here,
L = 3 m
weight of beam , W1 = 200 N
W2 = 300 N
theta = 30 degree
maximum tension , T = 500 N
a)
taking moment of force about point A
W1 * l/2 + W2 * x - T * sin(theta) * l = 0
200 * 3/2 + 300 * x - 500 * sin(30) * 3 = 0
solving for x
x = 1.5 m
the weight is placed at a maximum distance of 1.5 m from the point A
b)
the horizontal component of reaction force , Fx = T * cos(theta)
Fx = 500 * cos(30) = 433 N
c)
equating the forces vertically
the vertical component , Fy = W1 + W2 - T * sin(theta)
Fy = 300 + 200 - 500 * sin(30)
Fy = 250 N
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