The acceleration of the system of two bodies | vedclass

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(A) For a system of two masses $M$ connected by a string over a pulley on a wedge with angles $	heta_1 = 37^\circ$ and $	heta_2 = 53^\circ$,the acce

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(A) For a system of two masses $M$ connected by a string over a pulley on a wedge with angles $	heta_1 = 37^\circ$ and $	heta_2 = 53^\circ$,the acceleration $a$ is given by the net force divided by the total mass.The net driving force is $F_{net} = Mg \sin 53^\circ - Mg \sin 37^\circ$.The total mass of the system is $M + M = 2M$.Using Newton's second law,$F_{net} = (2M)a$.Therefore,$a = \frac{Mg \sin 53^\circ - Mg \sin 37^\circ}{2M} = \frac{g}{2} (\sin 53^\circ - \sin 37^\circ)$.Given $\sin 53^\circ \approx 0.8$ and $\sin 37^\circ \approx 0.6$,and $g = 10 \ m/s^2$:$a = \frac{10}{2} (0.8 - 0.6) = 5 	imes 0.2 = 1 \ m/s^2$.

The acceleration of the system of two bodies over the wedge as shown in the figure is ....... $m/s^2$. (Take $g = 10 \ m/s^2$)

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