4: 4-5

Problem: Determine the moment about point B of each of the three forces acting on the beam

Variable Solution:

Givens: 

F1 = 375 [lb]

F2 = 500 [lb]

F3 = 160 [lb]

a = 8 [ft]

b = 6 [ft]

c = 5 [ft]

d = 0.5 [ft] 

α = 30

Numerical Solution:

M_BF1 = F1(b+c) → (375 [lb])(6 [ft] + 5 [ft]) 

M_BF1 = 4125 [lb⋅ft] CCW

M_BF2 = cF2cos(tan^-1(3/4)) → (5 [ft])(500 [lb])cos(tan^-1(3/4)) 

M_BF2 = 2000 [lb⋅ft] CCW

M_BF3 = dF3sinα → (0.5 [ft])(160 [lb])sin(30) 

M_BF3 = 40 [lb⋅ft] CCW