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(C) $1$. The block of mass $m$ exerts a normal force $N$ on the wedge $M$ perpendicular to the inclined surface.$2$. The angle of the wedge is $	heta

Vedclass · Accepted Answer

$\frac{N \sin 	heta}{M}$ along $-x$-axis

Vedclass · Answer

(C) $1$. The block of mass $m$ exerts a normal force $N$ on the wedge $M$ perpendicular to the inclined surface.$2$. The angle of the wedge is $	heta$ with the horizontal. The normal force $N$ makes an angle $	heta$ with the vertical,or $(90^\circ - 	heta)$ with the horizontal.$3$. The horizontal component of the normal force $N$ acting on the wedge $M$ is $N_x = N \sin 	heta$.$4$. This horizontal force acts towards the left ($-x$-direction) because the block pushes the wedge to the left.$5$. According to Newton's second law,the horizontal acceleration $a_M$ of the wedge $M$ is given by $F_{net, x} = M a_M$.$6$. Therefore,$N \sin 	heta = M a_M$,which gives $a_M = \frac{N \sin 	heta}{M}$ along the $-x$-axis.

Consider the shown arrangement. Assume all surfaces to be smooth. If $N$ represents the magnitude of the normal reaction between the block of mass $m$ and the wedge of mass $M$,then the acceleration of $M$ along the horizontal equals:

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