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

Respuesta :

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)

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

                            M_AD = 1835.73 + 116.49528 - 593.0568

                            M_AD = 1359.17 lb-in

The moment of the force about the line joining points A and D is; 617.949 lb.in

What is the moment of the force?

We are given;

Force exerted by cable EF at E; T_EF = 46 lb.

From the diagram of the guy wire, we can draw a triangle and we will have the following coordinates;

A(0, 0, 0)

D(48, -12, 36)

E(E_x, 96, E_z)

Also, we can get that;

BC² = 48² + 36²

BC = √(48² + 36²)

BC = 60 in

Also, from similar triangles, we will have the coordinate of E as;

E(36, 96, 27)

Position of Vector of EF is;

EF = {(21 - 36)i + (-14 - 96)j + (57 - 27)k} in

EF = {-15i - 110j + 30k} in

Magnitude of EF from online calculation = 115 in

Force along cable EF is;

F_EF = 46{(-15i - 110j + 30k)/115}

F_EF = {-6i - 44j + 12k} lb

Position vector of AE is {36i + 96j + 27k} in

Position vector of AD is {48i - 12j + 36k} in

Magnitude of AD = 61.188 N

Unit vector of AD; λ_ad = {48i - 12j + 36k}/61.188

λ_ad = 0.7845i - 0.1961j + 0.5883k

M_ad = λ_ad × r_ea × T_EF

M_ad = [tex]\left[\begin{array}{ccc}0.7845&-0.1961&0.5883\\36&96&27\\6&-44&12\end{array}\right][/tex]

Solving this gives;

M_ad = 617.949 lb.in

Read more about moment of a force at; https://brainly.com/question/25329636

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