You open a restaurant and hope to entice customers by hanging out a sign (see th

Question

You open a restaurant and hope to entice customers by hanging out a sign (see the figure (Figure 1) ). The uniform horizontal beam supporting the sign is 1.60 mlong, has a mass of 19.0 kg , and is hinged to the wall. The sign itself is uniform with a mass of 25.0 kg and over-all length of 1.20 m . The two wires supporting the sign are each 37.0 cm long, are 95.0 cm apart, and are equally spaced from the middle of the sign. The cable supporting the beam is 2.00 m long.

Part A

What minimum tension must your cable be able to support without having your sign come crashing down?

Part B

What minimum vertical force must the hinge be able to support without pulling out of the wall?

Explanation / Answer

The cable is 2m. long, the beam is 1.6m.
The angle between the beam and the cable is therefore

arccos(1.6/2) = 36.86 degrees.
The torque about the hinge for the beam alone = (19 x g) x (1.6/2) = 148.96N/m.
The 2 wires supporting the sign are 0.95 metres apart, and the beam is 1.6 metres long with 1 of them attached to the end of the beam.
The CM of the sign is therefore 1.6 - (0.95/2) = 1.125m. from the hinge.
The torque due to the sign = 1.125 x (25 x g) = 275.625N/m.
Total torque = (275.625 + 148.96) = 424.585N/m.
Vertical force component at beam end = (424.585/1.6m) = 265.365N.
a) Tension in cable = (265.365/sin36.86) = 442.3779N.
b) The torque on the hinge about the cable attachment due to the beam mass = 148.96N/m.
The torque due to the sign about the cable attachment = (25 x g) x 0.475m = 116.375N/m.
Total torque at hinge = (116.375 + 148.96) = 265.335N/m.
Vertical force on hinge minimum = (265.335/1.6m) = 165.834375N.

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