### Video Transcript

Given that the coefficient of static friction between a body and a plane is square root of three over four, what is the angle of friction? Round your answer to the nearest minute if necessary.

Let’s begin our solution by agreeing on symbols that can represent the information given in the statement. The coefficient of static friction square root of three over four we can refer to as the Greek letter 𝜇 with a subscript 𝑠. We are asked to solve for the angle of friction in this scenario. We’ll call that angle 𝜃.

Let’s start by drawing a sketch of this body on a plane. In this situation, we can imagine that the body has a flat surface interfacing with the flat surface of the plane. The angle of friction we want to solve for is the angle 𝜃 at which the plane is inclined.

To determine this angle, we’ll want to look at the forces that are acting on this body. We know one force acting on it — that’s the force of gravity acting straight down with a magnitude equal to the weight — we’ll call it 𝑊 — of the body. Since the body is at rest, that’s why we have a coefficient of static friction since it’s not moving. We know there are two other forces that are acting on this body. One is called the normal force — named normal because it acts perpendicular to the surface of the plane. And there’s also a frictional force. We’ve called it 𝐹 sub 𝑓 that acts on the body to keep it from sliding down the plane.

Because the body is at rest and not in motion, we know that these three forces balance one another out. They’re in equilibrium. To investigate this equilibrium, we can define a set of coordinate axes, where positive 𝑦 points perpendicular to the plane surface and positive 𝑥 points up the incline. Let’s investigate forces in the 𝑥-direction first to see how they balance out.

The weight force due to gravity can be broken up into 𝑥- and 𝑦-components, where the 𝑥-component points down the plane and the 𝑦-component points into the plane. The triangle formed by these two components in the weight force itself is a right triangle and the top most angle of that triangle is equal to 𝜃. So as we focus on the forces solely in the 𝑥-direction, we can write that the force of friction which is positive according to our coordinate definition minus the 𝑥-component of the weight force — what we’ll call 𝑊 sub 𝑥 — is equal to zero. And again, it’s equal to zero because the body is at rest.

Looking at this equation, we can expand on the friction force and the 𝑥-component of the weight force to write them in terms of some of our given variables. For the weight force, looking back at the triangle we’ve drawn, we see that that component — the 𝑥-component of the weight force — is equal to the weight 𝑊 multiplied by the sine of the angle 𝜃.

Now, for the frictional force, we can expand that as well. And we’ll do it by recalling a mathematical definition for that force. The friction force 𝐹 sub 𝑓 is equal to the coefficient of friction whether static or kinetic multiplied by the normal force 𝐹 sub 𝑁. In our case, we write 𝜇 sub 𝑠 because our friction coefficient is static — the body is not in motion.

But what about 𝐹 sub 𝑁 — the normal force. Looking again at our diagram, we can see that again since our body is in equilibrium, the magnitude of the normal force pointing up must be counteracted by the 𝑦-component of the weight force pointing down into the plane. That means we can replace 𝐹 sub 𝑁 with 𝑊 times the cosine of 𝜃.

We now have an equation for the balance of forces in the 𝑥-direction of our diagram. If we rearrange this equation by adding 𝑊 sin 𝜃 to both sides, when we look at the result, we see that the weight force 𝑊 cancels from each side. So our result is independent of the weight of the body. To simplify further, let’s recall a tree in a metric identity. This identity says that the tangent of an angle 𝜃 is equal to the sine of that angle over the cosine of that angle. So if we divide both sides of our equation by the cosine of 𝜃, then that term cancels out on the left-hand side. And on the right-hand side, we find the tangent of 𝜃.

We’ve been told what 𝜇 sub 𝑠 — the coefficient of static friction — is in the problem statement and 𝜃 is what we want to solve for. So to get there, let’s take the inverse tangent of both sides of this equation. We find that 𝜃 is equal to the inverse or arctangent of 𝜇 sub 𝑠. And when we insert our given value for 𝜇 sub 𝑠 and enter this expression on our calculator, we find that 𝜃 to the nearest minute is 23 degrees and 25 minutes. That’s the angle of inclination of this plane — the angle of friction.