Question

A block of mass $m$ is placed on a smooth wedge of inclination $\theta $ . The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ( $g$ is acceleration due to gravity) will be

• A. $\frac{m g}{ cos \theta }$
• B. $m g \, cos \, \theta $
• C. $m g \, sin \, \theta $
• D. $\frac{m g}{ sin \theta }$

Answer 1

Answer: (A)

$N = m a \, \text{​} \, sin \, \theta + m g \, cos \theta $ $\ldots \left(\right.1\left.\right)$
Also, $m \, g \, sin \, \theta = m \, a \text{​} \, cos \, \theta $ $\ldots \left(\right.2\left.\right)$
From (2), $a = g \, tan \, \theta $
$\therefore \, \, N=mg\frac{\left(sin\right)^{2} \theta }{ cos \theta }+mg \, cos\theta \, \left[\right.substituting \, \, the \, \, value \, \, of \, a\left]\right. \\ \therefore \, \, N=\frac{m g}{cos \, \theta }\left(\right.\left(sin\right)^{2}\theta +\left(cos\right)^{2}\theta \left.\right)=\frac{m g}{cos \, \theta } \\ or, \, \, N=\frac{m g}{ cos \theta }$