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3 years ago

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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

nt of that force about the line joining points A and D.

1 answer:

Vikki [24]3 years ago

3 0

Answer:

M_AD = 1359.17 lb-in

Explanation:

Given:

- T_ef = 46 lb

Find:

- Moment of that force T_ef about the line joining points A and D.

Solution:

- Find the position of point E:

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

BE / BC = 45 / 60 = 0.75

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

- Find unit vector 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 >

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

- Find unit vector 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)

$M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right]$

M_AD = 1835.73 + 116.49528 - 593.0568

M_AD = 1359.17 lb-in

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