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Thread: Coefficient of friction, modulus of elasticity problem?

  1. #1
    Super Member fardeen_gen's Avatar
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    Coefficient of friction, modulus of elasticity problem?

    Two small rings, each of weight $\displaystyle W$, slide one upon each of two rods in a vertical plane, each inclined at an angle $\displaystyle \alpha$ to the vertical; the rings are connected by a fine elastic string of natural length $\displaystyle 2a$, and whose modulus of elasticity is $\displaystyle \lambda$; the coefficient of friction for each rod and ring is $\displaystyle \tan \beta$; show that, if string is horizontal, each ring will rest at any point of a segment of the rod whose length is:

    $\displaystyle W\lambda^{-1}a\csc \alpha\{\cot (\alpha - \beta) - \cot (\alpha + \beta)\}$.
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  2. #2
    MHF Contributor
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    Hello fardeen_gen
    Quote Originally Posted by fardeen_gen View Post
    Two small rings, each of weight $\displaystyle W$, slide one upon each of two rods in a vertical plane, each inclined at an angle $\displaystyle \alpha$ to the vertical; the rings are connected by a fine elastic string of natural length $\displaystyle 2a$, and whose modulus of elasticity is $\displaystyle \lambda$; the coefficient of friction for each rod and ring is $\displaystyle \tan \beta$; show that, if string is horizontal, each ring will rest at any point of a segment of the rod whose length is:

    $\displaystyle W\lambda^{-1}a\csc \alpha\{\cot (\alpha - \beta) - \cot (\alpha + \beta)\}$.
    There are two points on each rod where the rings are in limiting equilibrium: one (higher up the rod) where the ring is about to slide down, and one (lower down) where the ring is about to slide up.

    In the first case the friction force acts upwards along the rod; in the second it acts downwards.

    Draw a diagram showing the forces acting on the ring in the first case, where the friction force acts up the rod. The four forces are:

    $\displaystyle F:$ friction force
    $\displaystyle N:$ normal contact force between the rod and the ring
    $\displaystyle W:$ weight of ring
    $\displaystyle T_1:$ tension in string

    Then, since equilibrium is limiting, $\displaystyle F = N\tan\beta$

    Resolve vertically: $\displaystyle N\sin\alpha + F\cos\alpha=W$

    Resolve horizontally: $\displaystyle T_1 +F\sin\alpha = N\cos\alpha$

    From these, you can eliminate $\displaystyle F$ and $\displaystyle N$ to get:

    $\displaystyle T = \frac{W(\cos\alpha -\tan\beta\sin\alpha)}{\sin\alpha + \tan\beta\cos\alpha}$

    $\displaystyle = \frac{W\cos(\alpha+\beta)}{\sin(\alpha+\beta)}=W\c ot(\alpha + \beta)$

    Similarly if $\displaystyle T_2$ is the tension in the string when the ring is on the point of slipping upwards, you can show that $\displaystyle T_2 = W\cot(\alpha-\beta)$

    Now if the length of the segment of the rod between these two equilibrium positions is $\displaystyle d$, then the change in length of the half of the string acting upon one of the rings is $\displaystyle d\sin\alpha$.

    Using Hooke's Law:

    $\displaystyle T_2 - T_1 = \frac{\lambda d\sin\alpha}{a} = W(\cot(\alpha - \beta) - W\cot(\alpha + \beta)$

    $\displaystyle \Rightarrow d = W\lambda^{-1}a\csc \alpha\{\cot (\alpha - \beta) - \cot (\alpha + \beta)\}$

    Grandad
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