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?

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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T1 = 131.4 [N]

T2 = 261 [N]

Explanation:

To solve this problem we must make a sketch of how will be the semicircle, for this reason we conducted an internet search, to find the scheme of the problem. This scheme is attached in the first image.

Then we make a free body diagram, with this free body diagram, we raise the forces that act on the body. Since it is a problem involving static equilibrium, the sum of forces in any direction and moments must be equal to zero.

By performing a sum of forces on the Y axis equal to zero we can find an equation that relates the forces of tension T1 & T2.

The second equation can be determined by summing moments equal to zero, around the point of application of the T1 force. In this way we find the T2 force.

The value of T2, is replaced in the first equation and we can find the value for T1.

Therefore

T1 = 131.4 [N]

T2 = 261 [N]

The free body diagram and the developed equations can be seen in the second attached image.

50 inches should be the resultant vector

its c i just checked it