# Video: Finding the Coefficient of Friction between an Inclined Plane and a Resting Body

A body weighing 60 N rests on a rough plane inclined to the horizontal at an angle whose sine is 3/5. The body is pulled upward by a force of 63 N acting parallel to the line of the greatest slope. Given that the body is on the point of moving up the plane, find the coefficient of friction between the body and the plane.

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### Video Transcript

A body weighing 60 newtons rests on a rough plane inclined to the horizontal at an angle whose sin is three-fifths. The body is pulled upward by a force of 63 newtons acting parallel to the line of the greatest slope. Given that the body is on the point of moving up the plane, find the coefficient of friction between the body and the plane.

Now, there’s quite a lot of information in this question. So we’ll simply begin by sketching a diagram. We have a plane inclined to the horizontal at an angle whose sin is three-fifths. In other words, if we label this angle 𝜃, sin of 𝜃 is three-fifths. A body resting on the plane has a weight of 60 newtons. In other words, it exerts a downward force on the plane of 60 newtons. And of course, we know that this means there is a reaction force of the plane on the body that acts perpendicular to the plane. We’re told that the body is being pulled upward by a force of 63 newtons acting parallel to the line of greatest slope that’s shown.

The body is on the point of moving up the plane. That means there’s a frictional force acting in the opposite direction to the direction this wants to move. Now, because the body is on the point of moving, when we think about the forces parallel to the plane, we know that their sum is zero. In other words, the forces that act parallel to and up the plane are equal to the forces of that parallel and down the plane. Once we’ve considered all the relevant forces, our next job is to look at the forces perpendicular and parallel to the slope. Whenever we’re interested in friction, we tend to begin by considering the forces perpendicular to the slope. This will allow us to either calculate a value or an expression for 𝑅.

We know, perpendicular to the slope, the body is in equilibrium. And so the force 𝑅 must be equal to the component of the weight that acts perpendicular to the slope. And so we need to add in this right-angled triangle, the force we’re interested in as 𝑥. And we know since the body is in equilibrium in this direction, 𝑅 is equal to 𝑥. But what is 𝑥? Well, we notice we have a right-angled triangle. 𝑥 is adjacent to the included angle, and we know that the hypotenuse is 60 newtons. We can therefore use the cosine ratio, such that cos of 𝜃 is adjacent over hypotenuse, 𝑥 over 60. If we multiply through by 60, we see that 𝑥 is 60 cos 𝜃. And therefore, 𝑅 itself must also be 60 cos 𝜃.

Now, we can in fact find the exact value of cos 𝜃. We were told that sin 𝜃 is three-fifths. Now, we know that sin 𝜃 is opposite over hypotenuse. So if we add a little right-angled triangle, the side opposite to the angle 𝜃 could be three units and the side on the hypotenuse would be five units. Using the Pythagorean triple, three squared plus four squared equals five squared, we see that the third side in this triangle is four units. And this means that cos 𝜃 which is adjacent over hypotenuse must be four-fifths. 𝑅 is therefore 60 multiplied by four-fifths which is 48 newtons. Now, we know the magnitude of 𝑅. We’re ready to resolve forces parallel to the slope.

Remember, the body is on the point of moving up the plane. So the forces parallel and acting up the plane must be exactly equal to the forces acting parallel to and down the plane. We have 63 newtons pulling the body up the plane. We said that there is a frictional force acting in the opposite direction. And we also need to consider the component of the weight that’s parallel to the plane. Let’s call that 𝑦 this time. Now, 𝑦 itself is the side opposite the included angle 𝜃 on our diagram. So we use the sine ratio, such that sin 𝜃 is 𝑦 over 60.

By multiplying through by 60, we see that 𝑦 is equal to 60 sin 𝜃. And so we have an equation 63 is equal to friction plus 60 sin 𝜃. But what is friction? Well, we say that friction is 𝜇𝑅, where 𝜇 is the coefficient of friction and 𝑅 is the reaction force between the plane and the body. Well, we know that the reaction force is 48 newtons. So the frictional force is 𝜇 times 48 or 48𝜇.

Similarly, we saw earlier that sin 𝜃 is three-fifths. So our equation becomes 63 equals 48𝜇 plus 60 times three-fifths. 60 multiplied by three-fifths is 36. And then we’ll begin to solve for 𝜇 by subtracting 36 from both sides, so that 48𝜇 is 27. And then finally, we divide by 48. And we get 𝜇, the coefficient of friction, which is what we were looking to calculate, to be equal to nine sixteenths.