A mass, m, is resting on a slope being slowly tilted upwards from horizontal. The static friction co-efficient is 0.3 and the dynamic friction co-efficient is 0.2: at what angle will the mass begin to slip?

The static friction force holding the object on the slope is given by Fr where R = mg by Newton's second law of motion. We use a < sign as the static friction is a reaction force And the force dragging it down the slope is the component of the weight parrallel to the slope given by W(par)=mgsin(theta) While horizontal, the mass will not slip down the slope as sin(0)=0 so W(par)=0 - there is no force acting in this direction. The mass will slip when the component of the weight acting down the slope exceeds the force available from friction. So the angle we are looking for is when: Fr = W(par) Thus we re-write the equations above as: u(s)mgcos(theta) = mgsin(theta) if we divide by cos(theta) and move all the constants to the other side we have: tan(theta) = u(s)mg/(mg) the mg cancels out and we are left with: theta = tan^-1(u(s)) theta=11.3 degrees The dynamic friction will limit the movement of the mass once it moves from rest, and so the same calculation needs to be repeated with the dynamic friction co-effictient to check that it is lower. However, since the co-efficient is smaller we know that this is uneccessary and the final answer is 11.3 degrees