A long thin uniform rod lies flat on the table as shown. One end of the rod is slowly pulled up by a force that remains perpendicular to the rod at all times. The rod is to be brought to the vertical position without any slipping of the bottom end? The coefficient of static friction between rod and ground can be:

Step-by-step solution

Since the rod is raised slowly (quasistatically), the entire system remains in equilibrium at any moment in time. Thus, the torque about the point of rotation of the rod is zero about any axis, and the net external forces are 0 . F _ fric = F =-F I+ m g l 2 =0 F= m g 2 N=m g-F The F force causes the rod's bottom end to slip in one direction. The friction force counters this force to stop the slipping when F_ fric = _s N F . Hence, _ s F mg - F = 2- ^2 Maximum of s can be found by setting its derivative with respect