A beam AB whose weigh is M and is located at an angle of α from the wall and is hinged at point A to the wall, a cable is connected to the wall and the beam end at point B as shown in the figure. Find the direction and the value of the force acting on hinge A and the tension in the cable CB.
From the free body diagram, we can write the equilibrium equations in the x and y directions:
x direction: 
F_{x} − T sinθ = 0 
(1) 
y direction: 
F_{y} + T cosθ − mg − M g = 0 
(2) 
Another condition that should be fulfilled is that the sum of the moments at the hinge A is ΣM_{A} = 0
If α + θ is not 90 degree then we have to calculate the projection of T that is perpendicular to the beam which is T sin(α + θ)
Moments at A:
m g L sinα + M g sinα L / 2 − T sin(α + θ) L = 0 
(3) 
from eq. (3) we can find the value of T.
 (4) 
Substitute eq. (4) into eq. (1) and (2) we get the forces F_{x} and F_{y} : 
 (5) 
 (6) 
Once we found F_{x} and F_{y} we can find the reaction force F and the angle of the reaction γ on hinge A.
If α + θ = 90 degree then T F_{x} and F_{y} are:
The weight of a uniform horizontal beam is 10kgf a weight of m = 500kgf is hanged on point B . A cable holds the end of the beam to the wall at an angle of 60 degree. Find the reaction force on hinge A and its direction.
(run example).
Because α + θ = 150 degree we have to use eq. (4) (5) and (6).
Given the weight W which is equal to m g = W
Tension is: 
T = sin 90 (500 + 10 / 2)/cos(90 + 60 − 90) = 1010 kgf 
F_{x} is: 
F_{x} = sin90 sin60 (500 + 10 / 2) / sin150 = 874.7 kgf 
F_{y} is: 
F_{y} = 510 − sin90 cos60 (500 + 10 / 2) / sin150 = 5 kgf 
γ is: 
γ = arctan (874.7 / 5) = 89.7 degree 
