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Finding Friction's Magnitude

Two masses ...

Question

Two masses $M_{1}$ and $M_{2}$ are attached to the end of a light string which passes over a massless pulley attached to the top of a double inclined smooth plane of angles of inclination $α$ and $β$ . If $M_{2} > M_{1}$ and $β > α$ then the acceleration of block $M_{2}$ down the inclined will be:

\frac{M_{2} g (s i n β)}{M_{1} + M_{2}}

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\frac{M_{1} g (s i n α)}{M_{1} + M_{2}}

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(\frac{M_{2} s i n β - M_{1} s i n α}{M_{1} + M_{2}}) g

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Solution

The correct option is C $(\frac{M_{2} s i n β - M_{1} s i n α}{M_{1} + M_{2}}) g$
first draw the free body diagram of the given figure
note the following
1)mass $M_{1}$ has its acceleration in upward direction
2)mass $M_{2}$ has its acceleration in downward direction
3)tension will remain same in both the string
now resolving the forces for both the block we find
T- $M_{1} g s i n α$ = $M_{1}$ a
T- $M_{2} g s i n β$ =- $M_{2}$ a
hence we substitute the value of T in any of the following we get
T= $M_{1}$ a+ $M_{1} g s i n α$
$M_{1}$ a+ $M_{1} g s i n α$ - $M_{2} g s i n β$ =- $M_{2}$ a
$M_{1} g s i n α$ - $M_{2} g s i n β$ =-a( $M_{2} + M_{1}$ )
a= $\frac{M_{2} g s i n β - M_{1} g s i n α}{M_{2} + M_{1}}$

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Two masses M1 and M2 are attached to the end of a light string which passes over a massless pulley attached to the top of a double inclined smooth plane of angles of inclination α and β. If M2>M1 and β>α then the acceleration of block M2 down the inclined will be: