A body weighing five newtons rests on a rough horizontal plane. The coefficient of friction between the body and the plane is three-quarters. Given that 𝐹 is the magnitude of the friction force measured in newtons, express the range of its possible values as an interval.
Note firstly that we’ve been given the weight, not the mass of the body. That is, the force that the body exerts on the plane is five newtons directly downwards. Now, according to Newton’s third law of motion, the plane itself exerts a force equal in magnitude and opposite in direction on the body. We could call that 𝑅, the reaction force. But, of course, since the body is at rest, we can say that this is equal to five newtons.
Horizontally, we have two forces acting on the body. We have a force acting forwards — I’ve called that 𝐹 one. And then we have the force acting backwards. And that’s 𝐹; it’s the magnitude of the friction force. And we know friction is equal to 𝜇𝑅, where 𝜇 is the coefficient of friction and 𝑅 is the reaction force.
In order to solve this problem, the keyword we’re looking for is the fact that the body is resting on the rough horizontal plane. Let’s begin by imagining that the body itself is on the point of moving. At this point, the forward force is exactly equal to the frictional force. 𝐹 one is equal to 𝐹. Now, of course, we know that friction is equal to the coefficient of friction multiplied by the reaction force.
So at the point that the body is ready to move, 𝐹 and 𝐹 one themself must be equal to three-quarters times five, which is equal to 15 over four. If 𝐹 one is larger than this, then the body itself will begin moving. Of course, if no force is applied at all, there’s no way that the body can move. So the lowest value of 𝐹 one and therefore 𝐹 would be zero. And we can say that the range of all possible values of 𝐹 is therefore the closed interval from zero to 15 over four.