Video: Studying the Equilibrium of a Body Resting on a Rough Inclined Plane in Different Conditions

A body weighing 60 N rests on a rough inclined plane. A force ๐น acts on the body such that its line of action is up the line of greatest slope of the plane. When ๐น = 67 N, the body is on the point of moving up the plane, and when ๐น = 36 N, it is on the point of moving down. Find the angle of inclination of the plane rounding your answer to the nearest minute if necessary.

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Video Transcript

A body weighing 16 newtons rests on a rough inclined plane. A force ๐น acts on the body such that its line of action is up the line of greatest slope of the plane. When ๐น is equal to 67 newtons, the body is on the point of moving up the plane, and when ๐น equals 36 newtons, it is on the point of moving down. Find the angle of inclination of the plane rounding your answer to the nearest minute if necessary.

There is so much information here weโ€™re going to begin by sketching a diagram out. We have a body that rests on an inclined plane. Now, we donโ€™t know the angle that this plane is inclined to the horizontal. So, letโ€™s call that ๐œƒ or ๐œƒ degrees. We are told, however, that the body weighs 60 newtons. In other words, the force it exerts on the plane in a downward direction is 60 newtons. Now, of course, we know this means there is a reaction force of the plane on the body, and this acts perpendicular to the plane.

Then, we have this force ๐น that acts on the body such that its line of action is up the line of greatest slope of the plane. In other words, it acts in a direction parallel to the plane. Weโ€™re then told two pieces of information about the size or the magnitude of this force. When ๐น is 67 newtons, the body is about to move up the plane. This means two things. Itโ€™s in limiting equilibrium, so the vector sum of its forces is zero. But since the body is about to move up the plane or is on the point of moving up the plane, the friction acts in the opposite direction to this; it acts down the plane. So, letโ€™s call this scenario a. Weโ€™re going to begin by resolving our forces parallel to the plane, remembering, of course, we said that the vector sum of these forces is zero.

Another way to think about this is that the forces that act parallel and up the plane must be equal to the forces that act parallel and down the plane. We have ๐น, which is 67 newtons, acting up the plane. This is going to be equal to the frictional force, acting in the opposite direction, plus the component of the weight that acts parallel to the plane. If we add in a little right-angled triangle as shown, we see, we can use right angle trigonometry to find the value of ๐‘ฅ. ๐‘ฅ is the opposite side in this triangle. And โ„Ž, the hypotenuse, is 60 newtons. We can use, therefore, the sin ratio such that sin ๐œƒ is ๐‘ฅ over 60, and then we multiply by 60. So, ๐‘ฅ is 60 sin ๐œƒ.

Weโ€™re going to repeat this process for the other value of ๐น. Now, this time the force is 36 newtons, and the object is on the point of moving down the plane. This means friction is going to be acting in the opposite direction; itโ€™s going to be acting up the plane. So, once again, we resolve forces parallel to the plane. We have 36 newtons acting up the plane plus the frictional force. The sum of these is equal to the component of the weight that acts parallel to the plane, so 60 sin ๐œƒ again.

And what we might be inclined to do is to try and find a value for the friction. To do so, we would need to resolve the forces perpendicular to the plane and then use the fact that friction is ๐œ‡๐‘…. Itโ€™s the coefficient of friction times ๐‘…, the reaction force. But actually, if we look really carefully, we see that we have a pair of simultaneous equations in friction and ๐œƒ. And so, if we can eliminate friction from these simultaneous equations, weโ€™ll be able to solve for ๐œƒ.

Letโ€™s rearrange our first equation by subtracting 60 sin ๐œƒ from both sides. Then, we substitute our expression for friction into our second equation. So, 36 plus 67 minus 60 sin ๐œƒ equals 60 sin ๐œƒ. We simplify by adding 36 and 67, and then we also add 60 sin ๐œƒ to both sides of our equation. So, 103 is 120 sin ๐œƒ. And then, we divide through by 120. So, sin ๐œƒ is 103 divided by 120. Letโ€™s take the inverse sin or arc sin of both sides of our equation to solve for ๐œƒ. That gives us 59.12 and so on.

Now, the question asks us to find the angle of inclination of the plane giving our answer to the nearest minute, if necessary. Now, using the conversion function on our calculator, we get 59 degrees, seven minutes, and 47.8 and so on seconds. Correct to the nearest minute, thatโ€™s 57 degrees and eight minutes.

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