Question

A 1220-N uniform beam is hinged at a vertical wall at one end and is supported by a cable that is attached to the beam at the other end. A 200-kg load hangs at the end of the beam. The lower end of the beam makes an angle of 30° above the horizontal, while the supporting cable is attached to the wall at an angle of 40°, as shown.

(a) Calculate the magnitude of the tension in the supporting cable.

(b) Determine the vertical and horizontal components of the reaction force exerted on the beam at the hinge.

[(a) T = 2260 N; (b) 1450 N, vertically; 1450 N, horizontally]

Answer 1

Consider the following -

T = tension in the cable
Fx = horizontal reaction force of the wall on the lower pin
Fy = vertical reaction force of the wall on the lower pin
L = length of the beam

Also assume that -

up is positive
right is positive
clockwise is positive moment

sum forces in the horizontal direction to zero
so,
Fx - Tcos(90-40) = 0

=> Fx = Tcos50

sum forces in the vertical direction to zero

Fy + Tsin50 - 1220 - 200*9.8 = 0
Fy = 3180 - Tsin50

sum moments about the cable to beam attachment point to zero

Fy(Lcos30) - Fx(Lsin30) - 1220((Lcos30)/2) = 0
Fy(Lcos30) = Fx(Lsin30) + 1220((Lcos30)/2)
Fy(cos30) = Fx(sin30) + 1220(cos30)/2
Fy(cos30) = Fx(sin30) + 610(cos30)
(3180 - Tsin50)(cos30) = Tcos50(sin30) + 610(cos30)
3180cos30 - Tsin50cos30 = Tcos50sin30 + 610cos30
3180cos30 - 610cos30 = Tcos50sin30 + Tsin50cos30
2570cos30 = T(cos50sin30 + sin50cos30)
T = 2570cos30 / (cos50sin30 + sin50cos30)

=> T = 2226 / (0.321 + 0.663) = 2262 N

(b) Horizontal component of the reaction force , Fx = Tcos50

= 2262*0.643 = 1453 N

(c) Vertical componet, Fy = 3180 - Tsin50

=> Fy = 3180 - 2262*cos50
=> Fy = 1726 N

I think the answer of Fy is not matching with the values, you have mentioned. Please check your text book for any correction.