Chapter 4
LECTURE NOTES
The branch of mechanics dealing with the cause of motion is called dynamics. Aristotle held that rest is the
natural state of all matter and that a force was required to maintain motion. This accorded well with everyday
experience and was accepted for 2000 years until Galileo.
Around 1600, Galileo noted friction and air resistance had misled Aristotle. Galileo found that the natural state
of a body is rest or uniform motion.
In the Principia (1687) Newton stated his 3 laws of motion.
1.
Every body continues in its state of rest or uniform motion (motion in a straight line at constant
speed) unless acted upon by a nonzero net force.
This was essentially Galileo’s principle of inertia. By force, we mean a push or pull in some
direction. Each body has mass, which measures this quantity of matter. We measure mass in
kilograms (kg) and force in newtons (N).
2.
Fnet = ma where Fnet =  F, the vector sum of all forces acting on the body and a is the acceleration
of the body. We see 1N = 1kg m/s2.
In the British system of units, mass is measured in slugs and force in pounds: 1 lb = 1 slug • ft/s2.
A variant of the metric system defines forces in dynes: 1 dyne = 1g cm/s2.
Writing vector equations like Fnet = ma is shorthand for the system F netx = max, F net y = may ,
and so on.
3.
Whenever one object exerts a force on another, the second object exerts an equal but oppositely
directed force upon the first.
This is often given as: for every action there is an equal but opposite reaction. Note the action and
reaction forces act on different objects.
One of the most common forces we experience is the force of gravity. This force is commonly called weight,
points down toward the center of the earth, and has magnitude w = mg where g is the acceleration due to
gravity. Near the surface of the earth g is constant and is given by g = 9.80 m/s2 = 32.0 ft/s2. This forces acts
on a body whether it is at rest or in motion.
Another common type of force is a contact force or normal force (because it acts perpendicular or normal to a
surface). An object at rest on the ground feels the force of gravity W down and a normal force N up. The
normal force comes from the ground pushing up on the object and is really the sum of electrical forces between
atoms of the ground. We can draw a diagram of this situation. If we clearly show the forces acting on the body
under consideration, we have a force diagram (also called a free-body diagram).
N
Note: Although the earth pulls down on all atoms of the box and the
ground pushes up all over the bottom surface, we draw single vectors
representing the sum of these forces acting at the center of mass of the
box.
W
Chapter 4
LECTURE NOTES
In the y direction, we can apply Fnet= ma to this diagram. Let up be +; then N – W = may or N – mg = may.
Note: Normal forces are always non-negative, arising as they do in response to other forces. Note that N and W
are not an action-reaction pair of forces as they both act on the same body. The reaction force paired to W is
the force of gravity the box exerts upward on the earth; the reaction force paired to N is another normal force as
the box pushes down on the ground.
Examples with no friction or air resistance:
N
Example 1:
y
x
An object of mass m is being pulled to the right
by a rope with tension T. First, choose a
coordinate system. Second, draw a free-body
diagram showing only the forces acting on mass
m. Third, apply Fnet = ma to write a system of
reactions. Fourth, solve the equations.
T
W
x: T = max
y: N – W = may (note: we may substitute W = mg)
Now solve for the desired quantity; perhaps ax = T/m.
Note: A force exerted by a rope is called tension; if the rope has very little mass, the tension T is the same
all along the rope. Ropes can only pull, never push.
Note: If the box stays on the level ground ay = 0.
Example 2: Same as example 1, but the rope is at an angle  above horizontal.
y
N
T
x
Ty

Tx
W
Resolve the force T into components
Tx = T cos 
Ty = T sin 
w: Tx = max so ax = T cos  /m
y: Ty + N – mg = may
Note: We have not used y equation, but when we repeat this problem with friction it will be needed.
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Chapter 4
LECTURE NOTES
Example 3: Atwood's machine. Find the tension in the rope and the acceleration of the masses.
Assume m2 > m1
Here we combine 2 force diagrams and show the (massless)
pulley which only changes the direction of motion. The same T
acts on both masses and they have the same acceleration a.
T
T
+
T
T
m1g
m2g
m1 g
T-m1g = m1a
+
m2 g
m2 g – T = m2 a
Note: We have drawn separate force diagrams for each mass and written F = ma for each mass. It is best
to let the + direction be the direction of a so it is up for m1 and down for m2 ; this way we do not have to
worry about writing +a or –a.
Solving the first equation, T = m1 a + m1g; so the second equation becomes m2g – m1g = m1a + m2a or
 m  m1 
g
a =  2
 m 2  m1 
Example 4: Elevator problems. A body of mass m rests upon a scale in an elevator. The scale will
supply the normal force N, keeping the body from falling out of the elevator. The scale will record the
force N which is also called the apparent weight of the body. Draw the force diagrams showing forces
acting on the body.
+
When the elevator is at rest or has constant velocity up or down. a = 0.
N-mg = may = 0, so N = mg. The apparent weight is the same as the
actual weight.
N
scale
mg
You feel pressed to the floor when the elevator starts to accelerate
upward. The normal force N is holding you up and exerting an extra
force to accelerate you upward. Let up be the + direction.
N - mg = may
N = mg + may
The apparent weight is more than the actual weight.
When the elevator is accelerating downward, we must let down be the + direction (or write –ay).
mg – N = may
N = mg – may.
The apparent weight is less than the actual weight. You feel lighter when the elevator accelerates
downward.
Note: If ay = g, N = 0 for the last case. In an elevator falling freely, all objects fall at the same rate so
you do not press on the scale –- you are weightless.
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Chapter 4
LECTURE NOTES
FRICTION
Friction is resistance to motion across a surface or through a medium. Frictional forces are usually indicated by
lowercase f. For moving (or kinetic) friction, we write fk =  k N. The frictional force is proportional to the
normal force, where  k , the coefficient of kinetic friction, is the dimensionless proportionality constant. The
frictional force is independent of the surface area in contact or the speed of the body but does depend on how
hard the body is pressed against the surface (as measured by N).
Static friction fs ≤  s N is usually greater than kinetic friction (  s ≥  k). It is harder to set an object into motion
against friction than to keep it moving.
Examples with friction
Example 1: A box of mass m is dragged across a rough surface at constant speed by a horizontal rope
under tension T. Find T.
N
+y
f
T
+x
Force diagram
mg
x: T – fk = max = 0 constant speed means ax = 0
y: N – mg = may = 0 box stays on flat surface means, ay = 0.
fk =  k N.
Equation of Motion:
Solve: N = mg so fk =  k mg and T =  k mg .
Note: If ax ≠ 0, we could find ax =
T   k mg
if we were given T.
m
Example 2: same As example 1, but the rope is at an angle  above horizontal.
N
T

fk
+y
Ty
Tx
+x
mg
Resolve the tension into components in the x and y directions. Tx = T cos  and Ty = T sin  .
Equations of Motion
x: Tx – fk = max
y: Ty + N – mg = may
fk =  k N
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Chapter 4
LECTURE NOTES
If the box moves at constant speed and stays on the level surface ax = ay = 0. Solving,
N = mg – Ty = mg – T sin 
fk =  k N =  k (mg – T sin  )
Tx = fk or T cos  =  k mg -  k T sin 
 k mg
So T =
cos    k sin 
Example 3: Motion on an inclined plane. Take +x to be along the surface of the incline in the direction
of acceleration and +y perpendicular to the surface of the incline. Find the coefficient of kinetic friction
between a body of mass in and the surface of the plane if the body goes down the incline at constant
speed.
N
fk
+y
 mg cos 
mg
+x

mg sin

Resolve the weight mg into components parallel (mg sin  ) and perpendicular (mg cos  ) to the incline.
N is always perpendicular to the surface.
Equation of motion
x: mg sin  - fk = max = 0 (constant speed)
y: N – mg cos  = may = 0 (stays on surface)
fk =  k N
Solving,  k
=
f k mg sin 

 tan 
N mg cos 
Example 4:
m1
m2
Draw separate force diagrams. Find T, a. Note a1x
N
+y
fk
T
m1 g
+x
= a 2 y = a.
T – fk = m1a
N – m1g = m1a 1y = 0
fk =  k N
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Chapter 4
LECTURE NOTES
T
+x
m2 g – T = m2 a
Solve this system:
T -  k m1 g = m1 a
m2 g - T = m2 a
m g   k m1g
We find a = 2
m1  m 2
Now substitute to find T.
+y
m2 g
Example 5: An object falling against air resistance.
f air resistance
mg – fair = may
+y
mg
fair depends on the density of the body, its size and speed. Air resistance usually increases as the square of
the velocity, so we find a terminal velocity where fair = mg and ay = 0; real objects, falling through the air,
feel more resistance the faster they fall until they reach terminal velocity. They fall faster and faster until
they reach this speed, then fall at this speed the rest of the way to the ground. For a human falling from a
plane (before opening a parachute) terminal velocity is 120 mph.
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