Question

Horizontal beam AB is 200 kg, 2.4 m long, and is welded at point A. The man is 80 kg and applies a tension of 300 N on the cable. Diameter of the pulley is 300 mm and BC = 300 mm. Determine:
(a) horizontal and vertical components of force at A,
(b) magnitude and direction of the moment supported at A.

Answer 1

Answer:

$a)-3346.8\;N \\ b)-4937.04\;N-m$

Explanation:

a) - In Free-body diagram :

At point D, the free body diagram of a man :

$D = Fn\\Fn=T+mg\\put\;values\;in\;it\\ .\;\;\;\;=300+(80)(9.81)=1084.8\;N$

$Mg=200\times9.81=1962\;N$

$\sum Fx=0\; where\; Ax=0$

$\sum Fy=0\; where\; Ay-Mg-Fn-T=0$

Then, put the value in the equation.

$Ay=3346.8\;N$

b)-

$Ma=Mg(AE)+Fn(AD)+T=4937.04\;N-m$