Question

The door of an almirah is $6\text{ft}$ high, $1.5\text{ft}$ wide and weighs $8\text{kg}$. The door is supported by two hinges situated at a distance of $1\text{ft}$ from the top and bottom ends. Assuming force exerted on the hinges are equal, the magnitude of the force is
[Take $g=10{\text{m/s}}^2$]

• A. $15\text{N}$
• B. $10\text{N}$
• C. $28\text{N}$
• D. $43\text{N}$

Answer 1

The correct option is D $43\text{N}$

Given,
Height of almirah $\left(h\right)=6\text{ft}$
Width of almirah $\left(l\right)=1.5\text{ft}$
Weight of almirah $\left(W\right)=8g\text{N}$
Also given that hinges are at $1\text{ft}$ distance each from the top and bottom ends.
Assuming the door is uniform, the centre of gravity of the door lies exactly at the centre of mass.
FBD of the door is shown in the figure below. Both forces at the hinges ($F$) act in the plane of the door.


From the diagram shown below, we find
$\tan \theta =\frac{\text{opposite side}}{\text{adjacent side}}=\frac{2}{3/4}=\frac{8}{3}$
Using figure (b), we can write, $\cos \theta =\frac{3}{\sqrt{73}}$

Taking net torque $\left(\tau \right)$ about O:
$\frac{l}{2}\times W=r_1\times F_1+r_2\times F_2$
From the figure we can deduce that,
$\frac{l}{2}=\frac{1.5}{2}=\frac{3}{4}\text{ft}$
$r_1=2\text{ft}$ & $r_2=2\text{ft}$
$F_1=F\cos \theta$ & $F_2=F\cos \theta$
$\therefore 8g\times \left(\frac{3}{4}\right)=4\times F\cos \theta$
$\Rightarrow 6g=4F\left(\frac{3}{\sqrt{73}}\right)$
$\Rightarrow F=5\sqrt{73}\approx 43\text{N}$
[putting $g=10{\text{m/s}}^2$]
Thus, option (d) is the correct answer.