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Thread: [SOLVED] Friction - Uniform rod?

  1. #1
    Super Member fardeen_gen's Avatar
    Jun 2008

    [SOLVED] Friction - Uniform rod?

    A uniform rod, of weight W, can turn freely about a hinge at one end, and rests with the other against a rough vertical wall making an angle \alpha with the wall. Show that this end may rest anywhere on an arc of a circle of angle 2\arctan [\mu \tan \alpha], and that in either of the extreme positions the pressure on the wall is \frac{1}{2}W[\cot^2 a + \mu^2]^{-\frac{1}{2}}, where \mu is the coefficient of friction.
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  2. #2
    Mar 2009
    Take rod AB of length 2l hinged at A, B resting against wall, and let perpendicular from A to wall meet wall at O. Assume OB makes an angle \theta with vertical through O.

    Apart from forces at the hinge, forces acting are the weight at midpoint of AB, the normal reaction N at B perpendicular to wall, and the friction F at B perpendicular to OB along the wall.

    Choose axes so that \mathbf i is along AO, \mathbf j is horizontal and perpendicular to AO, and \mathbf k is vertically up. Then

    \overline{AO}=\begin{pmatrix}2l\sin\alpha\\0\\0\en  d{pmatrix}, \overline{OB}=\begin{pmatrix}0\\2l\cos\alpha\,\sin  \theta\\2l\cos\alpha\,\cos\theta\end{pmatrix} so that \mathbf p=\overline{AB}=\begin{pmatrix}2l\sin\alpha\\2l\co  s\alpha\,\sin\theta\\2l\cos\alpha\,\cos\theta\end{  pmatrix}.

    The relevant forces are the weight \mathbf W=\begin{pmatrix}0\\0\\-W\end{pmatrix}, friction \mathbf F=\begin{pmatrix}0\\-F\cos\theta\\F\sin\theta\end{pmatrix}, and normal reaction \mathbf N=\begin{pmatrix}-N\\0\\0\end{pmatrix}. The forces at A are irrelevant as we are about to take moments about A.

    So, on we go: {\textstyle\frac12}\mathbf p\times\mathbf W+\mathbf p\times\mathbf F+\mathbf p\times\mathbf N=\mathbf 0, which we can write as {\textstyle\frac12}\mathbf p\times(\mathbf W+2\mathbf F+2\mathbf N)=\mathbf 0.

    This gives \begin{pmatrix}l\sin\alpha\\l\cos\alpha\,\sin\thet  a\\l\cos\alpha\,\cos\theta\end{pmatrix}\times\begi  n{pmatrix}-2N\\-2F\cos\theta\\2F\sin\theta-W\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.

    Expanding the vector product on the left after cancelling l, \begin{pmatrix}2F\cos\alpha-W\sin\theta\,\cos\alpha\\W\sin\alpha-2F\sin\theta\,\sin\alpha-2N\cos\theta\,\cos\alpha\\2N\sin\theta\,\cos\alpha-2F\cos\theta\,\sin\alpha\end{pmatrix}=\begin{pmatr  ix}0\\0\\0\end{pmatrix}.

    The first equation gives F={\textstyle\frac12}W\sin\theta.

    The third equation is F=N\frac{\tan\theta}{\tan\alpha}, and since F\leq\mu N we see that \tan\theta\leq\mu\tan\alpha. So the maximum value of \theta=\arctan(\mu\tan\alpha) which leads by symmetry to the given answer.

    Eliminating F we see that N={\textstyle\frac12}W\tan\alpha\cos\theta, and at the extreme positions \cos\theta=\frac1{\sqrt{1+\tan^2\theta}}=\frac1{\s  qrt{1+\mu^2\tan^2\alpha}}. This gives your other answer, my friend.
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