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Chapter 5
Force and Motion
The Concept of Force
A force describes the action of a body on another
body.
A force is capable of changing an object’s state of
motion, or shape.
The Concept of Force
We distinguish two types of forces:
1. A contact force, such as a push or pull, friction,
tension from a rope or string, and so on.
2. A force that acts at a distance, such as gravity, the
magnetic force, or the electric force.
Contact forces
- Involve physical contact between
objects.
Field forces:
(acting at a distance)
-No physical contact between
objects
- Forces act through empty space
gravity
electric
magnetic
The Concept of Net Force
Net force—the vector sum of all forces acting on
the body. Perfectly equivalent to the group of
forces, causing the same effect.
ΣF = F1+F2+...Fi+...Fn
Fi
F2
ΣFx = F1x+F2x+...Fix+...Fnx
ΣFy = F1y+F2y+...Fiy+...Fny
Fn
F1
Example Net Force
Two tugboats pull a tanker with 10000 N each at 30o.
What is the net force?
  
F  F1  F2
Fx  F1x  F2 x
Fy  F1 y  F2 y
Fx  F1 cos   F2 cos 
Fy  F1 sin   F2 sin 
Fx  10000 N cos 30  10000 N cos 30  17300 N
Fy  10000 N sin 30  10000 N sin 30  0 N
ΣF=173000N
Newton’s First Law:
The Law of Inertia
Galileo’s Principle of Inertia
Inertia: property to maintain actual state of motion
or rest, or oppose to any change in motion.
Measured by mass (kg)
Inertial frame of reference:
-A frame (system) that is not accelerating.
- Newton’s laws hold only true in non-accelerating (inertial)
frames of reference!
Are the following inertial frames of reference:
- A cruising car?
- A braking car?
- The earth?
- Accelerating car?
Is Austin a good IRF?
• Is Austin accelerating?
• YES!
– Austin is on the Earth.
– The Earth is rotating.
• What is the centripetal acceleration of Austin?
2
2
– T = 1 day = 8.64 x 104 sec,
v
 2R 
aU 

 / R – R ~ R = 6.4 x 106 meters .
R  T 
E
• Plug this in: aU = .034 m/s2 ( ~ 1/300 g)
• Close enough to 0 that we will ignore it.
• Austin is a pretty good IRF.
Force, Mass, and Newton’s
Second Law
Fundamental Principle
Newton’s second law
(very important)
m
(mass)
a
(acceleration)
ΣF
(applied)
- A greater force causes, more
acceleration to an object (effect)
- The greater the mass of an object, the less
it accelerates under the action of an applied
force.

 F
a
m


a  F
1
a
m
Newton’s second law
(very important)
The acceleration of an object is directly proportional to
the net force acting on it and inversely proportional to its
mass.




F  ma
Fx  m  ax

Fy  m  a y
Example: Pushing a Box on Ice.
• A skater is pushing a heavy box (mass m =
100 kg) across a sheet of ice (horizontal &
frictionless). He applies a force of 50 N in the
i direction. If the box starts at rest, what is its
speed v after being pushed a distance d = 10
m?
v=0
F
m
a
i
Example: Pushing a Box on Ice.
• A skater is pushing a heavy box (mass m =
100 kg) across a sheet of ice (horizontal &
frictionless). He applies a force of 50 N in the
i direction. If the box starts at rest, what is its
speed v after being pushed a distance d = 10m
v
?
F
m
a
i
d
Example: Pushing a Box on Ice...
• Start with F = ma.
– a = F / m.
– Using Galileo’s formula: v2 - v02 = 2ad
2 Fd
– So v2 = 2Fd / m
v
v
m
F
m
a
i
d
Example: Pushing a Box on Ice...
2 Fd
m
v
• Plug in F = 50 N, d = 10 m, m = 100
kg:
– Find v = 3.2 m/s.
v
F
m
a
i
d
• A force F acting on a mass m1 results in
an acceleration a1.
The same force acting on a different mass
m2 results in an acceleration a2 = 2a1.
F
l
m1
a1
F
m2
a2 = 2a1
If m1 and m2 are glued together and the same force F acts
on this combination, what is the resulting acceleration?
F
(a) 2/3 a1
m1
m2
(b) 3/2 a1
a=?
(c) 3/4 a1
Unit of force:
• The unit of force is the Newton (N)
• One Newton: The force required to accelerate a 1 kg mass
to 1m/s2.
• 1N = 1kg·m/s2
• US Customary unit of force is a pound (lb):
1 N = 0.225 lb
The Force Due to Gravity:
Weight
Gravity=Weight Near Earth
The force of gravity and weight
• Objects are attracted to the Earth.
• This attractive force is the force of gravity Fw.


Fw  m  g
• The magnitude of this force is called the weight of the object.
• The weight of an object is, thus mg.
The weight of an object can very with location (less weight on the moon than
on earth, since g is smaller).
The mass of an object does not vary.
You’re stranded away from your
space ship. Fortunately you have a
propulsion unit that provides a
constant force F for 3 s. After 3 s you
moved 2.25 m. If your mass is 68 kg,
find F.
1. The constant force F provides the
required acceleration: F = ma.
2. Find acceleration from law
of motion: x = at2/2, a = 2x/t2
a = 2 * 2.25 m/(3s)2 = 0.5 m/s2
3. F = 68 kg * 0.5 m/s2 = 34 N
Newton’s third law
“For every action there is an
equal and opposite reaction.”
If two objects interact, the force F12 exerted by object 1 on object
2 is equal in magnitude and opposite in direction to the force F21
exerted by object 2 on object 1:


F12   F21
Action and reaction forces always act on different objects.
Normal Force N or FN
• Exerted on an object lying on a surface.
Always perpendicular to the surface.
N
N=-mg
mg
Tension T or FT
• Exerted on an object hanging on a string or
rope. Same if the string or rope is not
interrupted.
T
T = -mg
mg
Tension
When a cord is attached to a body and pulled taut, the cord pulls on the body
with a force T directed away from the body and along the cord.
Fig. 5-9 (a) The cord, pulled taut, is under tension. If its mass is negligible, the cord pulls on
the body and the hand with force T, even if the cord runs around a massless, frictionless pulley
as in (b) and (c).
Forces in Nature
4 Fundamental Forces
Gravitational Force
Electromagnetic Force
H bomb
Strong Nuclear b/w hadrons (protons, neutrons).
Muon (green) from a cosmic ray interacts
with an electron (red) knocked of an atom
Weak Nuclear b/w leptons (electrons, muons) and b/w
hadrons
Hooke’s Law
Molecular forces like spring forces
A 110 kg basketball player hangs on a rim
following a slam dunk, the rim being
deflected down by 15 cm. Calculate the
force constant of the rim.

F  0
Fy  ky  mg  0 
mg 110kg  9.8m / s 2
k

 7,190 N / m
y
0.15m
Problem Solving:
Free-Body Diagrams
Force Representation of Interactions
Dog Sled
Find acceleration of sled.
Show forces on sled only!
F: force exerted by the
dog on the sled.
w: weight of sled


F  ma
Fn: normal force
Dogsled Race
F = 150 N at 25o, weight of sled = 80
kg, find acceleration, normal force.
1. Draw FBD
2. Apply 2nd Law in Ox
Direction:
Fx  F cos   ma
a
3. Apply 2nd Law in Oy
Direction:
F cos  150 N  cos 25

 1.7m / s 2
m
80kg
Fy  Fn  F sin   mg  0
Fn  mg  F sin   80kg  9.8m / s 2  150 N  sin 25o  721N
Equilibrium Example
A picture weighing 8N
is supported by two
wires with tensions T1
and T2. Find the two
tensions
1. Draw FBD
2. Apply ΣF = 0 in this case:
T1 + T2 + w = 0
T1x + T2x + wx = 0
3. Break down into components:
T1y + T2y + wy = 0
T1cos30 - T2cos60 + 0 = 0
4. Solve for T2 in terms of T1 in the T1sin30 + T2sin60 –8 N = 0
first equation:
T2 =T1(cos30)/cos60 = √3T1
5. Plug in into second equation and
solve for T1:
6. Find T2
T1sin30 + √3T1 sin60 –8 = 0
T1 = 4 N
T2 = √3 x 4N = 6.93 N
Plane Example
As Mr. Avram’s jet plane takes of with his
physics class, you note that your 40 g yo-yo
is deflected at an angle of 22o with the
vertical.
a) What is the acceleration of the plane and
b) the tension in the string?
1. Draw a FBD.
2. Apply ΣFx = max,
and develop using trig.:
Tx + wx = max
Tsinθ + 0 = ma, or
Tsinθ = ma
*
Tcosθ -mg = 0, or
3. Apply ΣFy = may,
Ty + wy = mav
and develop using trig.:
Tcosθ = mg
(Tsinθ)/(Tcosθ) = ma/mg, or
4. Divide *
by **:
5. Find T:
**
tanθ = a/g or a = gtanθ , a = 9.8 x tan 22 = 3.96 m/s2
T = mg/cosθ = (0.04Kg x 9.8 m/s2)/cos 22 = 0.423 N
Suppose your mass is 80 Kg, and you’re standing on a
scale placed on the floor of an elevator.
What is the reading of the scale when:
a) the elevator is rising with an upward acceleration of a.
b) the elevator is descending with a downward
acceleration of a.
c) The elevator is rising at at 20m/s, and decelerating at 8
m/s2.
Inclined Plane
A crate of mass m is placed on a frictionless plane of incline .
y
O
xx
What is the largest angle θ, such that the
downward velocity at the bottom of the
plane won’t exceed 2.5 m/s. The height of
the ramp is 1 m.
Inclined Plane
1. Break down the weight into
components in the FBD.
2. Write Newton’s 2nd Law along the Ox axis.
wx  max
3. Use Galileo’s Formula:
4. Use right triangle geometry to find Δx
5. Substitute in Galileo’s Formula:
mg sin   ma  a  g sin 
v 2  2ax
sin  
h
h
 x 
x
sin 
v 2  2 g sin  
h
 2 gh
sin 
6. Find the downward component of the velocity: vd = vsinθ
7. Find max angle:
vd  2 gh sin 
2.5m / s  2  9.8m / s 2 1m  sin  max
34.4
Newton’s Third Law
Principle of Action and Reaction
Newton’s Third Law
When two bodies interact, the forces on the bodies from each other are always
equal in magnitude and opposite in direction.
• The minus sign means that these two
forces are in opposite directions
• The forces between two interacting
bodies are called a third-law force pair.
Problems With Two or
More Objects
Separate Objects!
Attwood’s machine.
Two objects of unequal mass (m1 and m2) are hung over a pulley.
(a) Determine the magnitude of the acceleration of the two objects and the tension in
the cord.
(b) Solve (a) for m1 = 2.00 kg and m2 = 4.00 kg.
Example Atwood
Body 1
Body 2
Subtract (1)-(2):
Factor:
Acceleration:
Tension:
T  m1 g  m1a
T  m2 g  m2 a
 m2 g  m1 g  m1a  m2a
 g (m2  m1 )  a(m1  m2 )
m2  m1
ag
m2  m1
4kg  2kg
a  9.8m / s 2
 3.27m / s 2
4kg  2kg
T  m1 g  m1a  m1 ( g  a)
T  2kg * (9.8m / s 2  3.27m / s 2 )  26.13N
Paul (mass mP),
accidentally falls of the
edge of a glacier,
dragging Steve (mass
mS) also. Find
acceleration and
tension.
Two Body Problem
1. Draw FBD and apply ΣFx = max for each person separately:
Steve: Fnx + mgx+T1x = mSaSx
Paul: T2x’+mpgx’= mpapx’
2. Because the rope does not stretch: asx= apx’= a
3. Because the rope has negligible mass, and friction is neglected T1 = T2 = T
4. Substituting and using components:
5. Adding the equations and solving
for a:
6. Solving for tension:
T
T  ms g sin   ms a
 T  m p g m p a
a
ms m p
ms  m p
ms sin   m p
ms  m p
(1  sin  ) g
g
Lecture 8, Act 2
Two-body dynamics

In which case does block m experience a larger acceleration?
In (1) there is a 10 kg mass hanging from a rope. In (2) a
hand is providing a constant downward force of 98.1 N. In
both cases the ropes and pulleys are massless.
m
a
m
a
10kg
F = 98.1 N
Case (1)
(a)
Case (2)
Case (1)
(b) Case (2)
(c)
same
Physics Engineering: Lecture 8, Pg 61
Lecture 8, Act 2
Solution
For case (1) draw FBD and write FNET = ma for each block:
(a)
T = ma
(a)

(10kg)g -T = (10kg)a

(b)
m
Add (a) and (b):
a
98.1 N = (m + 10kg)a
10kg
a

98.1N
m  10 kg
Note: T  98.1N 
(b)
m
m  10 kg
Physics Engineering: Lecture 8, Pg 62
Lecture 8, Act 2
Solution

For case (2) T = 98.1 N = ma
a
a
98.1N
m  10 kg
98.1N
m
a
98.1N
m
m
a
m
a
10kg
F = 98.1 N
Case (1)

Case (2)
The answer is (b) Case (2)
Physics Engineering: Lecture 8, Pg 63
Astronaut Pushing Boxes
An astronaut pushes m1
with FA. a) What is the
acceleration of the boxes?
b) What is the force
exerted by m1 on m2?
1. Draw FBD and apply ΣF = ma for each box separately:
FA-F21=m1a,
F12 = m2a
2. Since F12 = F21 = F, we get:
FA – F = m1a, F = m2a, or: FA – m2a = m1a
3 Solving for a we get:
a
4. Substituting we can find F:
FA
m1  m2
m2
F
FA
m1  m2