Friction on an inclined plane 

To show you how to calculate the friction on an inclined plane, we will use the diagram below. The diagram shows an object of mass m on an inclined plane.

Friction on an inclined plane

After trial and error, we realize that if the angle is slightly higher than 30 degrees, the object will slide down the incline. However, when the angle is 30 degrees, the object is on the verge of sliding. 

When the object is on the verge of moving, what is the coefficient of static friction μs between the object and the surface the object is placed on?

First important concept when dealing with friction on an inclined plane

Identify the forces

Since the object does not slide when the angle is 30 degrees, the force of friction is holding the object in place. It is usually directed opposite to the direction of movement. We show this force below but we use a green dot to represent the object to facilitate the reading. 

Friction on an inclined plane

When the angle is greater than 30 degrees, the object slides down. What force causes the object to slide? It is the force of gravity since no hands are pushing the object down. 

Friction on an inclined plane
Fg has two components and we show these with black vectors.The bigger black vector pointing down represents the weight.

Whenever there is a weight, there is a normal force and we show the normal force with an orange vector.

Notice also the other location of the angle. We are claiming this is the same angle as the one at the bottom right corner.

We will prove this in another lesson.

Friction on an inclined plane

Let's add a few more stuff that will help us to solve the problem.

Friction on an inclined plane

Second important concept when dealing with friction on an inclined plane

Apply Newton's second law:

Newton's second law is F = m × a

F is the sum of all forces.

If we apply Newton's second law along the y-axis, we get:

N - big black vector = m × a

When the object is on the verge of moving, it is not moving, so the acceleration is equal to 0.

The equation becomes

N - big black vector = m × 0

N - big black vector = 0

We can use trigonometric identities to find the big black vector.

cos θ =
big black vector / Fg
If 3 =
15 / 5
 , then 15 = 3 × 5

By the same fashion,

If cos θ =
big black vector / Fg
 ,then big black vector = cos θ × Fg

N - cos θ×Fg = 0

N = cos θ×Fg

If we apply Newton's second along the x-axis, we get:

Small black vector - Fs = 0

Since sin θ =
small black vector / Fg
, then small black vector = sin θ × Fg


sin θ× Fg - Fs = 0

sin θ× Fg = Fs

Now we can use the formula fs, max = μs × N
If 10 = 5 × 2, then 5 =
10 / 2
If fs, max = μs × N, then = μs =
fs, max / N

μs =
sin θ × Fg / cos θ× Fg

μs = tan θ

μs = 0.5773

This lesson about friction on an inclined plane is a little challenging.

Make sure you master trigonometric identities, vectors, and normal force before reading this lesson.

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