Video Transcript
A body weighing 25 newtons rests on a rough plane inclined to the horizontal at an angle whose cosine is four-fifths. The coefficient of friction between the body and the plane is one-fifth. A force of magnitude 𝐹 acts on the body, holding it in a state of equilibrium. Given that the force acts up the line of greatest slope of the plane, what can we know for certain about the value of 𝐹?
There’s rather a lot of information here. So we’re going to begin by drawing a sketch. We’re given a rough plane that’s inclined to the horizontal, as shown. We have a body resting on this plane with a weight of 25 newtons. That means the body exerts a downward force on the plane of 25 newtons. There’s a reaction force, a force of the plane on the body. This acts perpendicular to the plane as shown.
We’re told that the plane is inclined at an angle whose cosine is four-fifths. So let’s call that angle 𝛼. And we could say cos of 𝛼 is equal to four-fifths. And in fact, this allows us to work out the value of sin of 𝛼 too. If we construct a smaller right-angled triangle with an included angle of 𝛼, and we know that cos of 𝛼 is adjacent over hypotenuse, we can mark the hypotenuse as five units and the adjacent as four units.
The Pythagorean triple three squared plus four squared equals five squared tells us that the third side in this triangle must be three units. That’s the side opposite the included angle. Since sin of 𝛼 is opposite over hypotenuse, in this case, sin of 𝛼 must be equal to three over five or three-fifths. We’re very rarely interested in the value of tan of 𝛼. So we’ll get rid of our right-angled triangle.
We’re told that the coefficient of friction between the body and the plane, remember we define that as 𝜇, is equal to one-fifth. And then we have this force of magnitude 𝐹 acting on the body. We’re told that it acts up the line of greatest slope of the plane, so in this direction. But we’re also told that it acts in a state of equilibrium. So what does this mean?
Well, for a body to be in equilibrium, the vector sum of all the forces acting upon it must be equal to zero, in terms of this body which acts on a rough plane inclined to the horizontal. We can say that the sum of the forces acting perpendicular to the plane must be equal to zero. And the sum of the forces acting parallel to the plane must also be equal to zero.
And so we need to consider a further force acting on the body. And that’s the force of friction. There are two options here. If we call the frictional force 𝐹𝑟, if we assume the body is in limiting equilibrium, that is, it’s on the point of moving, it could be moving up the plane. In this case, the frictional force will be acting in the opposite direction. So it will be acting down the plane. Alternatively, it could be on the point of moving down the plane. If this is the case, the frictional force acts up the plane, acts in the same direction of 𝐹. And so we’ll consider these two scenarios.
Before we do though, we’re going to resolve forces perpendicular to the plane. This will allow us to find the value for 𝑅. Now, of course, the force of the weight of the body doesn’t act in either the parallel or perpendicular direction. And so we split it into its two components by adding a right-angled triangle. Note that the included angle is also 𝛼. The component of the force that acts perpendicular to the plane is the measurement of the adjacent side in our triangle.
And so we’re going to use the cosine ratio because, of course, we know the value of the hypotenuse. It’s 25 newtons. So we can say cos 𝛼 is adjacent over hypotenuse. Let’s replace cos of 𝛼 with four-fifths, because we were told that the cosine of the angle is four-fifths, and the hypotenuse with 25.
We can find the value of the adjacent side in this triangle by multiplying through by 25. When we do, we see that the component of the weight that acts perpendicular to the plane is four-fifths times 25, which is 20 or 20 newtons. We’re now able to resolve forces perpendicular to the plane. We know the vector sum of the forces is equal to zero because it’s in a state of equilibrium. In this case then, if we assume the force 𝑅 to be acting in the positive direction, we can say that 𝑅 minus 20 must be equal to zero. We’ll solve for 𝑅 by adding 20 to both sides. And when we do, we find 𝑅 is equal to 20 or 20 newtons.
We’ve now finished resolving forces perpendicular to the plane. So we’re going to move on to resolving parallel to the plane. We’re going to continue working on the assumption that the object is about to move up the plane. And so friction is acting in the opposite direction to 𝐹.
We’ll begin a little bit like we did when resolving perpendicular to the plane. We’ll find the component of the weight that acts in this direction, in the direction parallel to the plane. This is the opposite side in our triangle. We know the value of the hypotenuse. So we can use the sine ratio. sin 𝛼 is opposite over hypotenuse. And this is why we calculated the value of sin 𝛼. We can now replace sin 𝛼 with three-fifths and the hypotenuse with 25.
Let’s multiply through by 25 to find the value of the opposite side. When we do, we find that the component of the weight that acts parallel to the plane is three-fifths times 25, which is 15 or 15 newtons. We’ll add this to the diagram. And now we’re ready to resolve these forces parallel to the plane. The body is in equilibrium. So the sum of these forces is equal to zero.
We’ll take 𝐹 to be acting in the positive direction. And then we see friction, and the component of the weight that acts parallel to the plane is acting in the opposite direction. So the resultant force is 𝐹 minus 15 minus 𝐹𝑟 for friction. And that, of course, is equal to zero. But remember, friction is 𝜇𝑅. That’s the coefficient of friction times the reaction force. We can therefore replace 𝐹𝑟 with 𝜇 times 𝑅. That’s a fifth times 20.
Remember, we were told 𝜇, the coefficient of friction, is equal to one-fifth. And we calculated 𝑟 right at the beginning. And so 𝐹 minus 15 minus a fifth times 20 is zero. A fifth times 20 is four. And so our equation simplifies to 𝐹 minus 19 equals zero. We add 19 to both sides. And we find that in our first scenario, 𝐹 is equal to 19 or 19 newtons.
Let’s now move on to resolving forces parallel to the plane in our second scenario. That is, when the object is on the point of moving down the plane. So friction is acting in the same direction as 𝐹. Most of the forces remain unchanged. But since friction is acting in the same direction as 𝐹, we make it positive. So our overall force is 𝐹 minus 15 plus friction. That is, 𝐹 minus 15 plus a fifth times 20 is zero. This time, our equation is 𝐹 minus 15 plus four is zero or 𝐹 minus 11 is zero. We add 11 to both sides. And this time, we find 𝐹 is equal to 11.
And so we have two scenarios. When the body is in limiting equilibrium, that is, it’s on the point of either sliding up or down the plane, it can be as little as 11 newtons and as big as 19 newtons. Of course, the body might not be on the point of moving. It might just be in equilibrium. For this to be the case, 𝐹 can be anywhere between 11 and 19 newtons. We know for certain that the value of 𝐹 is greater than or equal to 11 newtons and less than or equal to 19 newtons.