Question Video: Analysis of a Body in Equilibrium on a Rough Inclined Plane Mathematics • Third Year of Secondary School

Video Transcript

A body of weight 𝑊 newtons rests on a rough plane that is inclined at an angle of 60 degrees to the horizontal. A force 𝐹 is acting on the body up the line of greatest slope of the plane. When 𝐹 is equal to 33 newtons, the body is on the point of moving down the plane. Whereas when 𝐹 is equal to 55 newtons, the body is on the point of moving up the plane. Find the value of 𝑊 and the coefficient of friction 𝜇 between the body and the plane.

Now, there’s an awful lot of going on here, so we’re simply going to begin by sketching a diagram. Now, in fact, we’re going to need to sketch two diagrams. We’re told two different bits of information about the force pushing the body up the slope. When the force is 33 newtons, the body is about to move down the plane. Whereas when it’s 55 newtons, it’s about to move up the plane. The fact that the body is about to move in different directions will affect the direction of the frictional force. And so, let’s label the first scenario, scenario a. And we’re going to begin by sketching this one out.

The body is resting on a plane at an angle of 60 degrees to the horizontal. We’re told that the weight of this body is 𝑊 newtons. In other words, the downward force that the body exerts on the plane is 𝑊. Now, of course, we know this means there’s a reaction force of the plane on the body. And this acts perpendicular to the plane. In scenario a, the force 𝐹 that acts on the body at the line of greatest slope of the plane is 33 newtons. And at this point, the body is on the point of moving down the plane. This means that frictional force is also acting upward; it’s acting against the direction in which the body wants to travel. Now that we have all relevant forces on our diagram, we’re going to resolve these forces perpendicular and parallel to the plane.

Now, since the weight doesn’t act in either of these directions, we split it up into its components parallel and perpendicular to the plane. We add a right-angled triangle as shown. We’ll call the component of the weight that acts perpendicular to the plane 𝑥 or 𝑥 newtons. And the component that acts parallel to the plane, we’ll call that 𝑦 newtons.

Now, 𝑥 is the side adjacent to the included angle in this triangle. And we know the hypotenuse is 𝑊 or 𝑊 newtons. We, therefore, use the cosine ratio. That is, cos 𝜃 is adjacent over hypotenuse. And we can say that cos 60 is equal to 𝑥 over 𝑊. So, 𝑥 is 𝑊 cos 60. cos 60 is one-half. So, we find 𝑥 is equal to a half 𝑊. Similarly, we can calculate the value of 𝑦 by using the sine ratio. sin 60 is 𝑦 divided by 𝑊. And if we multiply both sides by 𝑊, we get 𝑦 equals 𝑊 sin 60 or root three over two 𝑊.

We’ll now resolve forces perpendicular to the plane. We know that the body is in limiting equilibrium; it’s in the point of moving, but it’s not actually moving. This means the vector sum of its forces must be zero. We can say then that perpendicular to the plane, 𝑅 minus a half 𝑊 — remember, a half 𝑊 is acting in the opposite direction — is equal to zero. We add one-half 𝑊 to both sides, and we see that 𝑅 is one-half 𝑊. And next, we’ll resolve parallel to the plane for scenario a.

Once again, the vector sum of the forces is equal to zero. So, the frictional force plus the force acting up the line of greatest slope, that’s 33, minus root three over two 𝑊 must be equal to zero. But remember, friction is equal to 𝜇𝑅, where 𝜇 is the coefficient of friction and 𝑅 is the friction force labeled. This means that friction force is 𝜇 times a half 𝑊. Remember, we calculated 𝑅 to be a half 𝑊 earlier or a half 𝑊𝜇. We’ll simplify this equation just a little by multiplying through by two. So, 𝑊𝜇 plus 66 minus root three 𝑊 is equal to zero.

We’re now going to consider the second scenario. That’s when the force acting on the body is 55 newtons. This time, it’s on the point of moving up the plane. This means the frictional force acts in the opposite direction; it acts down and parallel to the plane as shown. Once again, the vector sum of our forces is zero. So, we have 55 minus the frictional force minus root three over two 𝑊 equals zero.

The reaction force remains unchanged, so we can rewrite friction as a half 𝑊𝜇. Once again, we multiply through by two. So, 110 minus 𝑊𝜇 minus root three 𝑊 equals zero. And we now see that we have a pair of simultaneous equations. Let’s clear some space and solve these.

Let’s begin by eliminating 𝑊𝜇. We’re going to add equation a and b. When we do, we get 66 plus 110. That’s 176. And negative root three 𝑊 plus negative root three 𝑊 is negative two root three 𝑊. We add two root three 𝑊 to both sides of our equation. And then, we divide through by two root three. So, 𝑊 is 176 over two root three. This simplifies to 88 root three over three or 88 root three over three newtons. We’re also looking to calculate the coefficient of friction 𝜇. So, let’s substitute 𝑊 back into one of our original equations.

Substituting into our equation a, and we get 88 root three over three 𝜇 plus 66 minus root three times 88 root three over three equals zero. Root three times root three and then divided by three is simply one. So, we get 88 root three over three 𝜇 plus 66 minus 88 equals zero. Well, 66 minus 88 is 22. And we’ll add 22 to both sides. So, 88 root three over three 𝜇 equals 22. And our final step will be to divide by 88 root three over three, giving us 𝜇 equals root three over four. So, 𝑊 is 88 root three over three or 88 root three over three newtons, and 𝜇 is root three over four.