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

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

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

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

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

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

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

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

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

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

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