A traffic light hangs from a pole. The uniform aluminum pole AB is 7.50 m long a

Question

A traffic light hangs from a pole. The uniform aluminum pole AB is 7.50 m long and has a mass of 12.0 kg. The mass of the traffic light is 21.5 kg. Determine (a) the tension in the horizontal massless cable CD, and (b) the vertical and horizontal components of the force exerted by the pivot A on the aluminum pole.

Further information, pole AC is 3.80m and the angle formed from the meeting of cable CD and pole AB is 37 degrees. I would greatly appreciate any help in order for me to get started. Not sure which direction to go in this one.

Explanation / Answer

I started by summing the torques at the hinge. There are three: T(light)=5*9.8*7.5*cos(37) T(cable)=-tension*6.31*sin(37) T(pole)=8*97.5/2)*cos(37) These all sum to zero. I computed tension=140.5 The sum of the horizontal forces must be zero. These are the tension in the cable and the horizontal reaction force at the hing. So the horizontal component is 140.5 The vertical forces are the weights of the pole and the light offset by the vertical reaction force at the hinge the vertical component is (8+5)*9.8 =127.4 I checked the answer by computing the magnitude of the reaction force two ways. Using Pythagorean theorem sqrt(140.5^2+127.4^2) =189 I then summed the forces parallel to the pole: 5*9.8*sin(37) 8*9.8*sin(37) 140.5*cos(37) =189 Please rate!

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