A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the mome

Answer:

M_AD = 1359.17 lb-in

Explanation:

Given:

- T_ef = 46 lb

Find:

- The moment of the force T_ef relative to the line connecting points A and D.

Solution:

- Determine the coordinates of point E:

                  mag(BC) = sqrt (48^2 + 36^2) = 60 in

                  BE / BC = 45 / 60 = 0.75

This means,           E = < 0.75*48, 96, 36*0.75> = < 36, 96, 27 > in

- Calculate the unit vector for EF:

                  mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in

                  vec(EF) = < -15, -110, 30 >

                  unit(EF) = (1/115) * < -15, -110, 30 >

- The tension on           T_EF = (46/115) * < -15, -110, 30 > = < -6, -44, 12 > lb

- Calculate the unit vector for AD:

                  mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in

                  vec(AD) = < 48, -12, 36 >

                  unit(AD) = (1/12*sqrt(26)) * < 48, -12, 36 >

                  unit(AD) = <0.7845, -0.19612, 0.58835 >

Next:

                  M_AD = unit(AD). ( E x T_EF)

                  M_AD = 1835.73 + 116.49528 - 593.0568

                  M_AD = 1359.17 lb-in