You open a restaurant and hope to entice customers by hanging out a sign (Figure 1)....

You open a restaurant and hope to entice customers by hanging out a sign (Figure 1). The uniform horizontal beam supporting the sign is 1.60 m long, has a mass of 20.0 kg , and is hinged to the wall. The sign itself is uniform with a mass of 33.0 kg and overall length of 1.20 m . The two wires supporting the sign are each 36.0 cm long, are 90.0 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?

Expert Solution

Solution:-

The cable is 2.0m. long,

And the beam is 1.6m.
Now let us find the angle between the beam and the cable arccos (1.6/2.0) = 36.87 degrees.
The torque about the hinge for the beam alone = (19 x 9.8) x (1.6/2) = 148.96N/m.
From the question, we have 2 wires supporting the sign are 0.9 meters apart, and the beam is 1.6 meters long with 1 of them attached to the end of the beam.
The CM of the sign is therefore 1.6 - (0.9/2) = 1.15m. from the hinge.
The torque due to the sign = 1.15 x (33 x 9.8) = 371.91N/m.
Total torque = (371.91 + 148.96) = 520.87 N/m.
Vertical force component at beam end = (520.87/1.6m) = 325.54N.
a) Tension in cable = (325.54/sin 36.87 ) = 542.56N.
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 = (33 x 9.8) x 0.45m = 145.53N/m.
Total torque at hinge = (145.53 + 148.96) = 294.49N/m.
Vertical force on hinge minimum = (294.49/1.6m) = 184.06N.

genius_generous answered 2 years ago

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