COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 59.
Free-Body Diagram:
Geometry:
Triangle ABC is isosceles. Thus distance CD = l cos
θ
2
,
Elongation of spring is equal to distance AB:
x = 2l sin
θ
2
,
and T = kx = 2kl sin
(a)
θ
2
.
Equilibrium for rod:
ΣM C = 0:
θ
P ( l cosθ ) − T l cos = 0
2
Pl cosθ − kl 2 (2sin
θ
2
P cosθ − kl sin θ = 0
tan θ =
(b)
θ
cos ) = 0
2
P
kl
P
or θ = tan −1
kl
For p = 2kl :
2kl
−1
= tan (2) = 63.435°
kl
θ = tan −1
or θ = 63.4°
COSMOS Chapter 4 Solution 59
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