A 175-kg utility pole is used to support at C the end of an electric wire. The tension in the wire is 600 N, and the wire forms an angle of 15° with the horizontal at C. Determine the largest and smallest allowable tensions in the guy cable BD if the magnitude of the couple at A may not exceed 500 N·m.

T_max = 2240N and T_min = 1522N

Explanation:

The component of the cable tension (T) perpendicular to the pole = 600 cos (15°) ≈ 580 N

The couple due to the cable is therefore (580 N*4.5 m) = 2608 N. m

The allowable couple due to the guy cable is therefore (2608 ± 500 N. m)

Between 2108 N. m and 3108 N. m

The component of the guy cable (Ts) perpendicular to the pole must therefore be between:

(2108 / 3.6) = 586 N and (3108 / 3.6) = 863 N

tan (θ) = (3.6 / 1.5) = 2.4

θ = 67.4°

after solving the calculation of regular equation, we get

therefore, T_max = 2240 N and T_min = 1522 N