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Cord AB of length a is attached to the end B of a spring having an unstretched length b. The other end of the spring is attached to a roller C so that the spring remains horizontal as it stretches. If...

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Cord AB of length a is attached to the end B of a spring having an unstretched length b. The other end of the spring is attached to a roller C so that the spring remains horizontal as it stretches. If a weight W is suspended from B, determine the angle ? of cord AB for equilibrium.

Given:
a = 5 ft
b = 5 ft
k = 10 lb/ft
W = 10 lb




Answered Same Day Dec 29, 2021

Solution

Robert answered on Dec 29 2021
129 Votes
Elongation of spring, x = a(1 – cos θ) = 5(1 – cos θ)
Tension in string = T
So, Tcos θ = kx
 Tcos θ = 10*5(1 – cos θ) = 50(1 – cos θ)
Also, Tsin θ = 10
cos θ/sin θ = 5(1 – cos θ)
 cos θ = 5sin θ – 5sin θcos θ
Hence, θ = 40.25
SOLUTION.PDF
SOLUTION.PDF

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