FBD of a cube moving with uniform velocity on a rough inclined plane

Question

I am a highschool student and I came across this physics numerical in which I had to calculate the torque due to the normal reaction force across the centre of a cube moving with uniform velocity on a rough inclined plane.

This torque had to be equal to mgsin∅ *a/2 where a = length of edge of cube since the torque by gravitational force is zero and the only force other than friction to counteract the toque due to friction is the normal reaction force. This logic fit well until I tried drawing a free body diagram of the forces acting on this cube.

I have always drawn the normal reaction force through its centre of mass without really thinking as to why I do so. If I do so in this case, it should not be able to produce any torque about its centre! From which point exactly do the friction force and the normal reaction force act ? I think the normal reaction force and friction force should be from the middle of the bottom-most edge of the cube perpendicular and parallel to the inclined plane respectively.

Can someone provide a proper FBD for this case and reasoning as to why we generally draw normal reaction force from centre of the cube generally but in this case, it isn't from the centre ?

Answer 1

The contact force (both normal and frictional components) do not act at a point on the base. They are distributed along the base of the cube, with higher force per unit area toward one side of the base and lower force per unit area toward the other side. If you are taking moments about the center of mass, the distribution of the frictional force does not affect the moment of the frictional force, but, for the normal force distribution, it does. One way of taking this into account is to add a couple to the resultant contact force on the surface. Another way is to say that the contact force distribution is effectively applied at a point not along the normal line from the center of mass to the side of the cube.