Search any question & find its solution
Question:
Answered & Verified by Expert
A block is between two surfaces as shown in the figure. Find the normal reaction at both surfaces. [Assume, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
Options:
Solution:
2550 Upvotes
Verified Answer
The correct answer is:
$N_1=37.2 \mathrm{~N}$ and $N_2=9.6 \mathrm{~N}$
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,
$F_1=12 \times \cos \theta=12 \times \frac{4}{5}=\frac{48}{5} \mathrm{~N}$
and vertical component,
$F_2=12 \times \sin \theta=12 \times \frac{3}{5}=\frac{36}{5} \mathrm{~N}$
So, total applied force on ground is
$10+\frac{36}{5}+2 \times 10=37.2 \mathrm{~N}$
So, reaction $N_1$ of ground $=37.2 \mathrm{~N}$ upwards Also, total force horizontally on wall is
$F_1=\frac{48}{5}=9.6 \mathrm{~N}$
So, reaction $N_2$ of wall $=9.6 \mathrm{~N}$ towards left
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,
$F_1=12 \times \cos \theta=12 \times \frac{4}{5}=\frac{48}{5} \mathrm{~N}$
and vertical component,
$F_2=12 \times \sin \theta=12 \times \frac{3}{5}=\frac{36}{5} \mathrm{~N}$
So, total applied force on ground is
$10+\frac{36}{5}+2 \times 10=37.2 \mathrm{~N}$
So, reaction $N_1$ of ground $=37.2 \mathrm{~N}$ upwards Also, total force horizontally on wall is
$F_1=\frac{48}{5}=9.6 \mathrm{~N}$
So, reaction $N_2$ of wall $=9.6 \mathrm{~N}$ towards left
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.