A uniform ladder rests against a vertical wall where there is negligible friction. The bottom of the ladder rests on rough ground where there is friction. The top of the ladder is at height h above the ground and the foot of the ladder is at a distance 2a from the wall. The diagram below shows the forces which act on the ladder. Which equation is formed by taking moments?
A) Wa+Fh=2Wa
B) Fa+Wa= Fh
C) Wa+2Wa=Fh
D) Wa-2Wa=2Fh

Question

A uniform ladder rests against a vertical wall where there is negligible friction. The bottom of the ladder rests on rough ground where there is friction. The top of the ladder is at height h above the ground and the foot of the ladder is at a distance 2a from the wall. The diagram below shows the forces which act on the ladder. Which equation is formed by taking moments? 
A) Wa+Fh=2Wa 
B) Fa+Wa= Fh 
C) Wa+2Wa=Fh  
D) Wa-2Wa=2Fh

anonymous · Answer

This is takes from cambridge as level physics code 9702/01/M/J/08

vincent-lyon.fr · Answer

You have not added the diagram, but I think moments about origin O would lead to equation A)
|dw:1401007861049:dw|

anonymous · Answer

@Vincent-Lyon.Fr thanks, althought i still couldn't get it

vincent-lyon.fr · Answer

At equilibrium, you must have:

clockwise torque = anticlockwise torque

F and W act clockwise, whereas –W acts anticolckwise