Video Transcript
The figure shows a body of mass 30 kilograms being pushed against a rough vertical wall by a horizontal force 𝐹. Given that the coefficient of static friction between the body and the wall is five-sixths, determine the minimal horizontal force 𝐹 that will cause the body to be in equilibrium.
We recall that if the body is in equilibrium, the sum of the net forces will equal zero. We will begin by adding our forces to the diagram. As the body has mass 30 kilograms, there will be a 30 kilogram-weight force acting vertically downwards. We would often convert this into newtons by multiplying the mass by gravity. However, in this case, we’ll keep the force in terms of kilogram-weight. There will be a frictional force, 𝐹 r, acting vertically upwards as this will prevent the body from sliding down the wall. Finally, we will have a normal reaction force acting perpendicular to the wall.
If we resolve vertically with the positive direction being upwards, the sum of the net forces is 𝐹 r minus 30. Adding 30 to both sides of this equation gives us a frictional force equal to 30 kilogram weights. Resolving horizontally, we have the equation 𝐹 minus 𝑅 is equal to zero. This means that the force 𝐹 must be equal to the normal reaction force. We know that the frictional force is equal to 𝜇, the coefficient of friction, multiplied by the normal reaction force. In this question, we know that 𝜇 is equal to five-sixths. Rearranging this equation by dividing by 𝜇, we see that the reaction force is equal to the frictional force divided by 𝜇. We need to divide 30 by five-sixths.
Dividing by a fraction is the same as multiplying by the reciprocal of the fraction. This means that we need to multiply 30 by six-fifths or six over five. We can divide the numerator and denominator by five. This leaves us with six multiplied by six, which is equal to 36. As the normal reaction force is equal to 36 kilogram-weight, the minimum horizontal force 𝐹 that will keep the body in equilibrium is 36 kilogram-weight.
