Video Transcript
A body weighing 47 newtons rests on a rough horizontal plane. Two horizontal forces of one newton and four newtons act on the body, causing it to be on the point of moving. Given that the angle between the lines of action of the two forces is 60 degrees, find the coefficient of static friction between the body and the plane.
Since the body is on the point of moving, we know it is in equilibrium. And this means that the sum of the forces in both the horizontal and vertical directions equal zero. We also know that the frictional force 𝐹 r is at its maximum when the body is on the point of moving. When this is the case, it is equal to 𝜇 multiplied by 𝑁, where 𝜇 is the coefficient of friction and 𝑁 is the normal reaction force. This is sometimes referred to as 𝑅. However, it will become clear while we have not done so in this question later.
We will begin by sketching the side view of the scenario in this question. We are told that the body weighs 47 newtons, so this force will act vertically downwards. By considering Newton’s third law, the normal reaction force 𝑁 will act in the opposite direction, in this question vertically upwards. As the surface is rough, there will be a frictional force 𝐹 r as already mentioned. Finally, we are told there are two horizontal forces of one newton and four newtons acting on the body.
We might be tempted to combine these so that a five-newton force acts on the body to the right, as shown in the diagram. However, this is not the case. We are told that the angle between the lines of action of the two forces is 60 degrees. In order to understand what this looks like, we will consider a top or bird’s eye view of the body. Looking down from above, the forces would look as shown. We can then create a parallelogram to help us calculate the resultant of these two forces 𝑅. It will be this force that acts to the right on our side view diagram.
Using our knowledge of supplementary angles which sum to 180 degrees, we have a triangle with side lengths 𝑅, four, and one. And the angle between the sides of length one and four is 120 degrees. To calculate the side length 𝑅, we can use the cosine rule. This states that 𝑎 squared is equal to 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 multiplied by the cos of angle 𝐴. Substituting the values from our triangle, we have 𝑅 squared is equal to one squared plus four squared minus two multiplied by one multiplied by four multiplied by the cos of 120 degrees.
The cos of 120 degrees is negative one-half. This means that the right-hand side simplifies to one plus 16 plus four. 𝑅 squared is therefore equal to 21. We can then take the square root of both sides of this equation. And since 𝑅 must be positive, 𝑅 is equal to root 21. The resultant of the one-newton and four-newton forces that act on the body causing it to be on the point of moving is root 21 newtons.
We will now return to our side view diagram and resolve vertically and horizontally. As already mentioned, the sum of the forces in these directions equals zero. Resolving vertically, where the positive direction is vertically upwards, the sum of our forces is 𝑁 minus 47. As this is equal to zero, we can calculate 𝑁 by adding 47 to both sides. The normal reaction force 𝑁 is therefore equal to 47 newtons.
Next, we can resolve horizontally. This time, we will let the positive direction be to the right. We have the equation root 21 minus 𝐹 r equals zero. This time, adding the frictional force to both sides of our equation, we see that this is equal to root 21 newtons. We are now in a position to calculate the coefficient of friction as required. Since 𝐹 r is equal to 𝜇𝑁, then 𝜇 is equal to 𝐹 r divided by 𝑁. The coefficient of friction is equal to the maximum frictional force divided by the normal reaction force, giving us a final answer of root 21 over 47. This is the coefficient of static friction between the body and the plane.
