NEWTON’S LESSON 12
INCLINE PLANES WITH FRICTION AND OTHER
FORCES
In the presence of friction or other forces (applied force,
tensional forces, etc.), the situation is slightly more
complicated.
The perpendicular component of force still balances the normal force since
objects do not accelerate perpendicular to the incline. The perpendicular
component and the normal force add to 0 N.
Yet the frictional force must also be considered when determining the net
force. The parallel component and the friction force add together to yield 5
N. The net force is 5 N, directed along the incline towards the floor.
EXAMPLE: This free-body diagram shows the
forces acting upon a 100-kg crate which is sliding
down an inclined plane. The plane is inclined at an
angle of 30 degrees. The coefficient of friction
between the crate and the incline is 0.3. Determine the
net force and acceleration of the crate.
First, find the force of gravity acting upon the crate
and the components of this force parallel and perpendicular to the incline.
Second, determine what you know:
The force of gravity is 980 N
The components of this force are Fgx = 490 N (980 N • sin 30 degrees)
and Fgy = 849 N (980 N • cos30 degrees).
Now the normal force can be determined to be 849 N (it must balance
the perpendicular component of the weight vector).
The force of friction can be determined from the value of the normal
force and the coefficient of friction; Ff is 255 N (Ff = μ*FN= 0.3 • 849
N).
Determine the net force: The net force is the vector sum of all the forces.
The forces directed perpendicular to the incline balance; the forces directed
parallel to the incline do not balance.
The net force is Fnet = Fgx - Ff
235 N = (490 N - 255 N).
The acceleration is 2.35 m/s2 down the incline (a = Fnet/m = 235 N/100 kg).
Example:
Diagram A
The Fg can be calculated from the mass of the object.
Fg = m • g = (1000 kg) • (9.8 m/s/s) = 9800 N
The parallel and perpendicular components of the gravity force can be
determined from their respective equations:
Fgx = m • g • sin (45 degrees) = 6930 N
Fgy = m • g • cos (45 degrees) = 6930 N
The forces directed perpendicular to the incline balance each other. Thus FN
is equal to Fy.
FN = 6930 N
There is no other force parallel to the incline to counteract the parallel
component of gravity. Thus, the net force is equal to the Fgx value.
Fnet = 6930 N, down the incline
The acceleration is can be found from a = Fnet / m :
a = Fnet / m = (6930 N) / (1000 kg)
a = 6.93 m/s2, down the incline
Diagram B
The Fg can be calculated from the mass of the object.
Fg = m • g = (1000 kg) • (10 m/s/s) = 9800 N
The parallel and perpendicular components of the gravity force can be
determined from their respective equations:
Fgx = m • g • sin (60 degrees) = 8487 N
Fgy = m • g • cos (60 degrees) = 4900 N
The forces directed perpendicular to the incline balance each other. Thus FN
is equal to Fy.
FN = 4900 N
There is no other force parallel to the incline to counteract the parallel
component of gravity. Thus, the net force is equal to the Fgx value.
Fnet = 8487 N, down the incline
The acceleration is can be found from a = Fnet / m :
a = Fnet / m = (8487 N) / (1000 kg)
a = 8.49 m/s2, down the incline
The affects of the incline angle on the acceleration of a roller coaster (or any
object on an incline) can be observed in the two previous problems.
As the angle is increased, the acceleration of the object is increased.
As the angle increases, the component of force parallel to the incline
increases and the component of force perpendicular to the incline decreases.
It is the parallel component of the weight vector which causes the
acceleration. Thus, accelerations are greater at greater angles of incline.
Roller coasters produce two thrills associated with the initial drop down a
steep incline.
The thrill of acceleration is produced by using large angles of incline on the
first drop; such large angles increase the value of the parallel component of
the weight vector (the component which causes acceleration).
The thrill of weightlessness is produced by reducing the magnitude of the
normal force to values less than their usual values. It is important to
recognize that the thrill of weightlessness is a feeling associated with a lower
than usual normal force. Typically, a person weighing 700 N will experience
a 700 N normal force when sitting in a chair. However, if the chair is
accelerating down a 60-degrees incline, then the person will experience a
350 Newton normal force. This value is less than normal and contributes to
the feeling of weighing less than one's normal weight - i.e., weightlessness.
HOMEWORK NEWTON’S LESSON 11
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Read the problem carefully and diagram the physical situation (free-body
diagram)
Identify known and unknown information in an organized manner (write on
diagram)
Take the time to plot out a strategy prior to beginning the solution
Identify an appropriate formula to use
Perform step-by-step algebraic manipulations
1. A student places two objects on a physics book and slowly tilts the cover. At a small
angle, object A starts to slide. At a large angle, object B starts to slide. Which object has
the greater coefficient of static friction? Explain.
2. Lon Scaper is doing some lawn work when a 2.0-kg tire escapes from
his wheelbarrow and begins rolling down a steep hill (a 30° incline) in San
Francisco. Sketch the parallel and perpendicular components of this
weight vector. Determine the magnitude of the components using
trigonometric functions. Then determine the acceleration of the tire.
Ignore resistance force.
Finally, determine which one of the velocity-time graph would represent
the motion of the tire as it rolls down the incline.
Explain your answer.
3. In each of the following diagrams, a 100.0-kg box is sliding down a frictional surface
at a constant speed of 0.200 m/s. The incline angle is different in each situation. Analyze
each diagram and fill in the blanks.
Diagram A
Diagram B
4. A 30.0 kg object is on an inclined plane set at 40.0° to the horizontal. What is the
acceleration experienced by the object if the coefficient of friction is 0.250?
5. Consider a 10.0 kg box on an inclined plane of 30o to the horizontal having a
coefficient of friction of μ= 0.50.
1. What is the force of friction on the box?
2. What is the acceleration of the box?
6. Suppose that you now want to drag a 5.0-kg mass up and down a frictionless 15º
inclined plane.
1. How much force must you apply to a string acting parallel to the incline's surface
to slide the 5.0-kg mass up or down the incline at a constant velocity (no
acceleration)?
2. How much upward force would be needed to accelerate the 5.0-kg mass up this
incline at 3.0 m/sec2?
3. How much upward force would be needed to restrict the 5.0-kg mass' downward
acceleration to 1.0 m/sec2?
4. If the 5.0-kg mass were allowed to slide down this incline without any additional
applied forces acting upon it, what would be its acceleration down the incline?
7. A 10.0 kg box accelerates at 2.0 m/s2 as it slides down a ramp that makes an angle of
30 degrees with the horizontal. Find the coefficient of friction.
8. A city planner is working on the redesign of a hilly portion of a city. An important
consideration is how steep the roads can be so that even low-powered cars can get up the
hills without slowing down. It is given that a particular small car, with a mass of 1100kg,
can accelerate on a level road from rest to 21 m/s in 14.0 s. Using this data, calculate the
maximum steepness of the hill.
HINT: The direction of motion depends not only upon the applied force, the weight of the
block, and also on the direction of the force of friction.
9. An object is being pulled up a 15° incline against a frictional coefficient of 0.15, and
requires a force of 835 N parallel to the surface of the ramp to move it at a constant
speed. What is the weight of the object?
HOMEWORK KEY INCLINE PLANE PROBLEMS WITH FRICTION AND
APPLIED FORCE
1.
2.
3.
4.
5.
6.
Object B
Graph B
A: Fg = 980 N, Ff = 490 N, FN = 849 N B: Fg = 980 N, Ff = 693 N, FN = 693 N
4.43 m/s2 down the incline
Ff = 42 N up incline, a = 0.66 m/s2 down incline
1.) 13 N up incline, 2.) 28 N up incline, 3.) 7.7 N up incline, 4.) 2.5 m/s2 down
incline
7. 0.34
8. 8.8 degrees
9. 2.1 x 103 N
HOMEWORK NEWTON’S LESSON 12
1. A student places two objects on a physics book and slowly tilts the cover. At a small
angle, object A starts to slide. At a large angle, object B starts to slide. Which object has
the greater coefficient of static friction? Explain.
Answer: Object B has a greater coefficient of static friction. As the incline increases, the
force of gravity in the x plane (parallel to surface of incline) increases. This parallel
force is in the opposite direction as the static friction force. As the incline increased, the
force of gravity (parallel to the incline) surpassed the static coefficient of friction
resulting in a downward acceleration. Therefore Object B has the greatest coefficient of
static friction as it required the largest incline angle.
2. Lon Scaper is doing some lawn work when a 2.0-kg tire escapes from
his wheelbarrow and begins rolling down a steep hill (a 30° incline) in San
Francisco. Sketch the parallel and perpendicular components of this
weight vector. Determine the magnitude of the components using
trigonometric functions. Then determine the acceleration of the tire.
Ignore resistance force.
Finally, determine which one of the velocity-time graph would represent
the motion of the tire as it rolls down the incline.
Explain your answer.
Answer: The Fg can be calculated from the mass of the object.
Fg = m • g = (2 kg) • (9.8 m/s/s) = 20 N
The parallel and perpendicular components of the gravity force can be determined from
their respective equations:
Fgx = 19.6 N • sin (30 degrees) = 9.8 N
Fgy = 19.6 N • cos (30 degrees) = 17 N
The acceleration of the tire is the ratio of net force to mass:
a = Fnet / m = (9.8 N) / (2.00 kg) = 4.9 m/s2
The correct graph is Graph B since the tire will accelerate (speed up) down the hill.
3. In each of the following diagrams, a 100.0-kg box is sliding down a frictional surface
at a constant speed of 0.200 m/s. The incline angle is different in each situation. Analyze
each diagram and fill in the blanks.
Diagram A
Diagram B
Answer:
Diagram A: The Fg can be calculated from the mass of the object.
Fg = m • g = (100 kg) • (9.8 m/s/s) = 980 N
The parallel and perpendicular components of the gravity force can be determined from
their respective equations:
Fgx = m • g • sin (30 degrees) = 490 N
Fgy = m • g • cos (30 degrees) = 849 N
For a constant speed motion, the forces parallel to the incline must balance each other.
Thus, the Ff is equal to the Fx.
Ff = 490 N
The forces directed perpendicular to the incline balance each other. Thus FN is equal to
Fy.
FN = 849 N
Diagram B: The Fg can be calculated from the mass of the object.
Fg = m • g = (100 kg) • (9.8 m/s/s) = 980 N
The parallel and perpendicular components of the gravity force can be determined from
their respective equations:
Fgx = m • g • sin (45 degrees) = 693 N
Fgx = m • g • cos (45 degrees) = 693 N
For a constant speed motion, the forces parallel to the incline must balance each other.
Thus, the Ff is equal to the Fx.
Ff = 693 N
The forces directed perpendicular to the incline balance each other. Thus FN is equal to
Fy.
FN = 693 N
4. A 30.0 kg object is on an inclined plane set at 40.0° to the horizontal. What is the
acceleration experienced by the object if the coefficient of friction is 0.250?
Answer:
5. Consider a 10.0 kg box on an inclined plane of 30o to the horizontal having a
coefficient of friction of μ= 0.50.
1. What is the force of friction on the box?
2. What is the acceleration of the box?
Answer:
a. First, Draw a free body diagram
The normal force acting on the box is mgcos30= 84.87 N, exactly the same as for the
frictionless inclined plane. The force of friction is defined as Ff
N, so plugging in the
appropriate values for and FN:
Ff = FN = 0.500*84.87 = 42.435 N = 42 N, upward.
b. The net force acting on the box is the difference between the downhill gravitational
force and the force of friction. From this balanced force equation you can determine
acceleration:
Fnet = mgsin30 - Ff
ma = mgsin30 - Ff
10.0*a = 10.0(9.8)(0.500) – 42.435
a = 0.6565 = 0.66 m/s2 downward
Because
, the direction of the acceleration is in the downhill direction.
6. Suppose that you now want to drag a 5.0-kg mass up and down a frictionless 15º
inclined plane.
1. How much force must you apply to a string acting parallel to the incline's surface
to slide the 5.0-kg mass up or down the incline at a constant velocity (no
acceleration)?
2. How much upward force would be needed to accelerate the 5.0-kg mass up this
incline at 3.0 m/sec2?
3. How much upward force would be needed to restrict the 5.0-kg mass' downward
acceleration to 1.0 m/sec2?
4. If the 5.0-kg mass were allowed to slide down this incline without any additional
applied forces acting upon it, what would be its acceleration down the incline?
Answer:
1. The parallel components of the gravity force:
Fgx = m • g • sin (45 degrees) = 5.0*9.8*sin (15 degrees) = 12.682 N
For a constant speed motion, the forces parallel to the incline must balance each other.
Fnet = Fgx - Fapp = 0. Thus, the Fapp is equal to the Fx.
Fapp = 12.682 N = 13 N upwards
2. If the object is accelerating up the incline at 3.0 m/s2 then the Fnet = ma =
5.0*3.0 = 15 N up the incline.
Determine an equation the represents the forces acting on the object in the x
plane.
Fnet = Fapp - Fgx
15 N = Fapp – 12.682
Fapp = 27.682 N = 28 N, up the incline
3. If the object is accelerating down the incline at 1.0 m/s2 then the Fnet = ma = 5.0*(1.0)
= 5.0 N
Determine an equation the represents the forces acting on the object in the x
plane.
Fnet = Fgx - Fapp
5.0 N =12.682 - Fapp
Fapp = 7.682 N = 7.7 N, up the incline
4. If no applied force, the object would accelerate down the incline due to the parallel
force of gravity alone:
Fgx = Fnet
12.682 = 5.0(a)
a = 2.536 = 2.5 m/s2 down incline
7. A 10.0 kg box accelerates at 2.0 m/s2 as it slides down a ramp that makes an angle of
30 degrees with the horizontal. Find the coefficient of friction.
The Fg can be calculated from the mass of the object.
Fg = m • g = (10 kg) • (9.8 m/s/s) = 98 N
The parallel and perpendicular components of the gravity force can be determined from
their respective equations:
Fgx = m • g • sin (30 degrees) = 49 N
Fgy = m • g • cos (30 degrees) = 84.9 N
For a box that accelerates at 2.00 m/s2, the Fnet = ma = 10.0*2.00 = 20.0 N
Fnet = Fgx - Ff
20.0 N = 49 - Ff
Ff = 29 N
The forces directed perpendicular to the incline balance each other. Thus FN is equal to
Fy.
FN = 84.9 N
Ff = FN
29 = 84.9

8. A city planner is working on the redesign of a hilly portion of a city. An important
consideration is how steep the roads can be so that even low-powered cars can get up the
hills without slowing down. It is given that a particular small car, with a mass of 1100kg,
can accelerate on a level road from rest to 21 m/s in 14.0 s. Using this data, calculate the
maximum steepness of the hill.
Answer:
First, think about the question, what is going on, and what are the forces acting on the
car?
Second, draw a free body diagram.
Third, determine the acceleration on the object:
vf = vi + at
21 = 0 + a(14)
a = 1.5 m/s up the hill.
This is the rate that a car can accelerate at. The Fapp on the object up the hill will
counteract the Fgx down the hill resulting in a Fnet of zero overall. If these two are equal
then the car will not slow down (accelerate) but rather maintain a constant velocity up the
hill.
Fapp = Fgx
ma= mgsinθ
1.5 = 9.8 sin θ
θ = 8.8 degrees
HINT: The direction of motion depends not only upon the applied force, the weight of the
block, and also on the direction of the force of friction.
9. An object is being pulled up a 15° incline against a frictional coefficient of 0.15, and
requires a force of 835 N parallel to the surface of the ramp to move it at a constant
speed. What is the weight of the object?
First, draw a free body diagram.
Second, since an Fapp of 835 N is required to move the object at a constant speed (no
acceleration), the Fnet must be zero and Fgx + Ff must add to Fapp:
0 = Fnet = Fapp - Fgx - Ff
Fapp = Fgx + Ff
835 N = mgsin 15 + mgcos 15(0.15)
recall: Ff = FN μ = Fgy* μ= mgcos 15*μ
FN = Fgy
835 = mg(sin 15 + (cos 15)*0.15)
mg = 835/(0.2588 + 0.144888)
mg = Fg = 2068.42 = 2.1 x 103 N
The weight of the object is 2.1 x 103 N.