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 gravity 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

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

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down

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