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Math Help - [SOLVED] Find coefficient of friction?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Find coefficient of friction?

    A picture is attached to a vertical wall by means of string AC with length l forming an angle α with the wall. The height of the picture BC = d. The bottom of the picture is not fastened. At what coefficient of friction between the picture and the wall will the picture be in equilibrium?

    For diagram, Mechanics on Flickr - Photo Sharing!

    I keep getting the value of μ in terms of weight of the body but that is not a known variable. How to find μ in terms of the lengths given?
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello fardeen_gen
    Quote Originally Posted by fardeen_gen View Post
    A picture is attached to a vertical wall by means of string AC with length l forming an angle α with the wall. The height of the picture BC = d. The bottom of the picture is not fastened. At what coefficient of friction between the picture and the wall will the picture be in equilibrium?

    For diagram, Mechanics on Flickr - Photo Sharing!

    I keep getting the value of μ in terms of weight of the body but that is not a known variable. How to find μ in terms of the lengths given?
    First, use the Sine Rule to find angle B:

    \frac{\sin B}{l}=\frac{\sin \alpha}{d}

    \Rightarrow \sin B = \frac{l\sin\alpha}{d}

    and \cos B = \sqrt{1-\frac{l^2\sin^2\alpha}{d^2}}

    Then, the Sine Rule again to find AB:

    AB=\frac{d\sin\alpha}{\sin C} = \frac{d\sin\alpha}{\sin(B+\alpha)}, noting that \sin C = \sin(B + \alpha)

    And if the perpendicular distance from B to AC produced = h, then h = AB\sin\alpha =\frac{d\sin^2\alpha}{\sin B\cos\alpha +\cos B\sin\alpha}

    Then, we assume that

    • the picture is on the point of slipping at B
    • the picture has mass m, and is uniform, so the centre of mass is at the mid-point of BC
    • the forces at B are N, \mu N
    • the tension in the string is T

    Resolve vertically: mg = \mu N + T\cos\alpha

    Resolve horizontally: N = T\sin\alpha

    Take moments about B: \tfrac12mgd\sin B = Th

    You now have all you need to eliminate mg, N and T and obtain an equation for \mu.

    Grandad
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  3. #3
    Super Member fardeen_gen's Avatar
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    Quote Originally Posted by Grandad View Post
    Hello fardeen_genFirst, use the Sine Rule to find angle B:

    \frac{\sin B}{l}=\frac{\sin \alpha}{d}

    \Rightarrow \sin B = \frac{l\sin\alpha}{d}

    and \cos B = \sqrt{1-\frac{l^2\sin^2\alpha}{d^2}}

    Then, the Sine Rule again to find AB:

    AB=\frac{d\sin\alpha}{\sin C} = \frac{d\sin\alpha}{\sin(B+\alpha)}, noting that \sin C = \sin(B + \alpha)

    And if the perpendicular distance from B to AC produced = h, then h = AB\sin\alpha =\frac{d\sin^2\alpha}{\sin B\cos\alpha +\cos B\sin\alpha}

    Then, we assume that

    • the picture is on the point of slipping at B
    • the picture has mass m, and is uniform, so the centre of mass is at the mid-point of BC
    • the forces at B are N, \mu N
    • the tension in the string is T
    Resolve vertically: mg = \mu N + T\cos\alpha

    Resolve horizontally: N = T\sin\alpha

    Take moments about B: \tfrac12mgd\sin B = Th

    You now have all you need to eliminate mg, N and T and obtain an equation for \mu.

    Grandad
    Shouldn't it be AB = d.sin C/sin α?

    Thank you Grandad!
    After eliminating mg, T and N,
    μ = {l cos α + 2√(d^2 - l^2 sin^2 α)}/l sin α, which is the required answer.

    Sine Rule For the Win!
    Last edited by fardeen_gen; April 15th 2009 at 11:28 PM. Reason: Final answer for future reference
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  4. #4
    MHF Contributor
    Grandad's Avatar
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    Hello fardeen_gen
    Quote Originally Posted by fardeen_gen View Post
    Shouldn't it be AB = d.sin C/sin α?
    Yes, of course. My apologies!

    Grandad
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