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. $\frac{mg}{cos\theta }$
• B. $mgcos\theta$
• C. $mgsin\theta$
• D. $\frac{mg}{sin\theta }$

Answer 1

Answer: (A)

$N=masin\theta +mgcos\theta$ $\dots (1)$
Also, $mgsin\theta =macos\theta$ $\dots (2)$
From (2), $a=gtan\theta$
$$\begin{gathered} \therefore N=mg\frac{{\left(sin\right)}^2\theta }{cos\theta }+mgcos\theta [substitutingthevalueofa] \\ \therefore N=\frac{mg}{cos\theta }({\left(sin\right)}^2\theta +{\left(cos\right)}^2\theta )=\frac{mg}{cos\theta } \\ or,N=\frac{mg}{cos\theta } \end{gathered}$$