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## Tuesday, 23 August 2011

### Physics GRE - #22

Two wedges, each of mass $m$, are placed next to each other on a flat floor. A cube of mass $M$ is balanced on the wedges as shown below.

Assume no friction between the cube and the wedges, but a coefficient of static friction $\mu < 1$ between the wedges and the floor. What is the largest $M$ that can be balanced as shown without motion of the wedges?

1. $\frac{m}{\sqrt{2}}$
2. $\mu\frac{m}{\sqrt{2}}$
3. $\mu\frac{m}{1-\mu}$
4. $2\mu\frac{m}{1-\mu}$
5. All $M$ will balance.

Solution :

We know immediately it's either choice 3 or 4 since any $M$ will balance if $\mu\rightarrow 1$, but it looks like we're going to have to do the work to figure out which one it is.
Due to the symmetry of the problem, we may assume that one wedge holds up a mass of $\frac{M}{2}$. By breaking down the force of the block on the wedge into components, we can see that the normal force on the wedge is $N = \left(m+\frac{M}{2}\right)g.$ So the friction force on the wedge is $F_{fr} = \mu\left(m+\frac{M}{2}\right)g.$ Since the friction force has to equal the mass of half the block (recall that we are only considering one wedge holding up a block of mass $M/2$), we know that $W_{1/2 - block} = \frac{M}{2}g = \mu\left(m+\frac{M}{2}\right)g.$ We can then solve this equation to obtain $M=2\mu\frac{m}{1-\mu}.$
What if the wedges weren't angled at 45 degrees? Can you find a general expression for $M$ at any angle?
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