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A uniform triangular lamina $ABC$ of weight $W$ newtons is right angled at $B$. $AB$ has length $0.6m$ and $BC$ has length $0.8m$ and the lamina rests in a vertical plane with $AB$ in contact with a rough horizontal table. A force of magnitude $P$ newtons is applied at $C$, in the plane of the lamina and at right angles to $AC$. Given that the lamina is in equilibrium and on the point of turning about $A$, find $P$ in terms of $W$, and the least possible value of the coefficient of friction between the lamina and the table.

For the first part of the question:

Length of $AC$ is: $$AC = \sqrt{0.8^2 + 0.4^2} = 1\text{m}$$

The center of mass of $ABC$ is : $$\frac{1}{3}(0.3+0.4+0.5)=0.4\text{m}$$

Taking moments about $A$:

$$0.4\cdot W = 1\cdot P$$ which gives $$P=0.4W$$

J132
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1 Answers1

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You meed to resolve the forces vertically and horizontally. If $N$ is the normal reaction at A and $F$ is the friction at A, you have$$N+P\cos CAB=W$$ and $$P\sin CAB=F$$ This leads to $$N=0.76W, F=0.32W$$ Then $$F\leq\mu N\Rightarrow \mu\geq \frac{8}{19}$$

David Quinn
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  • Thanks for that. I had resolved the forces correctly and thought that the angle CAB had to come into it as $P$ is not vertical. – J132 Sep 29 '15 at 22:09