contestada

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.

Respuesta :

Manetho

Answer:

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

Otras preguntas