A rock of mass 10 kg sits on a slope which makes an angle of 25 degrees with the
horizontal. If it does not slip down what is the magnitude of the frictional force?

Question

anonymous · Answer

The weight of the rock is 98 N, so the component that is parallel to the slope would be
$$98*\sin(25degrees)\approx41.4N$$
Thus the friction force must also be 41.4 N  (or maybe just 41, because of significant figures)

anonymous · Answer

ok cool, thats what i ended up with, i just wasn't sure if it'd be that simple.