5.2: Inclined Planes
Objectives
The student will:
 Understand how to analyze and work with forces on inclined planes
 Understand how to apply Newton’s Second Law to the inclined plane problems
Vocabulary
 inclined plane: Any flat surface tilted somewhere between horizontal and vertical.
Introduction
An inclined plane is any flat surface tilted somewhere between horizontal and vertical, like a ramp, a flat side of a hill, or a playground slide. Problems using inclined planes shows us how to divide up vector forces including gravity, normal force, and friction.
Using Inclined Planes
For objects in motion along inclined planes, it is convenient to establish the
As we will see, the normal force is always less than the weight when the object is placed on an inclined plane. To understand this, consider two extremes for an inclined plane:

A horizontal inclined plane
(θ=0 degrees) 
A vertical inclined plane
(θ=90 degrees) .
In the first case, the object is at rest because the net force on it is 0. In the second case the object has a net force on it of
Figure
below
shows an example between the two extremes. An object of weight 100 N is sliding down a frictionless 30degree inclined plane. We expect the acceleration of the object to be between 0 and
Consider the following questions.
1a. What normal force does an inclined plane provide in supporting a 100 N crate accelerating down a 30degree inclined plane?
1b. What acceleration, in the absence of friction, does the crate have along the inclined plane?
1c. What is the velocity of the crate after three seconds?
1d. What is the displacement of the crate after three seconds?
In order to answer these questions, we establish a coordinate system such that the
What to notice:

The components of
mg in the diagram above are dashed. 
The angle of the incline is the same as the angle between the
y− axis andmg . 
The
x− component ofmg ismgsin30∘ and they− component ofmg ismg cos30∘ .
We use Newton’s Second Law in the
1. To determine the normal force that the inclined plane provides we use Newton's Second Law for the forces in the
F_N &= mg \cos\theta\\
F_N &= 100\text{N} \cos 30^\circ\\
F_N &= 100\text{N} (0.87) = 87 \ \text{N}
2. To determine the acceleration of the crate down the plane, we use Newton's Second Law for the forces in the
a &= g\sin\theta,\\
a &= (10 \text{m/s}^2) \sin 30^\circ = 5.0 \ \text{m/s}^2
3. To determine the velocity of the crate at
Every second the crate gains another 4.9 m/s of velocity, so after 3.0 s:
In other words,
4. To determine the displacement of the crate at
Using the acceleration:
x_f &= \frac{1}{2}(5.0\text{m/s}^2)(3.0\text{s})^2+0+0 = 22.5 \ \text{m}
Using the average velocity:
http://demonstrations.wolfram.com/BlockOnAFrictionlessInclinedPlane/
Check Your Understanding
1. True or False: The direction of the normal force the inclined plane exerts on an object is opposite to the direction of
Answer: False. The normal force is perpendicular to the inclined plane.
2. True or False: The weight of a crate is 200 N. Its weight on an inclined plane will be smaller than 200 N.
Answer: False. Its weight is always 200 N.
3. True or False: The weight of a crate is 200 N. The normal force exerted on it when placed on an inclined plane will be smaller than 200 N.
Answer:
True. The normal force is
3. What is the acceleration of an object sliding down a frictionless inclined plane?
Answer:
Friction and Inclined Planes
There’s a quick activity that can be used to show the effect of static friction on an inclined plane. You will need a book and a coin (see picture below). Place the coin on the top of the cover of the book and slowly begin to lift the cover. You’ll notice that as you raise the cover the coin stays put. If the cover were a frictionless surface, the coin would slide down immediately as you began to lift the cover. Static friction keeps the coin stationary. However, there is a point at which the coin will begin to slide. There’s a critical height (angle) at which the component of the force of gravity down the inclined cover exceeds the force of static friction. If you play with this setup for a while, you’ll notice that once the coin begins to slide you can slightly lower the cover and the coin continues sliding, even though the cover is raised to a less extent than what was needed to have the coin begin its slide. This is another reminder that static friction tends to be greater than kinetic friction. We like to quantity how static and kinetic friction affect objects on inclined planes. We will take the case of static friction first.
Figure
above
shows the coin at the point where it is just about to slide. Up to this point static friction,
& \mu_s F_N*mg \sin\theta = 0, \ replacing \ F_N \ with \ mg \cos\theta,\\
& \mu_s (mg \cos\theta)mg \sin\theta = 0,\\
& \mu_s = \frac{\sin\theta}{\cos\theta},\\
& \mu_s = \tan\theta
Check Your Understanding
1. A coin rests on a 30degree inclined plane. It is determined that if the plane’s angle is increased further, the coin will slide down the plane. What is the coefficient of static friction between the coin and the plane?
Answer:
Since
2. True or False: Since the friction force is dependent upon the normal force, the coefficient of friction must have units of newtons.
Answer:
False. As you can see from the inclined plane problem, the coefficient of friction is dependent only upon the tangent of the angle. Recall that the definition of
If we exceed the maximum angle at which the static friction can hold the coin, the coin will slide and kinetic friction will act upon the coin. We consider this situation next. Figure below is almost identical to Figure above . The only differences are that the coin is represented as a dot and the kinetic friction vector has been drawn a bit smaller than the static friction vector.
In this case, the coin will accelerate down the book cover. We wish to determine the acceleration. We choose the direction down the book cover as positive. Using Newton's Second Law:
Check Your Understanding
1. A block slides down a 25degree incline with a constant velocity. What is the coefficient of kinetic friction between the block and the incline?
Answer:
2a. A skier of mass 60 kg skies down a slope of 18 degrees with an acceleration of
Answer:
f_k &= mg\sin\thetama \rightarrow (60)(9.8)\sin 18^\circ(60)(2.1)\\
f_k &= 55.7 = 56 \ N
2b. What is
Answer:
2c. The skier is hurt and a member of a rescue team pulls the skier up the slope at a constant velocity.
(i) What force must he exert?
(ii) Of this force, what percentage is the skier's weight?
Answers:
(i)
(ii)
2d. What is one benefit of using an inclined plane to lift heavy objects?
Answer: A smaller force is usually required to move objects up the plane than lifting against gravity.
An Incline PlanePulley System
Illustrative Example
How do we determine the acceleration and the tension between two masses when one of the masses is on the inclined plane and the other hangs over the plane, as in Figure below ?
Again, we call on Newton's Second Law in order to solve the system. Let’s assume there is no friction in the system and the rope between the masses does not stretch and has negligible mass. The values for the masses and the angle of the incline are given below.
Find, (A) the acceleration of the system and (B) the tension in the rope.
Finding the acceleration of the system
We always start by drawing FBDs and resolving vectors.
1. Draw an FBD for each mass.
2. Resolve any vectors that are not along the
The
The
Newton's Second Law equation for the mass
We assume that
Newton's Second Law equation for
Since
Solving the two equations above simultaneously for the acceleration gives:
Had the answer come out negative it would have indicated that we had chosen the direction of the acceleration incorrectly. It is also helpful to note that the tensions,
Finding the tension in the rope
Since the acceleration is known, we can substitute its value in either the Newton's Second Law equation for the mass
As a check for reasonableness we can determine if the tension is greater than
Check Your Understanding
1. True or False: In order for a hanging mass
Answer:
False. As we saw in the last problem,
2. For a given mass
Answer:
We go back to the equations for
Thus
A Pulley System with Friction along the Inclined Plane
Illustrative Example
Refer back to
Figure
above
. Assume that friction exists between the mass
Since friction is directed along the plane, it has only an
a. Find the acceleration of the system.
Newton's Second Law
NSL
& T(40)(9.8)\sin 25^\circ\mu_k F_N =40a, \ replacing \ F_N \ with \ mg \cos\theta,\\
& T(40)(9.8)\sin 25^\circ0.20(40)(9.8)\cos25^\circ = 40a \ Equation\ M
Solving Equations
b. Find the tension.
Using Equation
Check Your Understanding
In the absence of friction, if
Answer:
Yes. Try solving the equations with acceleration equal to zero, but keep
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Learning Objectives
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Date Created:
Jan 18, 2013Last Modified:
Jun 18, 2015Vocabulary
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