When a body of weight 262.5 newtons was resting on a rough plane inclined to the horizontal at an angle whose tangent is three-quarters, it was on the point of moving. The same body was later placed on a horizontal surface of the same roughness. A force 𝐹 acted on the body pulling upward at an angle of 𝜃 to the horizontal, where sin of 𝜃 is equal to three-fifths. Given that under these conditions the body was on the point of moving, determine the magnitude of 𝐹 and the normal reaction 𝑅.
Let’s begin by breaking down the information we’ve actually been given. The body has a weight of 262.5 newtons. And of course, that mustn’t be confused with the mass of the object. The weight is essentially the downwards force that the body exerts on the plane. Initially, the body is resting on a rough plane, which tells us that we’re going to need to consider the frictional force. And that plane is inclined to the horizontal at an angle whose tangent is three-quarters. Let’s let that angle be equal to 𝛼 so that tan of 𝛼 is three-quarters. And we’re told at this stage the body is on the point of moving.
So let’s add the body to our diagram and add any extra forces we’re interested in. We said that the downwards force that the body exerts on the plane is 262.5 newtons. There must be a normal reaction force of the plane on the body. Let’s call that 𝑅. Now, we know that the body is on the point of moving. There are no other forces, so we assume that it’s on the point of sliding down the slope. What’s keeping the body from sliding down the plane is the frictional force. And this will act against the direction in which the body is trying to move. So it’s acting upward and parallel to the plane as shown. We’re then told that the body is placed on a horizontal surface of the same roughness.
So how do we measure roughness? The measure of roughness is the coefficient of friction, which we defined to be equal to 𝜇. And so what we’re going to begin with doing is looking at our first scenario where the body is resting on the inclined plane and seeing if we can work out the value of 𝜇. Now to do this, we’re going to need to use the equation for friction. It’s friction is equal to 𝜇𝑅, where 𝑅 is the normal reaction force and we’ve seen that 𝜇 is the coefficient of friction. Now, since the body is on the point of moving, we can say that the vector sum of the forces must in fact be equal to zero. But then we can go ahead and split these forces into their parallel and perpendicular components when considering this with respect to the plane.
Let’s begin by thinking about the forces that act perpendicular to the plane. When we’re dealing with friction, this is a really sensible starting point cause it helps us work out our value for 𝑅. We have 𝑅 acting upwards and away from the plane. But what force acts downwards and perpendicular to the plane? We need to consider the component of the weight. And so we add a right-angled triangle, as shown. The included angle here is 𝛼. We want to find the adjacent side in this triangle since that’s perpendicular to the plane. And currently, we know that the hypotenuse itself is 262.5 newtons. And so we’re going to use the cosine ratio. Remember, this links the angle, the adjacent side, and the hypotenuse.
Since cos of 𝜃 is adjacent over hypotenuse, here we can say cos of 𝛼 is the unknown side divided by 262.5. But what is cos of 𝛼? Well, we know that tan of 𝛼 is three-quarters. And of course, the tangent ratio is the opposite side divided by the adjacent. So we can draw a right-angled triangle with an included angle of 𝛼, an opposite side of three units, and adjacent of four. Then by the Pythagorean triple, the other side must be five units. And so we can replace cos of 𝛼 with four-fifths. Let’s also define the side of the triangle that we’re trying to find to be 𝑥. And we’re now going to solve this equation for 𝑥. To do so, we multiplied both sides of this equation by 262.5. And so 𝑥 is equal to 210 or 210 newtons.
Resolving forces perpendicular to the plane then and taking the direction in which the reaction force is acting to be positive, the net sum of the forces acting perpendicular to the plane is 𝑅 minus 210. And of course, we know this is equal to zero. If we add 210 to both sides, we get 𝑅 is equal to 210 or 210 newtons. Now we do need to be careful because the normal reaction force 𝑅 will change when the body is on a horizontal surface. This is just our intermediate step. Now, what we can do is resolve the forces that act parallel to the plane to work out the value of 𝜇. We have friction acting up the plane, and then we need to consider the component of the weight that acts parallel to the plane. Now that’s the opposite side in that right-angled triangle we drew. And I’ve labeled that as 𝑦.
Since 𝑦 is the opposite side, this time, we’re going to use the sine ratio. Sine is opposite over hypotenuse. So sin 𝛼 here is 𝑦 over 262.5. But of course, once again, we can use our right-angled triangle to replace sin of 𝛼. This time, if we look at that triangle we drew, we see it’s equal to three-fifths. So now our equation is three-fifths equals 𝑦 over 262.5. We’re now going to multiply through by 262.5. And when we do, we get 𝑦 is 262.5 times three-fifths, which is 157.5 newtons. This is acting against the direction in which friction is. So the net sum of our forces parallel to the plane is friction minus 157.5. And of course, this is equal to zero.
But remember, we said friction is equal to 𝜇𝑅. And we just calculated the normal reaction force here to be 210. So we can replace friction with 𝜇 times 210 or 210𝜇. And we get 210𝜇 minus 157.5 equals zero. To solve, we begin by adding 157.5 to both sides. And finally, we’re going to divide by 210. 157.5 divided by 210 is three-quarters. So we find the coefficient of friction of the plane to be equal to three-quarters. And that’s great because we generally are expecting a value between zero and one for 𝜇. Now that we know the roughness of the plane, we’re going to clear some space and consider the second part of this question.
This time, the body is on a horizontal surface. But the downward force of the weight on the plane is exactly the same. We now have a force 𝐹 acting on the body, and it pulls it upward at an angle of 𝜃. We’re told, in fact, that sin of 𝜃 is equal to three-fifths. We saw earlier that when this was the case, cos of 𝜃 was four-fifths. That’s going to be useful in a moment. So let’s go back to the object on the plane. There’s one other force that we need to consider. And once again, that’s the normal reaction 𝑅. Now I’ve used the same letter 𝑅. But this isn’t to be confused with the reaction force we saw earlier. We’re now going to once again resolve forces perpendicular to the plane.
We have the reaction force acting upwards. Let’s take that to be the positive direction. Then in the opposite direction, we have 262.5 newtons. So we’re going to subtract that. We are going to need to consider the component of 𝐹 which acts perpendicular to the plane. So let’s add this right-angled triangle in. Since we’re looking to find the opposite, let’s call that 𝑥. And we already have an expression at least for the hypotenuse. We’re going to use the sine ratio. This time, we get sin of 𝜃 equals 𝑥 over 𝐹. But of course, we know sin 𝜃 is equal to three-fifths. So we get three-fifths is equal to 𝑥 over 𝐹, meaning that the component of the 𝐹 force that acts perpendicular to the plane is three-fifths 𝐹.
We’re going to add this to our expression because we said that this was the positive direction. And we know that it’s on the point of moving. So the vector sum of these forces is zero. We might in fact rearrange to make 𝑅 the subject. We could leave it as it is, but we’ll see why that might be useful in a moment. We’re now going to consider the forces parallel to the plane. We have 𝐹, the force acting on the body, pulling it upwards, acting in one direction. But of course, we know that this is a rough plane. So there’s the frictional force, and that is 𝜇𝑅. So let’s find the adjacent side of this triangle to find the component of 𝐹 that’s parallel to the plane. And since we have the adjacent and we have an expression for the hypotenuse, we’ll use the cosine ratio.
We get cos 𝜃 is equal to 𝑦 over 𝐹 and then we replace cos 𝜃 with four-fifths. And if we multiply both sides of this equation by 𝐹, we find the component of this force that acts parallel to the plane is four-fifths 𝐹. Let’s take this to be the positive direction, and then we’re going to subtract the frictional force. And of course, this is equal to zero. Now, we’re actually trying to find the magnitude of 𝐹 and the normal reaction 𝑅. So what we’re going to do is we’re going to replace 𝑅 in our new equation with this expression, 262.5 minus three-fifths 𝐹. That will give us an equation purely in terms of 𝐹.
At the same time, we replace 𝜇 with three-quarters, that’s the coefficient of friction we calculated earlier, and our equation becomes four-fifths 𝐹 minus three-quarters times 260.5 minus three-fifths 𝐹 equals zero. When we distribute these parentheses, this simplifies to five-quarters 𝐹 minus 196.875 equals zero. And so we add 196.875 to both sides. And finally, we’re going to divide by five-quarters. And that gives us 157.5 or 157.5 newtons.
And that’s great. We know the magnitude of 𝐹, and so we can use this to find the normal reaction 𝑅. We’re going to substitute this into this earlier equation. When we do, we get 𝑅 is 262.5 minus three-fifths times 157.5, which is 168 or 168 newtons. And so we’re done. The magnitude of 𝐹 is 157.5 newtons and the normal reaction 𝑅 is 168 newtons.