(Q) A uniform 100N ladder of length 5m rests on a rough floor for which the co-efficient of static friction is 0.8 and leans against a smooth wall. Determine the force P that must be applied at the center of such a rod in order to make it move.
All the equations of equilibrium including moments.
The Attempt at a Solution
The reaction at the bottom has to be 100N as well since there is no other verticle force. Thus, the frictional force will be 80N. Taking moments about the upper tip of the ladder yields:
P(2) - 100(1.5) -80(4) +100(3) =0.
Therefore, P = 85. However, the answer at the back is 75. Please help me with this. I don't know where I am going wrong.