A body weighing 68 kilogram-weight rests on a rough horizontal plane. When a horizontal force of 59.5 kilogram-weight acts on it, it is on the point of moving. Determine the coefficient of friction between the body and the plane.
We will begin by drawing a free-body diagram and adding on the forces from the question. We are told that the body has a weight of 68 kilogram-weight. As this acts vertically downwards, using Newton’s third law, we know there is a normal reaction force acting vertically upwards. We are told there is a horizontal force of 59.5 kilogram-weight acting on the body. And since the body is at rest on a rough horizontal plane, there will be a frictional force acting in the opposite direction to this. As the body is on the point of moving, we know it is in equilibrium. And this means that the sum of the forces in the horizontal and vertical directions will equal zero.
Resolving vertically, we have 𝑅 minus 68 is equal to zero, where the positive direction is vertically upwards. Adding 68 to both sides of this equation, we have 𝑅 equals 68. The normal reaction force is therefore equal to 68 kilogram-weight. Resolving horizontally, we have 59.5 minus the friction force 𝐹 𝑟 equals zero. If we add this friction force to both sides of our equation, we have 𝐹 𝑟 is equal to 59.5. The friction force acting on the body is 59.5 kilogram-weight. The maximum frictional force between a body and a rough surface satisfies the equation 𝐹 𝑟 is equal to 𝜇 multiplied by 𝑅, where 𝜇 is the coefficient of friction. This occurs when the body is on the point of moving.
Substituting in our values, we have 59.5 is equal to 𝜇 multiplied by 68. We can then divide both sides of this equation by 68. This simplifies to seven-eighths. If the 68-kilogram-weight body is on the point of moving, the coefficient of friction between the body and the plane is seven-eighths.