you want to hang a sign under the following conditions. A 40 kg beam is attached

Question

you want to hang a sign under the following conditions. A 40 kg beam is attached to a wall by a hinge. The beam is 1.5 m long. You want the beam to be set at an angle of 35 degrees from the vertical wall. the 50 kg sign hangs from the end of the beam by a wire. to keep the beam and sign in place, you attach another wire from the wall to the beam. this wire will be horizontal.

1-what is the closest wire can be to the hinge such that the tension in the wiredoes not go above 1500 N?

2-what is the magnitude of thr force of the hinge on the beam and in what direction does it point?

Explanation / Answer

Part A)

Solve by the sum of torques about the hinge...

The sign torque = mg(cos 55)(1.5) - clockwise

The weight of the beam torque = (mg)(cos 55)(.75) - clockwise

The torque from the horizontal wire = Tr(cos 35) - counterclockwise

Set the clockwise equal to the counterclockwise

(50)(9.8)(cos 55)(1.5) + (40)(9.8)(cos 55)(.75) = 1500r(cos 35)

r = .480 m (48 cm from the hinge)

Part B)

In the y direction, the force is 50(9.8) + 40(9.8) = 882 N

In the x direction the force is 1500 N

The net is from the Pythagorean Theorem

F2 = (882)2 + (1500)2

F = 1740 N

The direction is from the tangent function

tan(angle) = 882/1500

angle = 30.5o up and to the right above the x axis (or North of East)

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