A scaffold of mass 60 kg and length 5.0 m is supported in a horizontal position by a vertical cable at each end. Its center of g
ravity is at the middle of the scaffold. A window washer of mass 80 kg stands at a point 1.5 m from one end. What is the tension in (a) the nearer cable and (b) the farther cable
1 answer:
Answer:
a) 842.8 N
b) 529.2 N
Explanation:
Given
Mass of scaffold, m = 60 kg
Length of scaffold, l = 5 m
Mass of window washer, m(w) = 80 kg
Distance of the window washer, d = 1.5 m
From the attached image
a = 1.5 m
b = 2.5 m
c = 5 m
W = 60 * 9.8 = 588 N
W(w) = 80 * 9.8 = 784 N
Since the scaffold is in equilibrium, then
Στ(b) = 0
W * (c - b) + W(w) * (c - a) - Lc = 0
Lc = W * (c - b) + W(w) * (c - a)
L = [W * (c - b) + W(w) * (c - a)] / c
L = [588 * (5 - 2.5) + 784 * (5 - 1.5)] / 5
L = (588 * 2.5 + 784 * 3.5) / 5
L = (1470 + 2744) / 5
L = 4214 / 5
L = 842.8 N
Στ(a) = 0
Rc - W * b - W(w) * a = 0
R = [W * b + W(w) * a] / c
R = (588 * 2.5 + 784 * 1.5) / 5
R = (1470 + 1176) / 5
R = 2646 / 5
R = 529.2 N
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