Force
Chapter 4
Isaac Newton (1642 to 1727)
• Born 1642 (Galileo
dies)
• Invented calculus
• Three laws of motion
• Principia
Mathematica
Newton’s Three Law’s of Motion
1. All objects remain at rest or in uniform,
straight-line motion unless acted upon by
an outside force. (inertia)
2. Force = mass X acceleration
3. Every force has an equal and opposite
force.
The First Law
Inertia –tendency of an object to remain
at rest or in constant motion.
mass - measure of inertia.
Mass & inertia are directly proportional
The First Law
Ball in jar example:
The First Law
Direction of jar
The First Law
Now stop pushing the jar (same as not
wearing a seatbelt)
Jar gets stopped
Marble keeps going
The First Law
School bus example
This is the way you want to keep going
The bus has turned,
so you feel pulled to
the side.
The First Law
Bee in a car example
1. Bee is in air when car starts
2. Bee is on the seat when car starts
Does it take less force to push the elephant
(ignore friction) on earth or on the moon?
Does it take less
force to move the
elephant if he is
“weightless” in
space?
Inertial Reference Frames
• Non-accelerating (constant velocity)
reference frame
– All laws of physics are identical
– Cannot tell if you are moving in an Inertial
Reference Frame
– Speed of light question
The Second Law
Force = mass X acceleration
SF = ma
Sum of all the forces acting on a body
The Second Law: Use the Force
The Second Law
Situation One:
Non-moving Object
• Still has forces
Force of
the
material
of the
rock
Force of
gravity
The Second Law
Situation Two: Moving Object
SF = ma
ma = Fpedalling – Fair - Ffriction
Fair
Ffriction
Fpedalling
The Second Law
• Unit of Force = the Newton
SF=ma
SF = (kg)(m/s2)
1 N = 1 kg-m/s2
• A force of 1 Newton can accelerate a 1 kg
object from rest to 1 m/s in 1 s.
The Second Law: Example 1
A 60.0 kg bike and rider accelerates at 0.5
m/s2. How much extra force did the rider’s
legs have to provide?
SF = ma = (60.0 kg)(0.5 m/s2)
SF = 30 kg m/s2 = 30 N
The Second Law: Example 2
A force of 5 Newtons can accelerate a
watermelon 2.5 m/s2. What is the mass of
the watermelon?
(No actual watermelons were harmed in the production of this example problem)
The Second Law: Example 3
What Force is needed to accelerate a 5 kg
bowling ball from 0 to 20 m/s over a time
period of 2 seconds?
The Second Law: Example 4
Calculate the net force required to stop a
1500 kg car from a speed of 100 km/h
within a distance of 55 m.
100 km/h = 28 m/s
v2 = vo2 + 2a(x-xo)
a = (v2 - vo2)/2(x-xo)
a = 02 – (28 m/s)2/2(55m) = -7.1 m/s2
SF = ma = (1500 kg)(-7.1 m/s2) = -1.1 X 104 N
(negative sign tells us that the force is the in the
opposite direction of motion)
Fbrakes
Direction of
original motion
The 3rd Law
“For every force, there is an equal and
opposite force.”
The 3rd Law
Runner example:
• Does the runner push on the earth?
• Why does the runner move more?
• Does the earth move at all?
Backward force for the earth
Forward force for the runner
The 3rd Law
Fforward
Fgases
The 3rd Law
Sled of bricks on Ice:
• Would the sled move?
ICE
The 3rd Law
Why would Cyclops be in trouble?
Mass vs. Weight
Mass (kg)
– Amount of matter in an object
– Independent of gravity (the # of atoms in an
object does not change)
Weight (N)
– Force from gravity pulling on an object
– Weight = mg
– English unit is a pound
Cereal Boxes
• Metric unit of mass = kilogram
• English unit of mass = slug
1 lb = 4.45 N
2.2 lb equivalent to 1 kg
Weight Example 1
What is the weight of a 60.0 kg man?
Weight Example 2
If a freshman weighs 500 N, what is her
mass?
Weight Example 4
A 60.0 kg person weighs 100.0 N on the moon.
What is the acceleration of gravity on the
moon?
Weight = mgmoon
gmoon = Weight/m
gmoon = 100.0 N/60.0 kg = 1.67 m/s2
(Note that this is almost exactly 1/6th of earth’s
gravity)
Calculate the weight of a 56 kg person:
a) On earth (549 N)
b) On the moon (g=1.7 m/s2) (95.2 N)
c) Suppose that person is taken to a different
planet and has a weight of 672 N. Calculate
the acceleration of gravity on that planet.
d) If the person dropped a ball from a height of
1.80 m on that planet, calculate the speed
that it would hit the ground. (6.57 m/s)
The Normal Force
Normal Force – The force exerted by a
surface perpendicular to the contact
FN
FN
Fg = W
Fg = W
Normal Force: Example 1a
A 10.0 kg present is sitting on a table.
Calculate the weight and the normal force.
FN
Fg = W
Since the box is not moving SF = 0
SF = FN – W
0 = FN – 98.0 N
FN = +98.0 N
Normal Force: Example 1b
Suppose someone leans
on the box, adding an
additional 40.0 N of
force. Calculate the
normal force.
Again, the box is not moving so SF = 0
SF = FN – W – 40.0 N
0 = FN – 98.0 N – 40.0 N
FN = +138.0 N
Normal Force: Example 1c
Now your friend lifts up with
a string (but does not lift
the box off the table).
Calculate the normal
force.
Again, the box is not moving so SF = 0
SF = FN+40.0 N – W
0 = FN + 40.0 N – 98.0 N
FN = +58.0 N
Normal Force: Example 1d
What happen when the person pulls upward
with a force of 100 N?
SF = FN+ Fp – mg
SF = 0 +100.0N – 98N = 2.0N
ma = 2N
a = 2N/10.0 kg = 0.2 m/s2
Fp = 100.0 N
Fg = mg = 98.0 N
Example
A 2.0 kg ball is thrown into the air with a
force of 30.0 N.
a) Calculate the acceleration. (5.2 m/s2)
b) Calculate speed at the top of the throw,
2.00 m. (4.56 m/s)
c) Draw a free body diagram while the ball
is being tossed.
d) Draw a free body diagram while the ball
is above the thrower.
Free Body Diagrams: Ex. 1
Calculate the sum of the two forces acting
on the boat shown below. (53.3 N, +11.0o)
A basketball player throws a basketball through 1.2 m
from rest to a speed of 1.50 m/s. The basketball
has a mass of 0.56 kg and travels vertically.
a) Calculate the acceleration of the basketball.
(0.938 m/s2)
b) Calculate the force the player exerted on the ball.
(6.01 N)
c) Draw a free body diagram of the ball while being
thrown.
d) Draw a free body diagram of the ball in the air.
e) Calculate how long it will take the ball to reach the
peak height from when it leaves the player’s
hands. (0.153 s)
Free Body Diagrams: Ex. 2
A hockey puck slides at constant velocity
across the ice. Which of the following is
the correct free-body diagram?
A person drags the box (10.0 kg) at an angle
as shown below.
a. Calculate the acceleration of the box
(3.46 m/s2)
b. Calculate the normal force. (78.0 N)
Fp = 40.0 N
30o
FN
mg
Here is the free body diagram
Fpx = (40N)(cos30o)= 34.6N
Fpy = (40N)(sin30o) = 20.0 N
Fp = 40.0 N
30o
FN
mg
Your lawnmower has a mass of 18.0 kg and
the handle makes 60.0o angle with the
ground. You push with a force of 9.00 N.
a) Calculate acceleration. (0.25 m/s2)
b) Calculate the normal force. (184 N)
c) How far will the lawnmower have
travelled in 2.50 s. (78.1 cm)
You pull a child and sled (50.0 kg) with a rope
at an angle of 35.0o with the horizontal. The
child and sled experience an acceleration of
1.64 m/s2. in the horizontal direction.
a) Calculate the horizontal force on the sled. (82
N)
b) Calculate the Force of the pull (the
hypotenuse) (100N)
c) Calculate the vertical force of the pull on the
sled. (57.4 N)
d) Calculate the normal force on the sled. (433 N)
A man pushes a stroller (20.0 kg) across the
floor. He pushes at an angle of 65.0o, with a
horizontal acceleration of 0.50 m/s2.
a) Calculate the x component of the push. (10.0N)
b) Calculate the actual force of the push (22.8 N)
c) Calculate the y-component of the force (20.5 N)
d) Calculate the normal force on the stroller. (217
N)
e) Calculate the speed of the stroller (from rest) if
the stroller accelerates for 3.0 m (1.73 m/s)
Example 3
A 10.0 kg box is placed next to a 5.00 kg box.
The 10.0 kg box is pushed with a force of
20.0 N.
a) Calculate the acceleration of the boxes (1.33
m/s2)
b) Calculate the contact force between the two
boxes. (6.67 N)
20.0 N
10.0 kg
5.00 kg
Tension
• Flexible cord (can only pull)
• FT
• Neglect mass of the cord
Tension: Example 1
Two boxes are connected by a cord as shown.
They are then pulled by another short cord.
Find the acceleration of each box and the
tension in the cord between the boxes.
12.0 kg
10.0 kg
Fp= 40.0 N
First let’s find the acceleration (consider the
boxes as one mass)
SF = Fp (this is the only horizontal force)
ma = Fp
a= Fp/m = 40.0 N/(12.0 kg + 10.0 kg)
a=1.82 m/s2
To find the tension, let’s deal with each box
one at a time
SF = Fp – FT
10.0 kg
m1a = Fp – FT
FT
FN
m1g
Fp= 40.0 N
FT = m1a – Fp
FT= [(10 kg)(1.82 m/s2) – 40 N
FT= -21.8 N
The second box only has a horizontal pull
from the tension.
SF = FT
12.0 kg
m2a2 = FT
FN
m2g
FT
FT = (12.0 kg)(1.82 m/s2)
FT = 21.8 N
Note that the sign of the tension varies depending on
which box you consider.
A 5.00 kg box and a 12.0 kg box are
connected by a cord. The 12.0 kg box is
pulled horizontally with a force of 60.0 N.
a) Calculate the acceleration of both boxes.
(3.5 m/s2)
b) Calculate the force of tension in the cord.
(17.6 N)
a) Calculate the acceleration. (3.68 m/s2)
b) Calculate the tension in the cord (91.9 N)
b) Calculate how far the box will move in
1.50 s. (4.13 m)
25.0 kg
15.0 kg
Atwood’s machine
Calculate the
acceleration of the
elevator and the
tension in the
cable. (Atwood’s
Machine)
Draw free-body
diagrams for
both the
elevator and
counterweight
Ma = Mg -FT
ma = FT - mg
(a = 0.68 m/s2, FT = 10,500N)
M = 1150 kg
m = 1000 kg
A 10.00 kg and a 5.00 kg box
are connected by a cord.
The boxes are not moving.
a. Calculate the tension the
cords (49N, 147 N)
b. Suppose the boxes
accelerate at 0.80 m/s2
upward. Now calculate the
tensions. (53 N, 159 N)
A 5.00 kg and a 2.00 kg
box are connected by a
cord. The top box (5.00
kg) is lifted by a second
cord with an
acceleration of 0.750
m/s2.
a. Calculate the tension in
the bottom, connecting
cord. (21.1 N)
b. Calculate the tension in
the top cord. (73.85 N)
A 5.00 kg and a 2.00 kg
box are connected by a
cord. The top box (5.00
kg) is lifted by a second
cord with a Force of 100
N.
a. Calculate the tension in
the bottom, connecting
cord. (28.6 N)
b. Calculate the
acceleration. (4.5 m/s2)
Static Friction
• Friction that opposes motion before it
moves
• coefficient of static friction = ms
• Generally greater than kinetic friction (mk)
Ffr = msFN
Kinetic Friction
• Friction that opposes motion while it
moves
• coefficient of kinetic friction (unit-less) = ms
• Generally less than static friction
Ffr = mkFN
Friction: Example 1
A 10.0 kg box is on a table the coefficients
of friction are ms =0.40, mk = 0.30. What
is the force of friction if the force pulling
on the box is:
a) 0 N
b) 10N
F
c) 20N
F
d) 38N
e) 40N
fr
N
mg
Fp
First let’s calculate the maximum force from the
static friction.
Fmax = msFN
Fmax = msmg
Fmax = (0.40)(10.0 kg)(9.8 m/s2)
Fmax = 39 N
(the box will not move until at least 39 N are
applied)
SF = Fp – Ffr
a) Since the force of the pull is zero, the
box will not move:
SF = Fp – Ffr
0 = 0 N – Ffr
(Ffr = 0 N)
b) We are still under 39 N
SF = Fp – Ffr
0 = 10 N – Ffr (Ffr = 10 N)
c) Still under 39 N
SF = Fp – Ffr
0 = 20 N – Ffr (Ffr = 20 N)
d) Still under 39 N
SF = Fp – Ffr
0 = 38 N – Ffr (Ffr = 38 N)
e) Now we are over 39 N, so kinetic friction
takes over:
Ffr = mkFN
Ffr = (0.30)(10.0 kg)(9.8 m/s2) = 29 N
SF = Fp – Ffr
ma = 40 N – 29 N = 11N
(10.0 kg)(a) = 11 N
a= 1.1 m/s2
A 450 g wooden box is pulled at a constant
speed with a force of 1.10 N. Calculate
the value of mk.
(0.25)
A 2.50 kg sled is pulled at a constant speed.
What force is required if the mk is 0.11 on
ice?
Friction: Example 3
The coefficient
of kinetic
friction is 0.20.
a) Calculate
acceleration
(1.4 m/s2)
b) Calculate the
force of
tension in the
cord. (16.8 N)
m1=5.0 kg
m2=2.0 kg
Your little sister wants a ride on her sled.
Should you push or pull her?
Inclines
FN
mg
q
Inclines
FN
q mgcosq
q
mg
mgsinq
Inclines
FN = mgcosq
Ffr
mgsinq
q
Example 1
A 75.0 kg sled (and rider) slides down a hill
that has a 23.0o degree incline.
a) Calculate the force of acceleration. (287 N)
b) Calculate the value of acceleration.
(3.83m/s2)
c) Calculate the speed of the sled after it has
gone 5.00 m. (6.2 m/s)
d) Calculate the normal force on the sled. (677
N)
Example 2
A 20.0 kg child slides down a slide with a
45.0o incline.
a) Calculate the force of acceleration. (139 N)
b) Calculate the value of acceleration. (6.93
m/s2)
c) Calculate the speed of the child after 1.50s
(10.4 m/s)
d) Calculate the normal force on the child.
(139 N)
The coefficient of kinetic friction is 0.25.
a) Calculate the acceleration and FT.
b) Calculate how far the box will move in
0.50 s.
25.0 kg
15.0 kg
(2.14 m/s2, 115 N, 26.8 cm)
Two zombies (50.0 kg and 100.0 kg) are
pushed across a floor with a mk of 0.40. The
100.0 kg zombie is pushed with a force of
750.0 N.
a) Calculate the acceleration of the zombies.
(1.08 m/s2)
b) Calculate the contact force between the
zombies. (250 N)
750.0 N
A duck slides down a 20.0o bank. The bank
is 2.5 m long and very smooth.
a) Calculate the acceleration of the duck.
(3.35 m/s2)
b) Calculate the duck’s speed at the bottom
of the bank. (4.1 m/s)
c) The normal force of the duck is 9.2 N.
Calculate the mass of the duck. (1.00 kg)
Inclines: Example 3
a) What is the
acceleration of
the skier if
snow has a mk
of 0.10
b) What is her
speed after 4.0
s?
Free Body Diagram
First we need to
resolve the force
of gravity into x
and y
components:
FGy = mgcos30o
FGx = mgsin30o
The pull down the hill
is:
FGx = mgsin30o
The pull up the hill is:
Ffr= mkFN
Ffr= (0.10)(mgcos30o)
SF = pull down – pull up
SF = mgsin30o– (0.10)(mgcos30o)
ma = mgsin30o– (0.10)(mgcos30o)
ma = mgsin30o– (0.10)(mgcos30o)
a = gsin30o– (0.10)(gcos30o)
a = 4.0 m/s2
(note that this is independent of the skier’s
mass)
To find the speed after 4 seconds:
v = vo + at
v = 0 + (4.0 m/s2)(4.0 s) = 16 m/s
Inclines: Example 4
Suppose the snow is slushy and the skier
moves at a constant speed. Calculate mk
SF = pull down – pull up
ma = mgsin30o– (mk)(mgcos30o)
ma = mgsin30o– (mk)(mgcos30o)
a = gsin30o– (mk)(gcos30o)
Since the speed is constant, acceleration =0
0 = gsin30o– (mk)(gcos30o)
(mk)(gcos30o) = gsin30o
mk= gsin30o =
gcos30o
sin30o = tan30o =0.577
cos30o
Example 3
A 1.00 kg block is placed on a 37.0o incline
and does not slide down.
a) Draw a free body diagram
b) Calculate the normal force (7.83 N)
c) Calculate the force of acceleration (5.90 N)
d) Calculate the minimum value of ms. (0.753)
e) If the maximum value of ms is 0.800,
calculate the angle at which it will slide?
(38.7o)
Example 4
A 100.0 kg filing cabinet slides down a ramp
that has a coefficient of sliding friction (mk)
of 0.38. The ramp has an angle of 25.0o.
a) Calculate the normal force. (888 N)
b) Calculate the acceleration of the cabinet.
(0.766 m/s2)
c) If the coefficient of static friction is (0.60),
calculate the angle of the ramp at which it
will start slipping. (31.0o)
The acceleration of
the following system
is 1.00 m/s2. The
table is not smooth
and has friction.
a) Calculate the force
of tension in the
cord. (44.0 N)
b) Calculate mk(0.35)
c) How long will it
take the boxes to
move 1.50 m?
(1.73s)
10.0 kg
5.0 kg
Neglect friction for the following system.
Assume the 50.0 kg block will slide down
the incline.
a. Calculate the tension in the cord (199 N)
b. Calculate the acceleration of the blocks
(0.157 m/s2)
50.0 kg
20.0 kg
25.0o
For the following diagram, assume the mk = 0.18,
and the ball falls toward the ground.
a) Calculate the acceleration of the system (4.86
m/s2)
b) Calculate the tension in the cord. (494 N)
50.0 kg
100.0 kg
20 o
Formula Wrap-Up
SF=ma
Weight = mg
Fmax = msFN
Ffr = mkFN
(before it moves)
2. 116 kg
4. 1260 N
6. a) 196 N(both) b) 294 N
c) 98.0 N
8. 153 N
10. 1.29 X 103 N in direction of ball’s motion
12. 1.3 X 104 N
14. 5.04 X 104 N(max) 4.47 X 104 N (min)
16. -2.5 m/s2 (down)
18. a) 7.4 m/s2 (down)
b) 1176 N
24. Southwest
26. a) Fbat NE, mg S
b) mg S
28. b) 62.2 N c) 204 N c) 100N
30. a) 29 N b) 34 N and 68 N
32. a) 320 N b) 0.98 m/s2
34. q = 22o
38. 100 N, if mk=0 no force is required
40. a) static b) kinetic c) kinetic
42. 2.3
44.4.1 m
66. -21,000 N and 1.1 m
44. 4.1 m
46. a) xmin = v2/msg
b) 47 m c) 280 m
48. Bonnie’s (slips if angle is > 8.5o)
50. 0.32
52. a) 3.67 m/s2 b) 8.17 m/s
54. a) 1.85 m/s2 b) 0.82 m up, 1.49s
56. 101 N, mk = 0.719
58. a = 0.098 m/s2
time = 4.3 s
70. a) 16 m/s
b) 12 m/s
74.a) 75.0 kg
b) 75.0 kg
c) 75.0 kg
d) 98.0 kg
e) 52.0 kg
Force Table Lab Example
Calculate the resultant vector (size and
direction) of these three forces.