Question

In a city park a nonuniform wooden beam 5.00 m long is suspended horizontally by a light steel cable at each end. The cable at the left-hand end makes an angle of 30.0° with the vertical and has tension 620 N. The cable at the right-hand end of the beam makes an angle of 50.0° with the vertical.
As an employee of the Parks and Recreation Department, you are asked to find the weight of the beam and location of its center of gravity.

Answer 1

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