4-5 Newton’s Third Law of Motion
Any time a force is exerted on an object, that
force is caused by another object.
Newton’s third law:
Whenever one object exerts a force on a second
object, the second exerts an equal force in the
opposite direction on the first.
4-5 Newton’s Third Law of Motion
A key to the correct
application of the third
law is that the forces
are exerted on different
objects. Make sure you
don’t use them as if
they were acting on the
same object.
4-5 Newton’s Third Law of Motion
Rocket propulsion can also be explained using
Newton’s third law: hot gases from combustion
spew out of the tail of the rocket at high speeds.
The reaction force is what propels the rocket.
Note that the
rocket does not
need anything to
“push” against.
4-5 Newton’s Third Law of Motion
Helpful notation: the first subscript is the object
that the force is being exerted on; the second is
the source.
This need not be
done indefinitely, but
is a good idea until
you get used to
dealing with these
forces.
(4-2)
4-6 Weight – the Force of Gravity;
and the Normal Force
Weight is the force exerted on an
object by gravity. Close to the
surface of the Earth, where the
gravitational force is nearly
constant, the weight is:
4-6 Weight – the Force of Gravity;
and the Normal Force
An object at rest must have no net force on it. If
it is sitting on a table, the force of gravity is still
there; what other force is there?
The force exerted perpendicular to a surface is
called the normal force. It is
exactly as large as needed
to balance the force from
the object (if the required
force gets too big,
something breaks!)
Example 4-6
A friend has given you a special gift, a box of mass 10.0 kg with a
mystery surprise inside. The box is resting on the smooth
(frictionless) horizontal surface of a table. (a) Determine the weight
of the box and the normal force exerted on it by the table. (b) Now
your friend pushes down on the box with a force of 40.0 N. Again
determine the normal force exerted on the box by the table. (c) If
your friend pulls upward on the box with a force of 40.0 N, what now
is the normal force exerted on the box by the table?
(a) weight = mg = (10.0 kg)(9.80 m/s 2 ) = 98.0 N
Fy = FN - mg = 0  FN = mg
(b) Fy = FN - mg - 40.0 N = 0
FN = mg + 40.0 N = 138.0 N
(c) Fy = FN - mg + 40.0 N = 0
FN = mg - 40.0 N = 58.0 N
Example 4-7
What happens when a person pulls upward on the box in example
4-6 with a force equal to or greater than the box’s weight, say
FP=100.0 N rather than the 40.0 N used before?
Fy = FN - mg +FP = FN - 98.0 N +100.0 N = 0
FN = -2.0 N, but the table can' t pull down on the box!
Actually, the box is accelerating upward.
Fy = FP - mg =100.0 N - 98.0 N = 2.0 N
a=

Fy
2.0 N
=
= 0.20 m/s 2
m 10.0 kg
Example 4-8
A 65 kg woman descends in an elevator that briefly accelerates at
0.20g downward when leaving a floor. She stands on a scale that
reads in kg. (a) During this acceleration, what is her weight and what
does the scale read? (b) What does the scale read when the elevator
descends at a constant speed of 2.0 m/s?
(a) F = ma  mg - FN = ma  FN = mg - ma
FN = mg - 0.20mg = 0.80mg
Her weight is still mg = (65 kg)(9.80 m/s 2 ) =
640 N. The scale gives the normal force of
0.80mg, giving an apparent mass of 0.80m
or 52 kg.
(b) Since a = 0 now, the normal force is equal to
her weight, and the scale gives a mass of 65 kg.
4-7 Solving Problems with Newton’s Laws –
Free-Body Diagrams
1. Draw a sketch.
2. For one object, draw a free-body
diagram, showing all the forces acting
on the object. Make the magnitudes
and directions as accurate as you
can. Label each force. If there are
multiple objects, draw a separate
diagram for each one.
3. Resolve vectors into components.
4. Apply Newton’s second law to each
component.
5. Solve.
Example 4-9
Calculate the sum of the two forces exerted on the
boat by workers A and B (see figure).
FAX = FAcos45.0 = (40.0 N)(0.707) = 28.3 N
FBX = FBcos37.0 = (30.0 N)(0.799) = 24.0 N
FX = FAX + FBX = 28.3 N + 24.0 N = 52.3 N
FAY = FAsin45.0 = (40.0 N)(0.707) = 28.3 N
FBY = -FBsin37.0 = -(30.0 N)(0.602) = -18.1 N
FY = FAY + FBY = 28.3 N -18.1 N = 10.2 N
F = FX2 + FY2 = (52.3 N) 2 + (10.2 N) 2 = 53.3 N
 = tan -1
FY
10.2 N
= tan -1
= 11.0
FX
52.3 N
4-7 Solving Problems with Newton’s Laws –
Free-Body Diagrams
When a cord or rope pulls on
an object, it is said to be under
tension, and the force it exerts
is called a tension force.
Example 4-12
Two boxes, A and B, are connected
by a lightweight cord and are resting
on a smooth table. The boxes have
masses of 12.0 kg and 10.0 kg. A
horizontal force FP of 40.0 N is
applied to the 10.0 kg box. Find (a)
the acceleration of each box, and
(b) the tension in the cord
connecting the boxes.
For box A, we have Fx = FP - FT = mAa A
For box B, we have Fx = FT = mBa B
Assuming the cord remain taut, a A = a B = a
Adding the 2 equations above, we get
(mA + mB )a = FP - FT + FT = FP
FP
40.0 N
a=
=
= 1.82 m/s 2
mA + mB 22.0 N
The equation for box B gives us tension :
FT = mBa = (12.0 kg)(1.82 m/s 2 ) = 21.8 N
4-8 Applications Involving Friction, Inclines
On a microscopic scale, most
surfaces are rough. The exact
details are not yet known, but
the force can be modeled in a
simple way.
For kinetic – sliding –
friction, we write:
is the coefficient
of kinetic friction, and
is different for every
pair of surfaces.
4-8 Applications Involving Friction, Inclines
4-8 Applications Involving Friction, Inclines
Static friction is the frictional force between two
surfaces that are not moving along each other.
Static friction keeps objects on inclines from
sliding, and keeps objects from moving when a
force is first applied.
4-8 Applications Involving Friction, Inclines
The static frictional force increases as the applied
force increases, until it reaches its maximum.
Then the object starts to move, and the kinetic
frictional force takes over.
Example 4-16
Our 10.0 kg mystery box rests on a
horizontal floor. The coefficient of static
friction is µs=0.40 and the coefficient of
kinetic friction is µk=0.30. Determine the
force of friction acting on the box if a
horizontal external applied force FA is
exerted on it of magnitude (a) 0, (b) 10 N,
(c) 20 N, (d) 38 N, and (e) 40 N.
Fy = FN - mg = ma y = 0  FN = mg = (10.0 kg)(9.80 m/s 2 ) = 98 N
(a) No external force = no motion = no friction
(b) max static friction is sF = (0.40)(98 N) = 39 N
The box will not move until the external force is over 39 N.
Fx = 0  Ffr = 10 N
(c) Same reasoning as above, Ffr = 20 N
(d) Same reasoning as above, Ffr = 38 N
(e) Now we have motion.
Ffr = k FN = (0.30)(98 N) = 29 N
4-8 Applications Involving Friction, Inclines
An object sliding down an incline has three forces acting
on it: the normal force, gravity, and the frictional force.
• The normal force is always perpendicular to the surface.
• The friction force is parallel to it.
• The gravitational force points down.
If the object is at rest,
the forces are the same
except that we use the
static frictional force,
and the sum of the
forces is zero.
Example 4-21
The skier has just begun descending the 30.0 degree slope. Assuming the
coefficient of kinetic friction is 0.10, calculate (a) her acceleration and (b)
the speed she will reach after 4.0 s.
FGx = mgsin , FGy = -mgcos 
(a) Fy = FN - mgcos  = ma y = 0  FN = mgcos 
Fx = mgsin  - k FN = mgsin  - k mgcos  = ma x
a x = gsin30 - k gcos30 = 0.50g - (0.10)(0.866)g = 0.41g = 4.0 m/s 2
(b) v = v 0 +at = 0 +(4.0 m/s 2 )(4.0 s) =16 m/s

4-9 Problem Solving – A General Approach
1. Read the problem carefully; then read it again.
2. Draw a sketch, and then a free-body diagram.
3. Choose a convenient coordinate system.
4. List the known and unknown quantities; find
relationships between the knowns and the
unknowns.
5. Estimate the answer.
6. Solve the problem without putting in any numbers
(algebraically); once you are satisfied, put the
numbers in.
7. Keep track of dimensions.
8. Make sure your answer is reasonable.
Summary of Chapter 4
• Newton’s first law: If the net force on an object
is zero, it will remain either at rest or moving in a
straight line at constant speed.
• Newton’s second law:
• Newton’s third law:
• Weight is the gravitational force on an object.
• The frictional force can be written:
(kinetic friction) or
(static friction)
• Free-body diagrams are essential for problemsolving
Homework - Ch. 4
• Questions #’s 3, 5, 11, 13, 17
• Problems #’s 5, 7, 11, 15, 23, 25, 27,
31, 41, 43, 47, 53, 55