62 2
s
s
1
Practice:
What if we did force diagrams for individual masses?
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?
f
Choose downward = + direction. m1 = 80 kg, a = 2.5
m/s2, m2 = 5.0 kg, M = m1 + m2 = 85 kg
a) f = ?
 FM  W  f  M a
 f  W  Ma 
Chute &
person
+
Mg  Ma  M
g
 a
W = Mg
m
m 

  85 kg   9.8 2  2.5 2   621 N
s
s 

f
b) T = ?
 F2  T  W 2  f  m 2 a  T  f  W 2  m 2 a  f  m 2 g  m 2 a
33
m 
m 


 621 N   5.0 kg    9.8 2    5.0 kg    2.5 2   585 N T
s 
s 


Chute only
W2 = m2g
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?
f
Choose downward = + direction. m1 = 80 kg, a = 2.5
m/s2, m2 = 5.0 kg, M = m1 + m2 = 85 kg
a) f = ?
?
 F2  T  W 2  f  m 2 a
 f  T  W2  m2a  T  m2 g  m2a
 T  m2  g  a 

m  
m 
 584 N   5.0 kg     9.8 2    2.5 2    621 N
s  
s 

+
Chute only
W2 = m2g
T
T
Person only
 F1  W1  T  m1 a
 T  W1  m 1 a  m1 g  m1 a  m 1  g  a 
34
m
m 

 80 kg  9.8 2  2.5 2   584 N
s
s 

W1 = m1g
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.
+
N1
N2
3.0kg
Let right = +, m1 = 5.0 kg, m2 = 3.0 kg, F = 16.0 N
T
T
W2
5.0kg
W1
Consider only forces in the horizontal direction.
m1:  F1  F  T  m1 a 
m2:  F2  T  m 2 a
  F  m 2 a  m1 a

 F  m 2 a  m1a   m1  m 2  a
 a
 T  m 2 a  3.0 kg  2.0
35
m
s
2
F
m1  m 2
 6.0 N

16.0 N
3.0 kg  5.0 kg
Also applies
 2.0
m
s
2
F
16.0N
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.
N2
T
3.0kg
5.0kg
W2
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  F   m 1  m 2  a
 a
F
m1  m 2

16.0 N
3.0 kg  5.0 kg
 2.0
Then we can apply 2nd on either m1 or m2 to find T.
m2:  F2  T  m 2 a
 T  m 2 a  3.0 kg  2.0
36
m
s
2
 6.0 N
+
N
m
s
2
F 16.0N
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.
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 = ?
+
T
T
3
W2
5
+
W1
m1:  F  W  T  m1 a  T  W1  m1 a
1
m2:  F  T  W 2  m 2 a   W 1  m 1 a   W 2  m 2 a
 W1  W 2  m 1 a  m 2 a  m 1 g  m 2 g   m 1  m 2  a   m 1  m 2  g   m 1  m 2  a
 a 
 m1  m 2  g
m1  m 2
 5 kg

 3 kg   9.8
5 kg  3 kg
m
s
2
 2.45
m
s
2
m
m 

 T  W1  m1 a  m1 g  m1 a  m 1  g  a   5 kg   9.8 2  2.45 2   36.8 N
s
s 

37
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.
+
T
3
5
Take the two masses as a single system. Let clockwise as the
positive direction of the motions.
Then the net force accelerating the system is:
W2
 F  W1  W 2  m1 g  m 2 g   m1  m 2  g
+
W1
And this net force is acceleration a total mass of m  m1  m 2
So the acceleration of the system is
a 
F
m

 m1  m 2  g
m1  m 2
 5 kg

 3 kg   9.8
5 kg  3 kg
m
s
2
 2.45
m
s
2
Then we apply Newton’s second law on one of the mass to find the tension.
Let downward = + for 5kg. Then
 F  W1  T  m1 a
38
m
m 

 T  W1  m1 a  m1 g  m1 a  m 1  g  a   5 kg   9.8 2  2.45 2   36.8 N
s
s 

Example: 114-75
+
N1
F
v
In 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?
3.0 kg
37o
y
N2
T
F
x
W2x
T
W1

W2y
W2
Set up the coordinates as in the diagram. Also decompose W2
into the x and y directions as W2x and W2y.
m1  3.0 kg , m 2  1.0 kg ,   37 , F  12 N , W 2 x W 2 sin  , W 2 y  W 2 cos 
o
T ?
39
y
+
N1
N2
T
x
F
W2x
Continues …
T
W1
W2y

W2
  F1 x  T  m1 a
m1  
  F1 y  N 1  W1  0
a
m1
  F2 x  F  T  W 2 x  m 2 a
m2  
  F2 y  N 2  W 2 y  0
F  T  W2x  m2
T
m1
 T   F  W 2 sin 
40
Sub into the third eqn, we have
 F  W2x  T  m2

m1
m1  m 2
T
T
m1
 F  W 2 sin  
  F  m 2 g sin 

m1
m1  m 2
m
3.0 kg

o 
  12 N  1.0 kg  9.8 2  sin 37  
 4.6 N
s

 1.0 kg  3.0 kg
m1  m 2
m1
T
Practice: 110-41
A 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?
m  5 . 00 kg , F  12 . 0 N ,   25 . 0
y
o
a )a  ?
Ty
N T
 Fx  T x  T co s   ma
a
T cos 

Tx
12.0 N  cos 25.0
m
o
 2.18
5.00 kg
m
s
2
b) T = ?
Right before the force is large enough to lift off the block
 F y  T y  N  W  m a y  T sin   m g  0
 T 
41
mg
sin 
5.00 kg  9.8
m
s

sin 25.0
x

o
2
 116 N
W=mg
N = 0, ay = 0
Continues …
c) a = ?
Similar to part a),
a 
T cos 
m
42

116 N  cos 25.0
5.00 kg
o
 21.0
m
s
2