Question

A 3kg bar is laying on a smooth vertical wall of 3L of height and a floor of 2L. Determine the force that the wall exerts over the bar.

I have tried to get to length of the bar by using Pitagoras but no, then I tried using a free body diagram but I don't know the direction of the forces of the wall and floor.

Can someone help me?

Answer 1

well the length is L sqrt 13

but who cares?
force up from floor = m g = 3g
(because wall is smooth none up there)
force sideways from wall = F
thus force sideways at floor = F toward wall

Now moments about center
clockwise mg * 2L/2 = m g L
counter clockwise F*1.5 L + F*1.5 L
= 3 F L
so
m g L = 3 F L
F = m g /3 = 3 kg * 9.81 m/s^2 /3
= 9.81 Newtons