NEWTON’S LESSON 11 INCLINE PLANES
We see ramps that access many buildings; these are common and obvious
examples of
inclined planes.
What are some other examples of inclined planes that you come across
regularly?
A sloped road or driveway, a path up a hill, and the up or down
sections of a roller
Coaster, staircase to the top of the empire state building
Are there advantages to using an inclined plane? How does an inclined
plane make a job seem easier?
Force: An investigation
For this investigation we’ll need:
• A board
• A heavy block
• A rope
• A spring scale
First we:
1. Attach the spring scale to the block, then pull straight up by the rope (just
enough to lift the block).
2. How much force just sets the block in motion?
Then we:
1. Place the block at the bottom of the board.
2. Attach the spring scale to the block and pull it along the incline so that it
is just set into motion.
3. How much force just sets the block in motion?
What do you notice about these two forces?
It takes less force to slide the blocks along the incline than to lift them
straight up. But you must exert the smaller force through a larger
distance to get them to the top of the table.
1. Suppose you make the incline steeper, that is, make the angle of incline
greater.
How will that change the force you must exert to move the block along the
incline?
2. Now suppose you make the incline less steep, that is, lay the board flat on
the floor. How will that change the force you must exert to move the block
along the board?
The demonstrations show that as the angle of incline increases, the force
required to move the blocks at constant speed increases, also.
A thought investigation:
So, you want to get the block to the top of the table. You know that you need
to exert a smaller force to push or pull the blocks up the incline when the
angle of incline is smaller.
If the force to get the object to the table top decreases with a smaller incline
angle, what is the trade-off?
Distance is the trade off. The smaller the incline the longer the distance
to move the object.
The distance to move the block to the table top is shortest when the angle of
incline is the largest, 90o to the table.
The mechanical advantage of an inclined plane is the ratio of the length of
the sloped surface to the height it spans. When the distance is vertical,
mechanical advantage is 1. The higher the mechanical advantage, the less
force is needed to push or pull the object to the required height.
The ancient Egyptians figured this out over 3,000 years ago when they built
their pyramids. They used long, shallow ramps to help them move the heavy
stones to the top!
INCLINE PLANES AND ACCELERATION
An object placed on a incline plane will often slide down the surface.
DEMO:
Put a block on a plane and tilt it at varying degrees from horizontal to
vertical. What do you notice?
As the slope of the incline plane increases, what happens to the rate at
which the object will slide down it?
What happens when the incline is vertical? The acceleration on the
object would be equal to the acceleration due to gravity.
Objects are known to accelerate down inclined planes because of an
unbalanced force. Two forces acting upon a crate which is positioned
on an inclined plane (assumed to be friction-free).
 the force of gravity and the normal force. The force of gravity
(also known as weight) acts in a downward direction;
 the normal force acts in a direction perpendicular to the surface
(in fact, normal means "perpendicular").
The process of analyzing the forces acting upon objects on inclined planes
will involve resolving the weight vector (Fgrav) into two perpendicular
components.
- one directed parallel to the inclined surface and
- the other directed perpendicular to the inclined surface.
The perpendicular component of the force of gravity is directed opposite the
normal force and as such balances the normal force.
The parallel component of the force of gravity is not balanced by any other
force.
This object will subsequently accelerate down the inclined plane due
to the presence of an unbalanced force.
It is the parallel component of the force of gravity which causes this
acceleration. The parallel component of the force of gravity is the net
force.
EXAMPLE: A 5.0 kg mass is placed on an incline titled at 15º.
1. How much does the 5-kg mass weigh?
Fg = ma = 5*9.8 = 49 N
2. Calculate the magnitude of the component that is acting parallel to
the incline's surface?
Fx  mg sin   (5.0kg)(9.8m / s 2 ) sin( 15)  12.68 N
3. Calculate the magnitude of the component that is
acting perpendicular to the incline's surface?
Fy  mg cos  (5kg)(9.8m / s 2 ) cos(15)  47.33N
4. At what angle would these two components be equal in magnitude?
45o
5. In which range of angles would Fx > Fy? (force down the ramp be
greater than force into the surface of the ramp) (a) 0º<θ<45º (b)
45º<θ<90º
Answer: (b) 45º<θ<90º Note that the steeper the slope, the greater the
force pulling the box down the slope.
Example: A 75-kg skier skies down a snowy slope of 25 degrees. Find the
normal force, the net force, and the acceleration of the skier.
Given: m  75kg,  25, g  9.8 m
s2
Want: Fy  ?, Fnet  Fx  ?, a  ?
Draw: A free-body diagram like the one above and label it. (I’ll leave this to
you.)
Use equations from earlier:
Fy  mg cos  (75kg)(9.8m / s 2 ) cos(25)  666 N
Fx  mg sin   (75kg)(9.8m / s 2 ) sin( 25)  310.6 N
a  g sin   (9.8m / s 2 ) sin( 25)  4.14m / s 2
Example: Suppose you place a 10.0 kg box on a frictionless 30º inclined
plane and release your hold, allowing the box to slide to the ground, a
horizontal distance of d meters and a vertical distance of h meters.
1. What is the magnitude of the normal force?
2. What is the acceleration of the box?
3. What is the velocity of the box when it reaches the bottom of the slope
if it takes 1.2 seconds to reach the bottom and it starts from rest?
Answer:
First, draw a free body diagram:
1. The box is not moving in the y direction, so the normal force must be
equal to the y-component of the gravitational force. Calculating the normal
force is then just a matter of plugging a few numbers in for variables in order
to find the y-component of the gravitational force:
FN = Fgy = mgcos30
=(10)(9.8)(cos 30)
= 84.9 N
2. We know that the force pulling the box in the positive x direction has a
magnitude of mg sin 30. Using Newton’s Second Law, F = ma, we just need
to solve for a:
Fnet = Fgx
ma = mgsin 30
a = gsin30
= 9.8(0.500)
4.9 m/s2
3. We know the initial velocity, the time and acceleration:
v f = v i + at
v f = 0 + (5)(1.2) = 6.0 m/s
HOMEWORK NEWTON’S LESSON 11
1. Two boys are playing ice hockey on a neighborhood street. A stray puck travels across
the friction-free ice and then up the friction-free incline of a driveway. Which one of the
following ticker tapes (A, B, or C) accurately portrays the motion of the puck as it travels
across the level street and then up the driveway?
Explain your answer.
2. Little Johnny stands at the bottom of the driveway and kicks a soccer ball. The ball
rolls northward up the driveway and then rolls back to Johnny. Which one of the
following velocity-time graphs (A, B, C, or D) most accurately portrays the motion of the
ball as it rolls up the driveway and back down?
Explain your answer.
3. A golf ball is rolling across a horizontal section of the green on the 18th hole. It then
encounters a steep downward incline (see diagram). Friction is involved. Which of the
following ticker tape patterns (A, B, or C) might be an appropriate representation of the
ball's motion?
Explain why the inappropriate patterns are inappropriate.
4. Missy dePenn's eighth frame in the Wednesday night bowling league was a disaster.
The ball rolled off the lane, passed through the freight door in the building's
rear, and then down the driveway. Millie Meater (Missy's teammate), who was
spending every free moment studying for her physics test, began visualizing the
velocity-time graph for the ball's motion. Which one of the velocity-time graphs
(A, B, C, or D) would be an appropriate representation of the ball's motion as it
rolls across the horizontal surface and then down the incline? Consider frictional forces.
Explain your answer.
5. Three lab partners - Olive N. Glenveau, Glen Brook, and Warren Peace - are
discussing an incline problem (see diagram). They are debating the value of the normal
force. Olive claims that the normal force is 250 N; Glen claims that the normal force is
433 N; and Warren claims that the normal force is 500 N. While all three answers seem
reasonable, only one is correct. Indicate which two answers are wrong and explain why
they are wrong.
6. A car of mass 1100 kg rests upon an incline plane that is inclined at 10o to the
horizontal.
a. What is the weight of the car?
b. Assuming that there are no frictional forces acting between the car and the surface
on which it rests, draw a free-body diagram for the car.
c. What is the magnitude of the normal contact force between car and inclined
plane?
d. What is the magnitude of the force acting on the car parallel to the plane?
e. What is the acceleration of the car?
7. The weight of a book sliding down a friction-less inclined plane can be broken down
into two vector components: one acting parallel to the plane, and one acting
perpendicular to the plane.
a) At what angle will these two components be equal?
b) At what angle is the component parallel to the plane equal to zero?
c) At what angle is the component parallel to the plane equal to the weight?
d) As the angle goes up, how does the parallel component change? The perpendicular?
8. A 3.0 m long board has one end raised to a height of 60 cm to form an incline. A 4.0
kg mass is allowed to slide without friction down the entire length of the inclined plane.
a. What is the final speed of the mass when it reaches the bottom ?
b. If the mass is replaced with an 8.0 kg mass, what would be the new speed when it
reaches the bottom?
9. A thin clothesline strung between fixed points 10.0 meter apart supports a single suit
that weights 25 N. The suit is hung from the center of the line and this causes the line to
sag 10.0 cm below the horizontal at this point. What is the tension in the line?
10. Finding her car stuck in the mud, a bright graduate of a good physics course ties a
strong rope to the back bumper of the car, and the other end to a tree, as shown in the
diagram. She pushes at the midpoint of the rope with her maximum effort, which she
estimates to be a force Fapp = 300.0 N. The car just begins to budge with the rope at an
angle θ (see diagram) which she estimates to be 5.0o.
a. With what force is the rope pulling on the car?
b. Was this method a good idea and is she a good physics student?
c. What would happen if she continued to push on the rope?
HOMEWORK KEY
1. B
2. Graph D
3. B
4. Graph D
5. Glen is correct
6. a. 1.08 x 104 N, b. no answer, c. 1.06 x 104 N, d. 1.87 x 103 N, e. 1.70 m/s, down the
incline
7. a. 45o, b. 0o, c. 90o, d. increases, decreases
8. 3.4 m/s , no answer
9. 6.2 x 102 N
10. 1.7 x 103 N, see worked solution for b,c
HOMEWORK NEWTON’S LESSON 11
1. Two boys are playing ice hockey on a neighborhood street. A stray puck travels across
the friction-free ice and then up the friction-free incline of a driveway. Which one of the
following ticker tapes (A, B, or C) accurately portrays the motion of the puck as it travels
across the level street and then up the driveway?
Explain your answer.
Answer: B is the correct answer; it shows a constant velocity while traveling across the
level surface (which is not shown in C) and it shows the deceleration which would be
expected while traveling up a frictionless incline (which is not shown in A).
2. Little Johnny stands at the bottom of the driveway and kicks a soccer ball. The ball
rolls northward up the driveway and then rolls back to Johnny. Which one of the
following velocity-time graphs (A, B, C, or D) most accurately portrays the motion of the
ball as it rolls up the driveway and back down?
Explain your answer.
Answer: Graph D is the correct answer. Initially, the ball has a northward velocity and
is slowing down. For an instant in time, it has a zero velocity. Then the ball moves with a
southward velocity (i.e., in the opposite direction) and speeds up. At all times the ball has
a "negative" (southward) acceleration. These features are all depicted in Graph D.
3. A golf ball is rolling across a horizontal section of the green on the 18th hole. It then
encounters a steep downward incline (see diagram). Friction is involved. Which of the
following ticker tape patterns (A, B, or C) might be an appropriate representation of the
ball's motion?
Explain why the inappropriate patterns are inappropriate.
Answer: B is the correct answer.
Tapes A and C can both be ruled out since they show the golf ball moving with constant
velocity across the frictional surface. The ball should be slowing down. Tape B shows an
acceleration while moving down the hill which would be likely due to the presence of a
parallel component of the weight vector.
4. Missy dePenn's eighth frame in the Wednesday night bowling league was a disaster.
The ball rolled off the lane, passed through the freight door in the building's
rear, and then down the driveway. Millie Meater (Missy's teammate), who was
spending every free moment studying for her physics test, began visualizing the
velocity-time graph for the ball's motion. Which one of the velocity-time graphs
(A, B, C, or D) would be an appropriate representation of the ball's motion as it
rolls across the horizontal surface and then down the incline? Consider frictional forces.
Answer: Graph D is the appropriate representation.
The ball will slow down due to friction while moving across the level surface and it will
speed up due to the parallel component of the weight vector while moving down the
incline. Graph D depicts both of these features.
5. Three lab partners - Olive N. Glenveau, Glen Brook, and Warren Peace - are
discussing an incline problem (see diagram). They are debating the value of the normal
force. Olive claims that the normal force is 250 N; Glen claims that the normal force is
433 N; and Warren claims that the normal force is 500 N. While all three answers seem
reasonable, only one is correct. Indicate which two answers are wrong and explain why
they are wrong.
Answer: Only Glen is correct.
Warren is incorrect because he evidently believes that the normal force is equal to the
force of gravity. This is only true on level surfaces when the normal force is the only up
force. Here the normal force is perpendicular to the surface and equal to the
perpendicular component of gravity's pull.
Olive is incorrect because she has evidently used the wrong equation for computing the
perpendicular component of the weight vector. The parallel component is 250 N
(m*g*sine 30 degrees). But the normal force is equal to the perpendicular component of
the weight vector (m*g*cosine 30 degrees).
6. A car of mass 1100.0 kg rests upon an incline plane that is inclined at 10.0o to the
horizontal.
a. What is the weight of the car?
b. Assuming that there are no frictional forces acting between the car and the surface
on which it rests, draw a free-body diagram for the car.
c. What is the magnitude of the normal contact force between car and inclined
plane?
d. What is the magnitude of the force acting on the car parallel to the plane?
e. What is the acceleration of the car?
Answer:
a.
b.
c.
d.
e.
Fg = weight = 1100*9.8 = 10780 N = 1.08 x 104 N
No answer
Fgy = FN = mgcosθ = 1100*9.8*cos 10 = 10616.227 = 1.06 x 104 N
Fnet = Fx = mgsinθ = 1100*9.8*sin 10 = 1871.92 = 1.87 x 103 N
Fnet = Fx  ma = mgsinθ  a = 9.8sin 10 = 1.7017 = 1.70 m/s
7. The weight of a book sliding down a friction-less inclined plane can be broken down
into two vector components: one acting parallel to the plane, and one acting
perpendicular to the plane.
a) At what angle will these two components be equal?
b) At what angle is the component parallel to the plane equal to zero?
c) At what angle is the component parallel to the plane equal to the weight?
d) As the angle goes up, how does the parallel component change? The perpendicular?
Answer: a. 45o, b. 0o, c. 90o, d. increases, decreases
8. A 3.0 m long board has one end raised to a height of 60 cm to form an incline. A 4.0
kg mass is allowed to slide without friction down the entire length of the inclined plane.
a. What is the final speed of the mass when it reaches the bottom ?
b. If the mass is replaced with an 8.0 kg mass, what would be the new speed when it
reaches the bottom?
Answer
Since Ff = 0, and FN is cancelled by Fy, Fx is the net force
θ= sin -1( .6 / 3.0) = 11.3º
a.
Fnet = Fx
ma = mgsinθ
a = gsinθ
a =(9.8)sin11.3º = 1.92 m/s2
Applying vf2 = vi2+ 2ad gives:
vf2 = 0 + 2(1.92)(3.0)
vf = 3.4 m/s
b.In the equation a = gsinθ, we see that the acceleration of the object on the slope (in the
absence of friction) does not depend on the object's mass.
9. A thin clothesline strung between fixed points 10.0 meter apart supports a single
suit that weights 25 N. The suit is hung from the center of the line and this causes
the line to sag 10.0 cm below the horizontal at this point. What is the tension in
the line?
First, determine the angle the clothesline makes with the vertical:
θ = tan-1(5/0.1) = 88.85o
Since the weight is 25 N. Each cable must pull upwards with 12.5 N of force.
Thus, cos (88.85 degrees) = (12.5 N) / (Ftens).
Proper use of algebra leads to the equation
Ftens = (12.5 N) / [cos (88.85 degrees)] = 622.82 N = 6.2 x 102 N
10. Finding her car stuck in the mud, a bright graduate of a good physics course ties a
strong rope to the back bumper of the car, and the other end to a tree, as shown in the
diagram. She pushes at the midpoint of the rope with her maximum effort, which she
estimates to be a force Fapp = 300.0 N. The car just begins to budge with the rope at an
angle θ (see diagram) which she estimates to be 5.0o.
a. With what force is the rope pulling on the car?
b. Was this method a good idea and is she a good physics student?
c. What would happen if she continued to push on the rope?
Answer:
a. Since the force applied is 300 N. Each cable must pull with 150 N of force.
Thus, sine (5.0 degrees) = (150 N) / (Ftens).
Proper use of algebra leads to the equation
Ftens = (150 N) / [sin (5 degrees)] = 1721.056 N = 1.7 x 103 N
b. Great idea and awesome physics student! She was able to magnify her effort almost
six times using this technique!
c. As she continues to push on the rope, the angle will increase and the force of tension
on the rope will decrease. Therefore I would suggest that to maximize her results she
would continually retie the rope taut in order to keep her efforts as productive as possible.