Question

A traffic light hangs from a pole. The uniform aluminum pole AB is 7.3m long and has a mass of 10kg. The mass of the traffic light is 22.5kg.

A. Determine the tension in the horizontal massless cable. CD
B. Determine the vertical component of the force exerted by the pivot A on the aluminum pole.
C. Determine the horizontal component of the force exerted by the pivot A on the aluminum pole.

37° 3.80 m

Answer 1

Solution: 37° 3.80 m 9 F2 F1 First things are first: The tension from the cable is not acting at the tip of the pole, but only partially up the pole and before the hanging stop light. The distance that the cable connects to the hanging pole is the hypotenuese of a triangle which has a side of 3.8 m and opposite angle of 37 degrees, so the length of this hypotenuse is 3.80m l sin(37) - 3.80m -- 6.3772 sin(37)0.3m If we start by considering Newton's 2nd Law for angular rotation, and take our rotation through the base A, then we get T Tsin(37) (6.3m) (22.5kg) (g) cos(53) (7.3m) - 10.0kg) (g) cos (53) (7.3m/2) 0 We show the force of the weight of the pole acting from its center of mass (middle) since it is uniform. We can solve this T to find that T 312.6N Next we sum the forces in the x y directions, and since the system is static, these sums are zero, in which case: an

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