A body weighing 200 newtons rests on a rough plane inclined at an angle of 30 degrees to the horizontal. The body is pulled upward by a string making an angle of 30 degrees to the line of greatest slope of the plane. When the tension in the string is 80 root three newtons, the body is on the point of moving up the plane. Find the coefficient of friction between the body and the plane.
We’re looking to find the coefficient of friction between the body and the plane; that’s 𝜇. Now, there’s quite a lot of information here, so we’re going to begin by sketching a diagram. Firstly, we’re told that the body rests on a plane which is inclined at an angle of 30 degrees to the horizontal. The body weighs 200 newtons. In other words, the downward force that the body exerts on the plane is 200 newtons. And, of course, we know this means there is a reaction force of the plane on the body. This force acts perpendicular to and away from the plane.
We’re then told that the body is being pulled upward by a string which makes an angle of 30 degrees to the line of greatest slope of the plane. Now, we’ve added in that line. And we’ve added a force 𝑇 for tension since, really, it’s the tensional force that’s pulling this upwards. In fact, we’re told a little bit later on that the tension in the string is 80 root three newtons. And at that point, the body is on the point of moving up the plane.
Now, there is one force that hasn’t been added to our diagram. The body rests on a rough plane, so there’s a frictional force. That force acts against the direction in which the body is trying to move. So, we’ll assume at the moment that it acts parallel to the plane and in the opposite direction to the tensional force. Our job now is to resolve forces parallel and perpendicular to the plane.
We usually begin by resolving forces perpendicular to the plane. This will give us a value or an expression for 𝑅. Now, we know that the body is resting on the plane. For this to be the case, the forces acting upward on the body and perpendicular to the plane must be equal to the forces acting downward and perpendicular to the plane. Another way of thinking about this is that the vector sum of the forces must be equal to zero. Now, the problem that we have is that the force of the weight does not act either parallel or perpendicular to the plane. It acts at an angle to it. And so, we add in this right-angled triangle.
We’re looking to find the component of the weight that acts perpendicular to the plane. Let’s call that side of the triangle 𝑥. This side is adjacent to the included angle, so let’s use the cosine ratio. cos of 𝜃 is adjacent over hypotenuse, so here cos of 30 is 𝑥 over 200. We multiply through by 200. And we find 𝑥 is equal to 200 times cos of 30. And since cos of 30 is root three over two, this becomes 100 root three or 100 root three newtons.
Now, we’re going to need to repeat this process for the component of the tensional force that acts perpendicular to the plane. I’ve added a right-angled triangle and labelled that 𝑦. This time, 𝑦 is the opposite side in this triangle. And we know the hypotenuse is 80 root three. And we use the sine ratio. sin of 30 is 𝑦 over 80 root three. And so, we multiply through by 80 root three. So, 𝑦 is 80 root three sin of 30, which is 40 root three or 40 root three newtons.
Now, remember, we said that the vector sum of the forces must be zero. In the direction perpendicular to the plane then, the forces acting upwards must be equal to the forces acting downwards. That is, 𝑅 plus 40 root three, that’s the component of the tension that acts perpendicular to the plane, must be equal to the component for the weight that acts perpendicular to the plane. So, that’s 100 root three. And if we subtract 40 root three from both sides of our equation, we find 𝑅 is equal to 60 root three or 60 root three newtons. And that’s great! We’re now ready to work in a direction parallel to the plane.
Now, the key to resolving forces parallel to the plane is to realize that the body is in the point of moving. In other words, it’s in something that’s called limiting equilibrium. This means that frictional force will be at its absolute maximum, but that the overall force — that is, the forces acting parallel and up the plane — will be equal to the forces acting parallel and down the plane.
Once again, we’re going to need to consider the component of the tension and the weight that acts parallel to the plane. This time, we use the cosine ratio for the tensional force. And we get that it’s equal to 80 root three cos 30, which is 120 newtons. Similarly, the component of the weight that acts parallel to the plane is 200 sin 30, which is 100 newtons. And so, the component of the tension that pulls the body upwards is 120 or 120 newtons.
We have 100 newtons acting in the opposite direction. But remember, we also have the frictional force. And this is where the value of 𝑅 comes in. We know that friction is 𝜇𝑅. It’s the coefficient of friction multiplied by the reaction force. Now, we calculated 𝑅 to be equal to 60 root three, so we get 120 equals 100 plus 𝜇 times 60 root three. We subtract 100 from both sides so that 20 is 60 root three 𝜇. And then, we divide through by 60 root three.
Now, 20 over 60 simplifies to one-third. So, we have so far that 𝜇 is one over three root three. Remember, where possible, we want to rationalize that denominator. And so, we’re going to multiply the numerator and denominator of our fraction by root three. And that gives us root three over nine. And we remember that a good way to check whether we’ve calculated the correct value for the coefficient of friction is to make sure that it is between zero and one as root three over nine is.