Two wooden blocks are moving on a smooth hori | vedclass

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Question

Acceleration of the system:

$\mathrm{a}=\frac{\mathrm{P}}{\mathrm{M}+\mathrm{m}}$         $(1)$

The $FBD$ of mass $m$ is sh

Vedclass · Accepted Answer

$(M + m)g\,tan\beta $

Vedclass · Answer

Acceleration of the system:

$\mathrm{a}=\frac{\mathrm{P}}{\mathrm{M}+\mathrm{m}}$         $(1)$

The $FBD$ of mass $m$ is shown.

$\mathrm{R} \sin \beta=\mathrm{ma}$           $(2)$

$\mathrm{R} \cos \beta=\mathrm{mg}$           $(3)$

From $( 2)$ and $( 3)$

$a=g 	an \beta$

Putting the value of a in $(1),$ we get

$\mathrm{P}=(\mathrm{M}+\mathrm{m}) \mathrm{g} 	an \beta$

$O R$

$\mathrm{ma} \cos \beta=\mathrm{mg} \sin \beta$

$\Rightarrow a=g 	an \beta$

Two wooden blocks are moving on a smooth horizontal surface such that the block of mass $m$ remains stationary with respect to block of mass $M$ as shown in the figure. The magnitude of force $P$ is