Question

A block is between two surfaces as shown in the figure. Find the normal reaction at both surfaces.
[Assume $g =10 \,m / s ^{2}$ ]

• A. $N _{1}=37.2 N$ and $N _{2}=9.6 N$
• B. $N _{1}=38.2 N$ and $N _{2}=8.6 N$
• C. $N _{1}=40 N$ and $N _{2}=4 N$
• D. $N _{1}=37.5 N$ and $N _{2}=9.9 N$

Answer 1

Answer: (A)

The given situation is shown in the figure.Resolving applied force, we have following forces on block

Given, $\tan \theta=\frac{3}{4}, \sin \theta=\frac{3}{5}$ and $\cos \theta=\frac{4}{5}$

So, horizontal component,
$F_{1}=12 \times \cos \theta=12 \times \frac{4}{5}=\frac{48}{5} N$
and vertical component,
$F_{2}=12 \times \sin \theta=12 \times \frac{3}{5}=\frac{36}{5} N$
So, total applied force on ground is
$10+\frac{36}{5}+2 \times 10=37.2 N$
So, reaction $N_{1}$ of ground $=37.2 N$ upwards
Also, total force horizontally on wall is
$F_{1}=\frac{48}{5}=9.6 N$
So, reaction $N_{2}$ of wall $=9.6 \,N$ towards left