Calculate the magnitude of the force supported by the point at A under the action of the 2.5-kN load applied to the bracket. Neglect friction in the slot.
AB = .17 m
B to force = .16 m
angle between vertical and BtoForce = 24
The Attempt at a Solution
I found the x and y distance of the force from B
y = .16 cos 24 = .1462
x = .16 sin 24 = .0651
I tried multiplying the force by the y distance but it was incorrect.
I also found the distance from A to the force which is:
√((.17 + .0651)2 + (.1462)2) = .2768
and multiplied that by the force but that was also incorrect. Any ideas?
Any help appreciated!