A body weighing 20 newtons rests on a rough inclined plane. An upwardly inclined force 𝑝 acts on the body such that its line of action is parallel to the line of greatest slope of the plane. Given that when 𝑝 is equal to 22 newtons, the body is on the point of moving up the plane and when 𝑝 is equal to 10 newtons, it is on the point of moving down the plane, determine the coefficient of friction between the body and the plane.
And there’s an awful lot of information here, so we’ll begin by sketching out the first situation. We have a body that weighs 20 newtons resting on an inclined plane. Since the weight of the body is 20 newtons, that’s the force that acts vertically downwards on the slope. We don’t know the inclination of the plane, so let’s call that 𝜃. We have an upwardly inclined force 𝑝 the acts on the body. Now, it acts in a direction that’s parallel to the line of greatest slope of the plane, as shown.
Well, we’re actually told that when 𝑝 is equal to 22 newtons, the body is on the point of moving up the plane. Now, what’s stopping it moving up the plane is the friction. Remember, we were told it is a rough inclined plane. Friction acts against the direction in which the body is trying to move, as shown. There’s actually one other force we’re interested in, and that’s the reaction force of the plane on the body. This acts perpendicular to the plane.
Now, before we move on to the second bit of information, we’re going to use what we have to calculate 𝜇. That’s the coefficient of friction of our plane. To do so, we’ll resolve forces, both perpendicular and parallel to the plane. We usually begin by considering forces perpendicular. The problem is, the force representing the weight of the body doesn’t act in either a parallel or perpendicular direction to the plane. And so, we add in this right-angled triangle. And we think about its separate components; let’s call those 𝑥 and 𝑦.
We can label the triangle as shown and use right angle trigonometry to calculate expressions for 𝑥 and 𝑦 in terms of 𝜃. We use the cosine ratio to begin with. We know that cos 𝜃 is adjacent over hypotenuse. So here, cos 𝜃 is 𝑥 over 20 or 𝑥 equals 20 cos 𝜃. Using the same ratio in a similar way and we derive an expression for 𝑦 in terms of 𝜃; it’s 20 sin 𝜃. Now, we know the body is in limiting equilibrium. So, the vector sum total of its forces is zero. Perpendicular to the plane, we have the reaction force acting upwards and 20 cos 𝜃 acting in the opposite direction perpendicular to the plane. So, 𝑅 minus 20 cos 𝜃 equals zero, which means that 𝑅 is 20 cos 𝜃.
Armed with this information, we can resolve forces parallel to the plane. This time, we have 22 newtons acting up and parallel to the slope and then the friction force and the component of the weight. So, we say that 22 minus friction minus 20 sin 𝜃 equals zero. But we know that friction is 𝜇𝑅, where 𝜇 is the coefficient of friction. So, we replace friction with 𝜇𝑅. But then, we also recall that 𝑅 is 20 cos 𝜃. And so, our equation becomes 22 minus 20𝜇 cos 𝜃 minus 20 sin 𝜃 equals zero.
Now, we’ve done everything that we can with the first piece of information. We’re now going to consider what happens when 𝑝 is equal to 10 newtons. When 𝑝 is 10 newtons, the body is on the point of moving down the plane. This means friction acts up the plane. Remember, it acts in the opposite direction to that which the body is trying to move. So this time, when we resolve forces parallel to the plane, we get 10 newtons and friction upward and 20 sin 𝜃 downward parallel to the plane.
Replacing friction once again with 20𝜇 cos 𝜃 and our second equation becomes 10 plus 20𝜇 cos 𝜃 minus 20 sin 𝜃 equals zero. Notice now, we have a pair of simultaneous equations in 𝜇 and 𝜃. Let’s clear some space and solve them. We’re going to begin by eliminating 𝜇. That will give us an equation purely in terms of 𝜃. To eliminate 𝜇, we’re going to add our two equations. When we do, we find that 32 minus 40 sin 𝜃 equals zero. By rearranging, we find sin 𝜃 is 32 over 40. And so, 𝜃 is the inverse sin of this fraction That’s roughly 53 degrees.
Now, we will use the exact value for 𝜃 in the next part of our calculation. We now substitute 𝜃 into either one of our equations. Let’s choose the second equation. That’s 10 plus 20𝜇 cos of 53.13 minus 20 sin of 53.13 equals zero. Now, to get this 53.13, we calculated the inverse sine of 32 over 40. So, sin of 53.13 must be 32 over 40 or four-fifths. And so, our equation simplifies a little to 10 plus 20𝜇 cos of 53.13 and so on minus 16 equals zero. And cos of 53.13 is three-fifths.
Now, you might have noticed that we could have calculated this without using a calculator. By drawing a right-angled triangle and using the fact that sin 𝜃 was 32 over 40 or four-fifths, we see that the third side in this right-angled triangle is three. And then since cos is adjacent over hypotenuse, we get cos 𝜃 equals three-fifths. This equation then simplifies to negative six plus 12𝜇 equals zero. We add six to both sides and then divide through by 12. And we find 𝜇 to be equal to one-half. In other words, the coefficient of friction between the body and the plane is one-half.