Solution to 1992 Problem 6

The force on the mass M due to each of the masses m will be have a vertical component of
\begin{align*}\frac{1}{2}  M g\end{align*}
This force must be normal to the surface of the triangle, so the horizontal component is
\begin{align}\frac{1}{2} M g \label{eqn:1}\end{align}
The force of friction is
\begin{align*}\mu \left(m g + \frac{1}{2} \cdot M g\right)\end{align*}
Setting this equal to (1) and solving for M gives
\begin{align*}M = \frac{2 \mu m}{1 - \mu}\end{align*}
Therefore, answer (D) is correct.

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