Question

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

• A. $N_1=37.2N$ and $N_2=9.6N$
• B. $N_1=38.2N$ and $N_2=8.6N$
• C. $N_1=40N$ and $N_2=4N$
• D. $N_1=37.5N$ and $N_2=9.9N$

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
$10+\frac{36}{5}+2\times 10=37.2N$
So, reaction $N_1$ of ground $=37.2N$ upwards
Also, total force horizontally on wall is
$F_1=\frac{48}{5}=9.6N$
So, reaction $N_2$ of wall $=9.6N$ towards left