### Video Transcript

A body weighing 198 newtons rests on a rough inclined plane. When the plane is inclined at 30 degrees to the horizontal, the body is on the point of moving down the plane. The angle of inclination of the plane was then increased to 60 degrees. Find the minimum force required to keep the body in equilibrium given that the force is acting up the line of the greatest slope of the plane.

There is an awful lot of information here. So we’re simply going to begin by sketching a diagram. To begin with, this body that weighs 198 newtons rests on a plane inclined at 30 degrees to the horizontal. A weight of 198 newtons means that the body exerts a downward force on the plane of 198 newtons. And we know this means there is a reaction force 𝑅 of the plane on the body, acting perpendicular to the plane. And in this first scenario, we’re told that the body is on the point of moving down the plane. We need to ask ourself, what is stopping the body from moving down the plane?

Well, we’re told that it is a rough plane. And this means there is a frictional force that’s preventing the body from moving. And this force acts in the opposite direction to that which the body wants to move. Here, it acts up the plane. Now, we’re just going to work with this first bit of information. And we’re going to calculate 𝜇. That’s the coefficient of friction of our plane. To do this, we resolve forces perpendicular and parallel to the plane.

We’ll begin by looking at those perpendicular to the plane. We know that the body in this direction is in equilibrium. This means the sum of all forces acting perpendicular to the plane is zero. An alternative way to think about this is that the reaction force must be equal to the component for the weight that acts perpendicular to the plane. To find this component for the weight, we drop in a right-angled triangle. It has an included angle of 30 degrees, and we know its hypotenuse is 198 newtons.

We’re looking to calculate the component of this force that acts perpendicular to the plane. That’s 𝑥. And so we see we can use right-angle trigonometry. 𝑥 is the side adjacent to our included angle. And we know the length of the hypotenuse. We can therefore use the cosine ratio such that cos of 30 is 𝑥 divided by 198. By multiplying through by 198, we find 𝑥 is 198 cos of 30. And since cos of 30 is root three over two, we can simplify this and say that 𝑥 is 99 root three or 99 root three newtons.

We said that the sum of the forces acting perpendicular to the plane is zero. Well, if we assume that 𝑅 is acting in the positive direction, then 99 root three is acting in the negative direction. And we could say that 𝑅 minus 99 root three is equal to zero. And if we then add 99 root three to both sides, we find that 𝑅 is 99 root three newtons.

Now that we’ve calculated the value of 𝑅, we’re ready to work parallel to the plane. Once again, the body is in the point of moving down the plane. So we know that the forces that act upward and parallel to the plane must be exactly equal to those that act downwards and parallel to the plane. Well, we said we had friction acting upward. And this must be equal to 𝑦. 𝑦 is the component for the weight that acts parallel to the plane.

This time, 𝑦 is the side opposite the included angle. So we can say that sin 30 is 𝑦 over 198. This means that if we multiply through by 198, we find that 𝑦 is equal to 198 sin 30, which is simply 99 or 99 newtons. And so for the body to be at the point of moving down the plane, friction must be equal to 99.

But we might recall that friction is equal to 𝜇𝑅. It’s the coefficient of friction multiplied by the reaction force perpendicular to the plane. So our equation becomes 𝜇𝑅 equals 99. But we already saw that 𝑅 was equal to 99 root three. So we get 𝜇 times the 99 root three equals 99. We divide both sides of this equation by 99 and then divide through by root three. By rationalizing the denominator of our fraction here, we find that 𝜇 is equal to root three over three. Remember, this is the coefficient of friction of our plane, and this isn’t going to change.

Let’s clear some space and look at the second bit of information. This time, the angle of inclination of our plane is 60 degrees. So we change 30 degrees to 60 degrees in our diagram. The slope is now steeper, so the body wants to move down the plane. We still have a frictional force acting up the plane trying to prevent it. But we have an additional force. Let’s call that 𝐹. And we’re looking to find the minimum value of 𝐹 that ensures the body stays in equilibrium.

To do so, we need to begin by calculating the new value of 𝑅. Once again, we’ll resolve forces perpendicular to the plane. Remember, the sum of these forces is equal to zero. Or we can equivalently say that the forces that act upward, perpendicular to the plane, are equal to those that act downward and perpendicular to the plane. So 𝑅 is equal to 𝑥. Now this time, that’s 198 cos of 60. And since cos of 60 is one-half, we find that 𝑅 is now equal to 99 newtons.

We know that 𝜇 is still root three over three. And we’ve now calculated the new value of 𝑅. So we look at forces parallel to the plane. Remember, we’re looking to calculate the minimum value of 𝐹 to keep our body in equilibrium. So this means the sum of the forces that act parallel to the plane is zero. That is, the force plus the frictional force minus 198 sin 60 — that’s the component of the weight that acts parallel to the plane — must be equal to zero. And remember, of course, this means the forces that act upward and parallel to the plane must be equal to those that act downward and parallel to the plane.

We’re going to replace 198 sin 60 with 90 root three over two. 198 times sin 60 is 99 root three. And remember, we can replace friction with 𝜇𝑅. But of course, we calculated 𝑅 to be equal to 99 newtons. And we know the coefficient of friction of our plane is root three over three. So our equation becomes 𝐹 plus root three over three times 99 equals 99 root three, or 𝐹 plus 33 root three equals 99 root three. We’re going to subtract 33 root three from both sides so that 𝐹 is equal to 66 root three or 66 root three newtons.

And we can now say that the minimum force that is required to keep the body in equilibrium is 66 root three newtons up the line of greater slope of the plane.