[SOLVED] Friction - Uniform rod?

**fardeen_gen** · June 15th 2009, 12:23 PM

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.

**halbard** · July 5th 2009, 09:38 AM

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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