Answer:
Explanation:
System of forces in balance
ΣFx = 0
ΣFy = 0
∑MA = 0
MA = F*d
Where:
∑MA : Algebraic sum of moments in the the point (A)
MA : moment in the point A ( N*m)
F : Force ( N)
d : Horizontal distance of the force to the point A ( N*m
Forces acting on the beam
T₁ = 620 N : Tension in cable 1 ,at angle of 30° with the vertical on the left
T₂ : Tension in cable 2, at angle of 50.0° with the vertical on the right.
W : Weight of the beam
x-y T₁ and T₂ components
T₁x= 620*sin30° = 310 N
T₁y= 620*cos30° = 536.94 N
T₂x= T₂*sin50°
T₂y= T₂*cos50°
Calculation of the T₂
ΣFx = 0
T₂x-T₁x = 0
T₂x=T₁x
T₂*sin50° = 310 N
T₂ = 310 N /sin50°
T₂ = 404.67 N
Calculation of the W
ΣFy = 0
T₂y+T₁y-W = 0
(404.67) *cos50° + 536.94 = W
W= 260.12+ 536.94
W= 797.06 N
Location of the center of gravity of the beam
∑MA = 0 , point (A) (point where the cable 2 of the right is located on the beam)
T₁y(5)-W(d) = 0
T₁y(5) = W(d)
d = T₁y(5)/W
d = 536.94(5) / 797.06
d = 3.37m
The center of gravity is located at 3.37m measured from the right end of the beam
