Video: Finding the Friction Force Acting on a Body on a Rough Horizontal Plane Where the Body Is about to Move

A body weighing 25.5 N rests on a rough horizontal plane. A horizontal force acts on the body causing it to be on the point of moving. Given that the coefficient of friction between the body and the plane is 3/17, determine the magnitude of the force.

02:00

Video Transcript

A body weighing 25.5 newtons rests on a rough horizontal plane. A horizontal force acts on the body, causing it to be on the point of moving. Given that the coefficient of friction between the body and the plane is three seventeenths, determine the magnitude of the force.

We can begin by sketching a diagram. We know that the body weighs 25.5 newtons. There will be a normal reaction force going vertically upwards of 𝑅 newtons. We will call the horizontal force that is applied to the body 𝐹. As the horizontal plane is rough, there will be a frictional force 𝐹 r newtons acting in the opposite direction to this. We are told that the body is on the point of moving, which means it is in limiting equilibrium. This tells us that the sum of the vertical and horizontal forces equals zero. Resolving vertically, the forces up are equal to the forces down. This means that 𝑅 is equal to 25.5. The normal reaction force is equal to 25.5 newtons. Resolving horizontally, the force 𝐹 is equal to the frictional force.

We are also told that the coefficient of friction is equal to three seventeenths. In limiting equilibrium, the frictional force is equal to the coefficient of friction multiplied by the reaction force. We can therefore calculate the frictional force by multiplying three seventeenths by 25.5. This is equal to 4.5 newtons. As the horizontal force acting on the body is equal to this, we know that this is equal to 4.5 newtons.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.