What’s the horizontal force exerted by the wall on rod XY?
First, I calculate the length of ZX and I get 0.288675m. I choose Z as pivot and apply the equation clockwise moment equal to anti-clockwise moment which is 0.288675 x F cos 30 = 75(0.5) and I get 150N for the answer. However, the answer is 130N. Anyone knows how to solve it?
- Wayne DeguManLv 73 months agoFavorite Answer
Resolving horizontally in the direction XY we have:
Tcos30 = X....where T is the tension in rod ZY and X is the horizontal force exerted on the rod XY
Resolving vertically at Y we have:
Tsin30 = 75
so, T = 75/sin30 => 150 N
Hence, X = 150cos30 => 129.9 N
or, 130 N...to 3 s.f.