The figure shows a body resting on a rough inclined plane where the coefficient of static friction between the body and the plane, 𝜇 sub 𝑠, equals 0.487. Given that the body is on the point of sliding down the plane, find the angle of inclination 𝜃, rounding your answer to the nearest minute, if necessary.
We’re told in this statement that the coefficient of static friction is 0.487. And we’re asked to solve for the angle of inclination 𝜃 at which this inclined plane is raised. We can start on our solution by considering the forces that are acting on the body. We know that gravity acts on this body. We can draw that in as a vector pointing straight down and assign that force of magnitude equal to the weight of the body. We’ll call it capital 𝑊. There is also a normal force that acts on the body perpendicular to the surface of the plane. We can refer to that force as 𝐹 sub 𝑁. And finally, there is a frictional force acting on this body that points up the plane. We can call that force 𝐹 sub 𝑓.
We’re told in the statement that this body is at rest. It’s not in motion. That means the three forces we’ve identified all balance one another out. Their sum is zero. To look more closely at those forces, let’s put two coordinate axes on this scenario. We’ll define motion in the positive 𝑦-direction as motion up perpendicular to the plane and motion in the positive 𝑥-direction is up the plane and parallel to it. Two of our forces, 𝐹 sub 𝑁 and 𝐹 sub 𝑓, are already completely aligned with these axes. 𝐹 sub 𝑊, however, has components in the 𝑥- and 𝑦-direction.
If we sketch in those 𝑥- and 𝑦-components, we see that those components, along with the magnitude of the vector itself, form a right triangle. And the topmost angle in that triangle is equal to 𝜃.
Now that all three of our forces are broken up into 𝑥- and 𝑦-components, let’s consider those components along the 𝑥-axis. We have the frictional force acting in the positive 𝑥-direction minus the 𝑥-component of the weight force which we’ll call 𝑊 sub 𝑥. That difference is equal to zero because these are all the forces that act in the 𝑥-direction. And our body is in equilibrium. We can expand on both these expressions, 𝐹 sub 𝑓 and 𝑊 sub 𝑥. Let’s consider 𝑊 sub 𝑥 first.
Looking at our diagram, we see that this component of the weight force is equal to 𝑊 times the sine of 𝜃. So we insert that expanded expression in our equation. For 𝐹 sub 𝑓, we can recall a definition for frictional force. Friction force 𝐹 sub 𝑓 is equal to the coefficient of friction 𝜇, whether static or kinetic, times the normal force 𝐹 sub 𝑁. In our case, we can write 𝜇 sub 𝑠 because our body is at rest not in motion. We can expand our 𝐹 sub 𝑁 term further based on our diagram. If we consider only forces in the 𝑦-direction on our diagram, we see that there are two, 𝐹 sub 𝑁, the normal force, and the 𝑦-component of the weight force. Since our body is at rest, that must mean that the magnitude of these two forces are equal. In other words, 𝐹 sub 𝑁 is equal to 𝑊 times the cosine of 𝜃.
We now have a fully expanded equation for forces in the 𝑥-direction of our scenario. If we add 𝑊 times the sign of 𝜃 to both sides, then we see we can cancel out the weight force, 𝑊, of our body from this expression. Our result is independent of it. At this stage, it will be helpful to recall a trigonometric identity. The tangent of an angle 𝜃 is equal to the sine of that same angle divided by 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, using our identity, we have the tangent of 𝜃. If we then take the inverse tangent of both sides of this equation, we see that 𝜃 equals the arc tangent of our coefficient of static friction, 𝜇 sub 𝑠. We’re given the value of 𝜇 sub 𝑠 and can insert that now.
When we evaluate this expression on our calculator, we find that, to the nearest minute, 𝜃 is 25 degrees and 58 minutes. That’s the maximum angle of inclination of our plane at which the body is still stationary.