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    • Thread Starter

    Recently started out on springs, few problems, help greatly appreciated

    1) Particle of mass 3kg is attached to one end of a light elastic string, natural length is 1m, modulus of elasticity is 14.7N. The other end of the string is attached to a fixed point. Particle is held in equilibrium by a horizontal force of 9.8N, string inclined to verticle at an angle x.
    a) value of x
    b) extension of string
    c) If the horizontal force is removed, find the magnitude of the least force that will keep the string inclined at the same angle?

    I have worked through a and b, however I am not sure how to approach part c. Here are my workings of a and b... ( I have also attached a diagram)

    a) 9.8 = sin x T and T cos x = 3g, therefore tan x = 1/3
    b) choosing 9.8 = sin x T, T = 9.8/sin x.
    Setting T = 14.7s/1 gives s = 2.11 ( where s is the extension in the string)
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    • Study Helper

    Study Helper
    (Original post by rainbowsss)

    Triangle of forces.

    You know one force (3g)

    You know the direction of the tension.

    Draw a diagram - you need this!

    What's the third force that creates equilibrium, and when will it be a minimum.
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Updated: April 1, 2011
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