You are asked to hang a uniform beam and sign using a cable that has a breaking

Question

You are asked to hang a uniform beam and sign using a cable that has a breaking strength of 368 N. The store owner desires that it hang out over the sidewalk as shown. The sign has a weight of 170 N and the beam's weight is 47 N. The beam's length is 1.30 m and the sign's dimensions are 1.00 m horizontally 0.80 m vertically. What is the minimum angle ? that you can have between the beam and cable? (Please work out numerically so I can follow the numbers) http://www.webassign.net/grr/p8-37.gif (for picture)

Explanation / Answer

ANS: The centre of mass of the sign is 0.5m. from the end of the beam. The centre of mass of the beam is its centre, 0.75m. Thus the normal force of the beam will be 1/2 its weight, = 28.5N. on each end. The sign CM is 0.5m. from the cable attachment point, so is 1m. from the wall. 217/ ((1/.5) +1) = 72.33N. is applied normally to the wall, and 144.67N. normally to end of the beam. Total normal weight on end of beam = (144.67 + 28.5) = 173.17N. Now, sketch a triangle. The "height" is the normal weight, 173.17N. We know the hypotenuse (the cable) cannot exceed 363N. Sin L = (173.17/ 363), so angle = 28.5 degrees. I must explain the derivation of the 217/ ((1/.5) + 1) line. 1: 0.5 is the ratio of the position of the CM of the sign to the wall and to the beam end, as it is 1m. from the wall, and 0.5m. from the end. (1/.5) = 2:1 ratio of distances. Because this is 3 "parts" of the beam (2 on one side of the CM, 1 on the other), you add the 2 and 1 together, making 3. You divide the normal force of the sign by 3. (217/3) = 72.33N. is the normal applied to the wall end, and the remainder (144.67N) is the normal weight proportion at the end of the beam. This, plus the normal of the beam itself, is the total of the normal weight at the end of the beam

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