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? It is not 28.5 degrees, I've tried that. (Please work out with numbers so I may be able to follow the logic.)

Explanation / Answer

I should ask: Where is the sign's Center of gravity? It makes a huge difference. Sign against the building and the CG is 0.5 from the building Sign centered on the beam and the CG is 0.75 from the building This one===> Sign at the end of the beam and the CG is 1.0 from building Sign some place else? Which problem should I solve? Draw the FBD. The attachment to the building is a pin on a roller with a force up for the reaction force at the building. This forces no moment at the building. OOPS: Where is the cable attached to the beam? Lets press on and maybe you can fill in the rest. Else, where is the cable attached? Sum moments about the pin at the wall Beam: Mo = Force * Distance = 50 N * 0.75 m = 37.5 N-m Sign: Mo = 200N * 1 m = 200 N-m Total CCW moment = 200 + 37.5 = 237.5 N-m We now need a CW moment (Vertical component of the tension in the cable) = 237.5 N-m We put the cable at distance "L" from the building L * vertical component of tension in cable = 237.5 N-m L*V = 237,5 N-m V = 237.5 N-m / L Now it is all Trig sine A = V/T = V/417 (We know V) ===> A = arcsin V/417

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