Physics, 01.04.2020 07:28 eliwas555

A 40-kg uniform semicircular sign 1.6 m in diameter is supported by two wires as shown. What
is the tension in each of the wires supporting the sign?

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Answer from: electrofy456

The tension in each of the wires supporting the sign are; T1 = 130.67 and T2 = 261.33 N

From online sources, the length from the right hand wire to the right end of the sign is 0.4 m.

Let the tension in the wire on the left be T1 and let the tension in the wire on the right be T2. These two tension forces are acting upwards.

Now, the sign will have a weight(W) acting at the centre in the downward direction.

Now, we know from equilibrium that;

Sum of downward forces = Sum of upward forces.

Thus;

T1 + T2 = W - - - (eq 1)

Taking moments about the left hand wire gives;

W(1.6)/2 = T2 × 1.2

Where W = mg = 40 × 9.8

W = 392 N

Thus;

392 × 0.8 = T2(1.2)

T2 = (392 × 0.8)/1.2

T2 = 261.33 N

Thus, Plugging the values into eq 1 gives;

T1 + 261.33 = 392 N

T1 = 392 - 261.33

T1 = 130.67

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A 40-kg uniform semicircular sign 1.6 m in diameter is supported by two wires as shown. What
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