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In the figure shown C is a fixed wedge. A block B is kept on the inclined surface of the wedge C. Another block A is inserted in a slot in the block B as shown in figure. A light inextensible string passes over a light pulley which is fixed to the block B through a light rod. One end of the string is fixed and other end of the string is fixed to A.S is a fixed support on the wedge. All the surfaces are smooth. Masses of A and B are same. Find the magnitude of acceleration of A
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Dear Student,

Please find below the solution to the asked query:

Consider the situation shown in the below figure.


The length of the string should be constant. Therefore we can find the relation between the accelerations of two blocks by using the length constraint.

$$\begin{gathered} i.e. x+y=L \\ differentiate twice on both sides with respect to time, \\ \frac{dx}{dt}+\frac{dy}{dt}=0\Rightarrow v_A+v_B=0 \\ \frac{d{v}_A}{dt}+\frac{d{v}_B}{dt}=0\Rightarrow a_A+a_B=0 \\ So, a_A=a_B in magnitude. \\ let a_A=a_B=a \end{gathered}$$

Now if we write the equations for the free body diagrams of the two blocks,

$$\begin{gathered} for block B, \\ 2mg \sin \theta -T=2ma \\ for block A, \\ T-mg \cos \theta =ma \\ Therefore, \\ 2mg sin \theta -T=2\left\{T-mg cos \theta \right\}\Rightarrow 2mg sin \theta -T=2T-2mg cos \theta \\ \Rightarrow 3T=2mg sin {37}^0+2mg cos {37}^0\Rightarrow 3T=2mg\times \frac{3}{5}+2mg\times \frac{4}{5} \\ \Rightarrow 3T=\frac{14mg}{5}\Rightarrow T=\frac{14mg}{15} \\ Therefore the acceleration of block A is, \\ T-mg cos {37}^0=ma\Rightarrow \frac{14mg}{15}-\frac{4mg}{5}=ma \\ \Rightarrow a=\frac{14g-12g}{15}\Rightarrow a=\frac{2g}{15}=\frac{20}{15} \\ \Rightarrow a=\frac{4}{3} \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \\ Block B acceleration is, \\ a=\frac{4}{3} \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \\ Therefore, the resul\tan t acceleration of the block A is, \\ {a}_A=\sqrt{a^2+a^2}=\sqrt{{\left(\frac{4}{3}\right)}^2+{\left(\frac{4}{3}\right)}^2}=\sqrt{\frac{16}{9}+\frac{16}{9}} \\ \Rightarrow a_A=\frac{\sqrt{32}}{3} \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

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