# Video: Equilibrium of Bodies Resting on Rough Inclined Planes

A body weighing 195 N is on a rough plane that is inclined at an angle of 45° to the horizontal. If the coefficient of friction between the body and the plane is (√(3))/(3), which of the following is true about the body? [A] The body is in a stable state of equilibrium. [B] The body is sliding down the plane. [C] The body is on the point of sliding down the plane.

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### Video Transcript

A body weighing 195 newtons is on a rough plane that is inclined at an angle of 45 degrees to the horizontal. If the coefficient of friction between the body and the plane is root three over three, which of the following is true about the body? Is it (A) the body is in a stable state of equilibrium, (B) the body is sliding down the plane, or (C) the body is on the point of sliding down the plane?

And let’s begin by simply sketching a diagram of all of our forces. Here is our plane, which makes an angle of 45 degrees to the horizontal. A body rests on the plane that weighs 195 newtons. That means the force that acts vertically downwards on the plane is 195 newtons. We know this means that there will be a reaction force. The reaction force acts perpendicular to the plane in the opposite direction to the weight of the body.

Now, we’re told that the body rests on a rough plane. This means there is a frictional force that’s trying to stop the body from moving down the plane; let’s call that 𝐹r. But of course, we know that friction is equal to 𝜇𝑅, where 𝜇 is the coefficient of friction and 𝑅 is the reaction force of the plane on the body perpendicular to the plane. In this case, we know 𝜇 is equal to root three over three. Our job is going to be to find 𝑅. And to do so, we add in this right-angled triangle. By including this triangle, we’ll be able to calculate the components for the weight that act parallel to the plane and perpendicular to the plane.

To decide whether the body is sliding down the plane or on the point of sliding down the plane or, in fact, in equilibrium, we need to establish whether the force of the weight that acts parallel to the plane is greater than or equal to the force of friction which acts in the opposite direction. So, let’s resolve perpendicular to the plane to find 𝑅.

We know that the body isn’t jumping off the plane. And so, this means that the sum total of the forces acting perpendicular to the plane is zero. Alternatively, we can say that the reaction force is equal to the force of the weight that acts perpendicular to the plane. But what is that force? Well, let’s label the part of the right-angled triangle that we’re interested in; let’s label it 𝑥. Then, we can see we can use right angle trigonometry to find the value of 𝑥. 𝑥 is the side adjacent to the included angle of 45 degrees, whereas 𝐻 is the hypotenuse. We can, therefore, use the cos ratio.

Remember, cos 𝜃 is adjacent over hypotenuse. And we can say that cos of 45 is 𝑥 over 195. We rearrange by multiplying both sides by 195. And we find 𝑥 is 195 cos of 45. This acts in the opposite direction to the reaction force. So, we can say that 𝑅 minus 195 cos of 45 equals zero. We can add 195 cos of 45 to both sides of our equation. So, 𝑅 is 195 cos of 45 or 195 root two over two. And of course, this is newtons.

We now know what 𝑅 is. So, next, we resolve parallel to the plane. Let’s begin by calculating the size of the frictional force. Remember that’s 𝜇𝑅. We know that 𝜇 is the coefficient of friction. It’s root three over three, and 𝑅 is 195 root two over two. Correct to one decimal place, that gives us 79.6 newtons.

Our next job is to consider the component of the weight that acts parallel to the slope. If we go back to our right-angled triangle, that’s 𝑦. This time, that’s the side opposite the included angle. By using the sine ratio, we see that sin of 45 equals 𝑦 over 195. 𝑦 is equal to 195 sin 45 then. This time, correct to one decimal place, we get 137.9 newtons. It should be quite clear to us that 137.9 newtons is greater than 79.6 newtons. This means the force acting down and parallel to the plane is greater than the force acting up and parallel to the plane.

Since this is the case, the body can’t be in equilibrium or even at the point of sliding down the plane. It must be moving in the direction of the greater force; it must be sliding down the plane. So, the answer is (B) the body is sliding down the plane.