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Problem: Determine the magnitude of force F in cable AB in order to produce a moment of 500 lb * ft about the hinged axis CD, which is needed to hold the panel in the position shown.

Variable Solution:

Givens: 
a = 4 [ft]
b = 6 [ft]
c = 6 [ft]
M_CD = 500 [lb⋅ft]

Numerical Solution:
F = M_CD⋅(4a²+b²)^1/2⋅(csin(tan^-1((4b²+a²)^1/2/c)))^-1⋅(2acos(tan^-1(a/(2b)))-bsin(tan^-1(a/(2b))))^-1
→(500 [lb⋅ft])⋅(4(4 [ft])²+(6 [ft])²)^1/2⋅((6 [ft])sin(tan^-1((4(6 [ft])²+(4 [ft])²)^1/2/(6 [ft]))))^-1⋅(2(4 [ft])cos(tan^-1((4 [ft])/(2(6 [ft]))))-(6 [ft])sin(tan^-1((4 [ft])/(2(6 [ft])))))^-1
F = 162.0 [lb]