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4-1.Newton's Laws of Motion
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In the figure shown $'P'$ is a plate on which a wedge $B$ is placed and on $B$ a block $A$ of mass $m$ is placed. The plate is suddenly removed and system of $B$ and $A$ is allowed to fall under gravity. Neglecting any force due to air on $A$ and $B$, the normal force on $A$ due to $B$ is
A$\frac{{mg}}{{\cos \,\theta }}\,$
B$mg\,\cos \,\theta $
Czero
D$\frac{{2mg}}{{\cos\, \theta }}\,$
Solution
$F.B.D.$ of block $A$
Applying Newton's second law for block $A$ in vertical direction
$\mathrm{m}_{\mathrm{A}} \mathrm{g}-\mathrm{N} \cos \theta=\mathrm{m}_{\mathrm{A}} \mathrm{g}$
where $\theta$ is the angle of the wedge
$\Rightarrow \mathrm{N} \cos \theta=0$
as $\theta<90^{\circ} \quad \therefore \mathrm{N}=0$
Applying Newton's second law for block $A$ in vertical direction
$\mathrm{m}_{\mathrm{A}} \mathrm{g}-\mathrm{N} \cos \theta=\mathrm{m}_{\mathrm{A}} \mathrm{g}$
where $\theta$ is the angle of the wedge
$\Rightarrow \mathrm{N} \cos \theta=0$
as $\theta<90^{\circ} \quad \therefore \mathrm{N}=0$
Standard 11
Physics
