APPLICATION OF
NEWTON’S LAWS OF
MOTION
✓Visualize the problem.
Is the object at rest, in constant speed or accelerating?
✓Create a free-body diagram.
Identify which object would you need the FBD for and the forces acting ON that object.
✓Identify which law of motion would apply.
✓In the FBD, resolve the x and y components of the forces
separately.
This time, always use the absolute value of the acceleration due to gravity.
✓Solve for what is asked. Note that the magnitude of force is always
positive.
APPLICATION: LAW OF INERTIA
Recall:
Consider, a mug (M) on top of a table (T):
Visualize the problem.
Law of inertia governs all particles in
equilibrium (bodies at rest or moving at
constant speed), as in:
Create an FBD for the mug.
σ 𝐹Ԧ𝑥 = 0
σ 𝐹Ԧ𝑦 = 0
Component
Form
Identify which law will apply.
The mug is at rest, therefore it
follows the law of inertia
(equilibrium).
such that,
σ 𝐹Ԧ =
𝐹𝑥2
+
𝐹𝑦2
= 0
Vector Form
All references for the images are placed in the author’s notes.
Resolve the components
𝑛
𝐹റ𝑇 𝑜 𝑀
𝑤
𝐹റ𝐸 𝑜 𝑀
σ 𝐹Ԧ𝑥 = 0
σ 𝐹Ԧ𝑦 = 𝑛 + (−𝑤) = 0
Create an FBD for the
vehicle.
Consider a vehicle moving steadily across a
bridge. If the truck has a mass of 4.8 ×
103 𝑘𝑔, how much force is exerted by the
bridge (B) on the vehicle (V)?
Solve for:
Force exerted by the bridge on the vehicle
(𝑛)
Visualize the
problem.
Identify which law will apply.
The vehicle is at constant
speed; therefore, it follows the
law of inertia (equilibrium).
Resolve the components.
𝑛
𝐹റ𝐵 𝑜 𝑉
𝑤
𝐹റ𝐸 𝑜 𝑉
෍ 𝐹Ԧ𝑥 = 0
෍ 𝐹Ԧ𝑦 = 𝑛 + (−𝑤) = 0
𝑛 + (−𝑚𝑔) = 0
𝑛 = 𝑚𝑔
𝑚
𝑛 = 4.8 × 10 𝑘𝑔 ∙ 9.8 2
𝑠
3
𝒏 = 𝟒. 𝟕 × 𝟏𝟎𝟒 𝑵, upwards
A gymnast with mass, mG = 50.0 kg, suspends
herself from the lower end of a hanging rope of
negligible mass. The upper end of the rope is
attached to the gymnasium ceiling.
(a) What is the magnitude of the gymnast’s
weight?
(b) What force (magnitude and direction) does
the rope exert on her?
(c) What is the tension at the top of the rope?
Visualize the
problem.
(a) What is the magnitude of the gymnast’s weight?
𝑤 = 𝑚𝑔
𝑚
𝑤 = 50.0 𝑘𝑔 × 9.8 2
𝑠
𝒘 = 𝟒. 𝟗𝟎 × 𝟏𝟎𝟐 𝑵
(b) What force (magnitude and direction) does the rope exert on her?
A gymnast with mass, mG = 50.0 kg, suspends
herself from the lower end of a hanging rope of
negligible mass. The upper end of the rope is
attached to the gymnasium ceiling.
(a) What is the magnitude of the gymnast’s
weight?
(b) What force (magnitude and direction) does
the rope exert on her?
(c) What is the tension at the top of the rope?
Visualize the
problem.
Solve for: Force exerted by rope (R) on gymnast (G).
Create an FBD for the
gymnast.
Identify which law will apply.
The gymnast is at rest, therefore
it follows the law of inertia
(equilibrium).
Resolve the components.
𝑇
𝐹റ𝑅 𝑜 𝐺
𝑤
𝐹റ𝐸 𝑜 𝐺
෍ 𝐹Ԧ𝑥 = 0
෍ 𝐹Ԧ𝑦 = 𝑇 + (−𝑤) = 0
𝑇=𝑤
𝑻 = 𝟒. 𝟗𝟎 × 𝟏𝟎𝟐 𝑵, upwards
(b) What is the tension at the top of the rope?
A gymnast with mass, mG = 50.0 kg, suspends
herself from the lower end of a hanging rope of
negligible mass. The upper end of the rope is
attached to the gymnasium ceiling.
(a) What is the magnitude of the gymnast’s
weight?
(b) What force (magnitude and direction) does
the rope exert on her?
(c) What is the tension at the top of the rope?
Visualize the
problem.
Solve for: Force acting on the rope.
Create an FBD for the rope.
𝐹റ𝐶 𝑜 𝑅
𝑤
Identify which law will apply.
The rope is in equilibrium,
therefore, all forces equate to
zero.
Resolve the components.
𝐹റ𝐺 𝑜 𝑅
෍ 𝐹Ԧ𝑥 = 0
෍ 𝐹Ԧ𝑦 = 𝐹റ𝐶 𝑜 𝑅 + (−𝑤) = 0
𝐹റ𝐶 𝑜 𝑅 = 𝑤
𝑭𝑪 𝒐 𝑹 = 𝟒. 𝟗𝟎 × 𝟏𝟎𝟐 𝑵, upwards
Solve for 𝑇1 , 𝑇2 , and 𝑇3 .
The diagram shows a car engine with weight
5.50 × 103 𝑘𝑔 hangs from a chain that is
linked at ring O to two other chains: one
fastened to the ceiling and the other to the
wall. The weights of the ring and chains are
negligible compared with the weight of the
engine.
Solve for the tension in each cable.(T1, T2, and
T3)
Create an FBD for the
engine.
𝑇1 - Force exerted by the chain on the engine
𝑇2 - Force exerted by the chain on the O-ring
𝑇3 - Force exerted by the chain on the O-ring
𝑇1
𝑤
Identify which law will apply.
The engine is in equilibrium,
therefore, all forces equate
to zero.
෍ 𝐹Ԧ𝑥 = 0
෍ 𝐹Ԧ𝑦 = 𝑇1 + (−𝑤) = 0
Resolve the components.
Visualize the
problem.
𝑇1 + (−𝑚𝑔) = 0
𝑇1 = 𝑚𝑔
𝑚
𝑇1 = 5.50 × 10 𝑘𝑔 ∙ 9.8 2
𝑠
3
𝑻𝟏 = 𝟓. 𝟑𝟗 × 𝟏𝟎𝟒 𝑵, upwards
Solve for 𝑇2 and 𝑇3 .
The diagram shows a car engine with weight
5.50 × 103 𝑘𝑔 hangs from a chain that is
linked at ring O to two other chains: one
fastened to the ceiling and the other to the
wall. The weights of the ring and chains are
negligible compared with the weight of the
engine.
Create an FBD
for the O-ring
Identify which law will apply. The O-ring is in equilibrium,
therefore, all forces equate to zero.
𝑇3
𝑇3𝑦
𝑇2
𝑇3𝑥
෍ 𝐹Ԧ𝑦 = 𝑇3𝑦 + (− 𝑇1 ) = 0
Solve for the y-component first.
60.0°
Visualize the
problem.
Resolve the components.
෍ 𝐹Ԧ𝑥 = 𝑇3𝑥 + − 𝑇2 = 0
𝑇1
Solve for the tension in each cable.(T1, T2, and
T3)
𝑇1 - 𝟓. 𝟑𝟗 × 𝟏𝟎𝟒 𝑵, upwards
𝑇2 - Force exerted by the chain on the O-ring
𝑇3 - Force exerted by the chain on the O-ring
෍ 𝐹Ԧ𝑦 = 𝑇3 sin 60.0 + (− 𝑇1 ) = 0
𝑇3 sin 60.0 = 𝑇1
𝑇3𝑦
𝑇3 sin 60.0 = 𝟓. 𝟑𝟗 × 𝟏𝟎𝟒 𝑵
𝑇3𝑥
𝟓. 𝟑𝟗 × 𝟏𝟎𝟒 𝑵
𝑇3 =
sin 60.0
𝑻𝟑 = 𝟔. 𝟐𝟐 × 𝟏𝟎𝟒 𝑵, 𝟔𝟎. 𝟎° N o E
Solve for 𝑇2 .
The diagram shows a car engine with weight
5.50 × 103 𝑘𝑔 hangs from a chain that is
linked at ring O to two other chains: one
fastened to the ceiling and the other to the
wall. The weights of the ring and chains are
negligible compared with the weight of the
engine.
60.0°
෍ 𝐹Ԧ𝑦 = 𝑇3𝑦 + (− 𝑇1 ) = 0
Solve for the x-component this time.
𝑇3𝑦
෍ 𝐹Ԧ𝑥 = 𝑇3 cos(60.0) + −𝑇2 = 0
Solve for the tension in each cable.(T1, T2, and
T3)
𝑇3 cos 60.0 = 𝑇2
𝑇3𝑥
Visualize the
problem.
𝑇1 - 𝟓. 𝟑𝟗 × 𝟏𝟎𝟒 𝑵, upwards
𝑇2 - Force exerted by the chain on the O-ring
𝑇3 - 𝟔. 𝟐𝟐 × 𝟏𝟎𝟒 𝑵, 𝟔𝟎. 𝟎° N o E
(𝟔. 𝟐𝟐 × 𝟏𝟎𝟒 𝑵) sin 60.0 = 𝑇2
𝑻𝟐 = 𝟑. 𝟏𝟏 × 𝟏𝟎𝟒 𝑵, -x-axis
A picture frame hung against
a wall is suspended by two
wires attached to its upper
corners. If the two wires
make the same
angle with the vertical, what
must this angle be if the
tension in
each wire is equal to 0.75 of
the weight of the frame?
(Ignore any
friction between the wall and
the picture frame.)
A man pushes on a piano with mass
180 kg so that
it slides at constant velocity down a
ramp that is inclined at 11 °above the
horizontal floor. Neglect any friction
acting on the piano.
Calculate the magnitude of the force
applied by the man if he pushes
parallel to the incline.
APPLICATION: LAW OF ACCELERATION
If a net external force acts on a body, the body
accelerates. The direction of acceleration is the
same as the direction of the net force.
σ 𝐹Ԧ𝑥 = 𝑚𝑎𝑥
σ 𝐹Ԧ =
σ 𝐹Ԧ𝑦 = 𝑚𝑎𝑦
𝐹𝑥2 + 𝐹𝑦2 = 𝑚𝑎റ
Component Form
Vector Form
𝑛
𝑎𝑥
Create an FBD for the box.
A worker (W) applies a constant
horizontal force with magnitude
20.0 N to a box with mass 40.0 kg
resting on a level floor with
negligible friction.
What is the acceleration of the
box (B) ?
Visualize the
problem.
𝐹റ𝑊 𝑜 𝐵
𝑤
Identify which law will apply.
The worker applied a constant
force (external force); therefore,
it follows the law of
acceleration.
Resolve the components.
෍ 𝐹Ԧ𝑦 = 𝑛 + (−𝑤) = 0
෍ 𝐹Ԧ𝑥 = 𝐹റ𝑊 𝑜 𝐵 = 𝑚𝑎𝑥
𝐹റ𝑊 𝑜 𝐵
= 𝑎𝑥
𝑚
20.0 𝑘𝑔 𝑚/𝑠 2
= 𝑎𝑥
40.0 𝑘𝑔
𝒂𝒙 = 𝟎. 𝟓𝟎𝟎 𝒎/𝒔𝟐 , +x-axis
𝑛
Create an FBD for the box.
𝑎𝑥
𝐹റ𝑊 𝑜 𝐼
An iceboat (I) is at rest on a frictionless
horizontal surface. A wind (W) is
blowing along the direction of the
runners so that 4.0 s after the iceboat
is released, it is moving at 6.0 m/s. The
combined mass of iceboat and rider is
2.0 x 102 kg.
What constant horizontal force does
the wind
exert on the iceboat?
Visualize
the
problem.
𝑤
Identify which law will apply.
The wind applied a constant
force (external force); therefore,
it follows the law of
acceleration.
Resolve the components.
෍ 𝐹Ԧ𝑦 = 𝑛 + (−𝑤) = 0
෍ 𝐹Ԧ𝑥 = 𝐹റ𝑊 𝑜 𝐼 = 𝑚𝑎𝑥
𝐹റ𝑊 𝑜 𝐼 = (2.0 × 102 𝑘𝑔) 𝑎𝑥
The acceleration is also missing, but we know that the Force exerted by
the wind on the iceboat is a constant force, meaning acceleration is
constant as well. Recall:
To get acceleration, we
have the following:
An iceboat (I) is at rest on a frictionless
horizontal surface. A wind (W) is
blowing along the direction of the
runners so that 4.0 s after the iceboat
is released, it is moving at 6.0 m/s. The
combined mass of iceboat and rider is
2.0 x 102 kg.
What constant horizontal force does
the wind
exert on the iceboat?
Visualize
the
problem.
𝑡 = 4.0 𝑠
𝑣𝑥 = 6.0 𝑚/𝑠
𝑣0𝑥 = 0.0 𝑚/𝑠
𝑎𝑥 = ?
𝑣𝑥 = 𝑣0𝑥 + 𝑎𝑥 𝑡
𝑣𝑥 − 𝑣0𝑥 = 𝑎𝑥 𝑡
𝑣𝑥 − 𝑣0𝑥
= 𝑎𝑥
𝑡
6.0 − 0.0
𝑚
= 𝑎𝑥 = 1.5 2
4.0
𝑠
෍ 𝐹Ԧ𝑥 = 𝐹റ𝑊 𝑜 𝐼 = 𝑚𝑎𝑥
𝐹റ𝑊 𝑜 𝐼 = (2.0 × 102 𝑘𝑔)(1.5
𝑚
)
𝑠2
𝑭𝑾 𝒐 𝑰 = 𝟑. 𝟎 × 𝟏𝟎𝟐 𝑵, +x-axis
𝑇
Create an FBD for the
elevator.
An elevator (E) and its load have a
combined mass of 8.00 x 102 kg . The
elevator is initially moving downward at
10.0 m/s as it slows to a stop with
constant acceleration in a distance of
25.0 m.
What is the tension T in the supporting
cable while the elevator is being
brought
Visualize
the to rest?
𝑎𝑦
𝑤
problem.
Identify which law will apply.
Resolve the components.
෍ 𝐹Ԧ𝑥 = 0
෍ 𝐹Ԧ𝑦 = 𝑇 + −𝑤 = 𝑚𝑎𝑦
𝑇 = 𝑚𝑎𝑦 + 𝑚𝑔
𝑇 = 𝑚(𝑎𝑦 + 𝑔)
The acceleration is missing, but we know that the elevator is slowing down with
constant acceleration.
An elevator (E) and its load have a combined
mass of 8.00 x 102 kg . The elevator is initially
moving downward at 10.0 m/s as it slows to a
stop with constant acceleration in a distance of
25.0 m.
What is the acceleration of the elevator?
What is the tension T in the supporting cable
while the elevator is being brought to rest?
To get acceleration, we
have the following:
𝑦0 = 0.00 𝑚
𝑦 = −25.0 𝑚
𝑣0𝑦 = −10.0 𝑚/𝑠
𝑣𝑦 = 0.00 𝑚/𝑠
𝑎𝑦 = ?
2
𝑣𝑦2 = 𝑣0𝑦
+ 2𝑎𝑦 𝑦 − 𝑦0
Visualize the
problem.
2
𝑣𝑦2 − 𝑣0𝑦
= 2𝑎𝑦 𝑦 − 𝑦0
2
𝑣𝑦2 − 𝑣0𝑦
= 𝑎𝑦
2 𝑦 − 𝑦0
02 − −10.0 2
= 𝑎𝑦
2 −25.0 − 0
𝒎
𝒂𝒚 = 𝟐. 𝟎𝟎 𝟐 , +𝐲 − 𝐚𝐱𝐢𝐬
𝒔
𝑇 = 𝑚(𝑎𝑦 + 𝑔)
𝑇 = 8.00 x 102 𝑘𝑔 (2.00
𝑚
𝑚
+
9.8
)
2
2
𝑠
𝑠
𝑻 = 𝟗. 𝟒𝟒 × 𝟏𝟎𝟑 𝑵, upwards
The acceleration is missing, but we know that the elevator is slowing down with
constant acceleration.
An elevator (E) and its load have a combined
mass of 8.00 x 102 kg . The elevator is initially
moving downward at 10.0 m/s as it slows to a
stop with constant acceleration in a distance of
25.0 m.
What is the acceleration of the elevator?
What is the tension T in the supporting cable
while the elevator is being brought to rest?
To get acceleration, we
have the following:
𝑦0 = 0.00 𝑚
𝑦 = −25.0 𝑚
𝑣0𝑦 = −10.0 𝑚/𝑠
𝑣𝑦 = 0.00 𝑚/𝑠
𝑎𝑦 = ?
2
𝑣𝑦2 = 𝑣0𝑦
+ 2𝑎𝑦 𝑦 − 𝑦0
Visualize the
problem.
2
𝑣𝑦2 − 𝑣0𝑦
= 2𝑎𝑦 𝑦 − 𝑦0
2
𝑣𝑦2 − 𝑣0𝑦
= 𝑎𝑦
2 𝑦 − 𝑦0
02 − −10.0 2
= 𝑎𝑦
2 −25.0 − 0
𝒎
𝒂𝒚 = 𝟐. 𝟎𝟎 𝟐 , +𝐲 − 𝐚𝐱𝐢𝐬
𝒔
𝑇 = 𝑚(𝑎𝑦 + 𝑔)
𝑇 = 8.00 x 102 𝑘𝑔 (2.00
𝑚
𝑚
+
9.8
)
2
2
𝑠
𝑠
𝑻 = 𝟗. 𝟒𝟒 × 𝟏𝟎𝟑 𝑵, upwards
A stockroom worker pushes a box
with mass 11.2 kg on a horizontal
surface with a constant speed of
3.50 m/s. The coefficient of
kinetic friction between the box
and the surface is 0.20. (a) What
horizontal force must the worker
apply to maintain the motion?
(b) If the force calculated in part
(a) is removed, how
far does the box slide before
coming to rest?
An 8.00-kg block of ice, released
from rest at the top of
a 1.50-m-long frictionless ramp,
slides downhill, reaching a speed
of 2.50 m/s at the bottom.
(a)What is the angle between the
ramp and the horizontal?
(b)What would be the speed of the
ice at the bottom if the motion
were opposed by a constant
friction force of 10.0 N parallel
to the surface of the ramp?
A 15.0-kg load of bricks hangs from one
end of a rope that passes over a small,
frictionless
pulley.
A
28.0kg
counterweight is suspended from the other
end of the rope. The system is released
from rest.
(a)Draw two free-body diagrams, one for
the load of bricks and one for the
counterweight.
(b)What is the magnitude of the upward
acceleration of the load of bricks?
(c)What is the tension in the rope while the
load is moving? How does the tension
compare to the weight of the load of
bricks? To weight of the counterweight?
You are lowering two boxes, one on
top of the other down the ramp by
pulling on a rope parallel to
the surface of the ramp. Both boxes
move together at a constant
speed of 15.0 cm/s. The coefficient
of kinetic friction between the
ramp and the lower box is 0.444, and
the coefficient of static friction
between the two boxes is 0.800.
(a) What force do you need to exert
to accomplish this?
(b) What are the magnitude and
direction of the friction force on
the upper box?
An inventor designs a pendulum
clock using a bob with mass m at
the end of a thin wire of length L.
Instead of swinging back and
forth, the bob is to move in a
horizontal circle with constant
speed
with the wire making a fixed
angle 𝛽 with the vertical direction.
This is called a conical pendulum
because the suspending wire
traces out a cone. Find the
tension F in the wire and
the period T (the time for one
revolution of the bob).
A passenger on a carnival Ferris wheel moves
in a vertical circle of radius R with constant
speed v. The seat remains upright during the
motion. Find expressions for the force the
seat exerts on the
passenger at the top of the circle and at the
bottom.
For a car traveling at a certain
speed, it is possible to bank a
curve at just the right angle so
that no friction at all is needed to
maintain the car’s turning radius.
Then a car can safely round the
curve even on wet ice. Your
engineering
firm plans to rebuild the curve so
that a car moving at a chosen
speed can safely make the turn
even with no friction. At what
angle 𝛽 should the curve be
banked? Assume that the radius
of the curve is 230 m and the
speed of the objects is 25 m/s.
APPLICATION: LAW OF INTERACTION
For every action there is an equal and
opposite reaction