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Inclined Planes

A block of ma...

Question

A block of mass m =5 kg is placed on the wedge of mass M =32 kg as shown in the figure. Find the acceleration of wedge with respect to ground. (Neglect any type of friction. String and pulley are ideal):

\frac{1}{2} m / s^{2}

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\frac{3}{4} m / s^{2}

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\frac{4}{3} m / s^{2}

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\frac{3}{5} m / s^{2}

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Solution

The correct option is D $\frac{3}{5} m / s^{2}$
$T (1 + c o s θ) - N s i n θ) = M a$ .....(i)
FBD of wedge from ground frame
$m g s i n θ - m a c o s θ - T = m a$
$\Rightarrow m g s i n θ - T = m a (1 + c o s θ)$ .......(ii)
$N = m (g c o s θ + a s i n θ)$ ......(iii)
Using (i) + (ii) $(1 + c o s θ) + (i i i) s i n θ$
$m g s i n θ + m g s i n θ c o s θ =$
$M a + m a (1 + c o s θ)^{2} + m g s i n θ c o s θ + m a s i n^{2} θ$
$\Rightarrow a = \frac{m g s i n θ}{M + 2 m (1 + c o s θ)}$
given $θ = 37^{o}, m = 5 k g$ and $M = 32 k g$
so, $a = \frac{3}{5} m / s^{2}$

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