# Video: Finding the Force That Makes a Body on a Rough Inclined Plane on the Point of Moving

A body weighing 407 N was resting on a rough plane inclined at an angle of 30° to the horizontal in limiting equilibrium. The angle of inclination of the plane was increased to 45° and a force 𝐹 was applied to the body acting parallel to the line of greatest slope of the plane. As a result of the action of the force, the body was on the point of moving up the plane. Find the magnitude of the force 𝐹 rounding your answer to the nearest newton if necessary.

07:25

### Video Transcript

A body weighing 407 newtons was resting on a rough plane inclined at an angle of 30 degrees to the horizontal in limiting equilibrium. The angle of inclination of the plane was increased to 45 degrees. And a force 𝐹 was applied to the body acting parallel to the line of greatest slope of the plane. As a result of the action of the force, the body was on the point of moving up the plane. Find the magnitude of the force 𝐹 rounding your answer to the nearest newton if necessary.

There’s an awful lot of information here. So we’re going to split it into two parts. The first bit of information we’re going to use is about the body resting on the plane inclined at an angle of 30 degrees. We’ll resolve forces in this scenario both parallel and perpendicular to the plane. In doing so, this will give us a value of 𝜇. It will tell us the coefficient or friction of our plane. We’ll then carry the information about the coefficient of the plane forward to help us resolve forces in the second scenario. Let’s begin by sketching the first scenario out.

We’re told that the weight of the body is 407 newtons. That means the force that the body exerts on the plane in a downward direction is 407 newtons. We know this means there is a reaction force that acts in the opposite direction perpendicular to the plane. Let’s call that 𝑅. Now, in this first scenario, the body is in limiting equilibrium. Essentially, it’s on the point of sliding. Now, in this case, it will be on the point of sliding down the slope. What’s stopping it sliding down the slope is the friction. Remember, it’s resting on a rough plane. We know that friction acts in the opposite direction to that which the body wants to move. And so, in this case, friction acts up the plane.

Now, we said we’re going to resolve forces parallel and perpendicular to our plane. The problem is the force that represents the weight of the body is not acting in either of those directions. And so we add in a right-angled triangle on this force. The included angle in this right-angled triangle is 30 degrees. We’ll label the two sides of our triangle 𝑥 and 𝑦, where 𝑥 is the component of the weight that acts perpendicular to the plane. Whereas 𝑦 is the component of the weight that acts parallel to the plane. We’re also going to use at some point the fact that friction is equal to 𝜇𝑅, where 𝜇 is the coefficient of friction and 𝑅 is that reaction force.

Now, generally, we begin by resolving forces perpendicular to the plane. This will allow us to either find a value or an expression for 𝑅. Since the body is in equilibrium, we know that the vector sum total of its forces is equal to zero. In other words, 𝑅 minus the component of the weight that acts perpendicular to the plane must be equal to zero. Now, of course, an alternative way of saying this is that 𝑅 must be equal to the component of the weight that acts perpendicular to the plane. 𝑅 must be equal to 𝑥.

But of course, this is a right-angled triangle, so we can use right-angle trigonometry to find the value of 𝑥. 𝑥 is the side of our triangle that’s adjacent to the included angle. And the hypotenuse is 407 newtons. This means we can use the cosine ratio, cos of 30 equals 𝑥 over 407. By multiplying through by 407, we find that 𝑅 is equal to 407 cos 30, which is 407 root three over two. Notice, we now have a value for 𝑅. And we’re ready to resolve forces parallel to the plane. Remember, we said that the body is in limiting equilibrium. This means that the forces acting up and parallel to the plane must be exactly equal to those acting down and parallel to the plane.

Well, we have friction that acts up the plane. Then we have the component of the weight that acts parallel to the plane which acts in the opposite direction. So friction must be equal to 𝑦. Remember, we could have alternatively said that friction minus 𝑦 equals zero. This time, though, we can use the right-angled triangle to find 𝑦. And we get sin 30 equals 𝑦 over 407. So 𝑦 must be equal to 407 sin 30, which is 407 over two. Now, remember, we said friction is equal to 𝜇𝑅. We replace 𝑅 with the value that we calculated as 407 root three over two. And we get 𝜇 times 407 root three over two equals 407 over two.

We notice we can divide through by a common factor of 407 over two so that root three 𝜇 equals one. We then divide through by root three to solve for 𝜇. One over root three is equivalent to root three over three. And we found our value for 𝜇. Remember, 𝜇, the coefficient of friction, does need to be between zero and one. One over root three or root three over three is indeed within this range. So that’s a good indication that we’re on the right track. And now that we’ve calculated our value for 𝜇, let’s go on to scenario two, where the angle of inclination of the plane is increased.

The downward force of the body on the plane is still 407 newtons. There’s still a reaction force that acts perpendicular to the plane. But because the angle of the plane has changed, this reaction force will also have changed. We’re told now that a force 𝐹 is applied to the body that acts parallel to the line of greatest slope of the plane. And as a result of the action of this force, the body is on the point of moving up the plane. This must mean that the force is acting up the plane. And so the friction is acting in the opposite direction; it’s acting down the plane.

Using a similar technique to earlier, we add in the components of the weight that act parallel and perpendicular to the plane. They are 407 sin 45 and 407 cos 45, respectively. Let’s now resolve, once again, perpendicular to the plane. We know it’s in equilibrium in this direction. So 𝑅 is 407 cos of 45. That’s 407 root two over two. Next, we resolve parallel to the plane again. Remember, it’s still in limiting equilibrium. But this time it’s in the point of moving up the plane. And so the sum total of the forces acting parallel to the plane is zero. Alternatively, we can say that this means that the forces that act up the plane must be exactly equal to those that act down the plane.

Well, we have our force 𝐹 acting up the plane. Then we have friction and the component of the weight that acts parallel to the plane. So 𝐹 is equal to friction plus 407 sin 45. Remember, though, we know friction is 𝜇𝑅 and we calculated 𝜇, the coefficient of friction, to be root three over three. So friction becomes root three over three multiplied by 407 root two over two. 407 sin 45 is also 407 root two over two. All that’s left is to calculate the value of 𝐹. It’s 453.94 and so on. But the question tells us to round our answer to the nearest newton if necessary. And when we do, we get 454 newtons. And so we see that the magnitude of the force 𝐹 is 454 newtons.