Originally Posted by **osalselaka** I tried to solve the problem.
I took the force which is exert by the nail on the hammer as Fn ,
The perpendicular distance to the nail force (Fn) from the single point of contact-5sin60
when we take the torque on the single point of contact,
150*30=5sin60*Fn
150*30=5*√3/2*Fn
Fn=1800/√3
Am I correct? |

Are you?

You didn't comment on my suggestion.

Consider horizontal force equilibrium.

Does F = the horizontal component of Fn,

Which gives a different equation from your torque equation.

There must be one more force acting that you have not considered, can you tell me what it is?