A box of mass \(2\ \text{kg}\) is placed on an inclined plane that makes \(30^\circ\) with the horizontal. The coefficient of friction between the box and inclined plane is \(0.2\). A force \(F\) is applied on the box perpendicular to the incline to prevent the box from sliding down. The minimum value of \(F\) is
\[
(\text{acceleration due to gravity }=10\ \text{m s}^{-2})
\]
Show Hint
When an external force is applied perpendicular to an inclined plane, it changes the normal reaction. Since friction is
\[
f=\mu N,
\]
increasing \(N\) increases the maximum friction available to prevent sliding.
Step 1: Write the given data.
Mass of the box is
\[
m=2\ \text{kg}.
\]
Angle of inclination is
\[
\theta=30^\circ.
\]
Coefficient of friction is
\[
\mu=0.2.
\]
Acceleration due to gravity is
\[
g=10\ \text{m s}^{-2}.
\]
Therefore,
\[
mg=2\times 10=20\ \text{N}.
\]
Step 2: Find the component of weight down the plane.
The component of weight along the inclined plane is
\[
mg\sin\theta.
\]
So,
\[
mg\sin30^\circ=20\times \frac{1}{2}.
\]
\[
mg\sin30^\circ=10\ \text{N}.
\]
This force tends to slide the box down the plane.
Step 3: Find the normal reaction.
The normal component of weight is
\[
mg\cos\theta.
\]
Since force \(F\) is applied perpendicular to the incline, it increases the normal reaction.
Therefore,
\[
N=mg\cos30^\circ+F.
\]
\[
N=20\times \frac{\sqrt3}{2}+F.
\]
\[
N=10\sqrt3+F.
\]
Step 4: Apply the condition for just preventing sliding.
To prevent the box from sliding down, friction must act up the plane.
For minimum \(F\), limiting friction is just equal to the downward component of weight.
So,
\[
\mu N=mg\sin30^\circ.
\]
Substituting the values,
\[
0.2(10\sqrt3+F)=10.
\]
Step 5: Solve for \(F\).
\[
10\sqrt3+F=\frac{10}{0.2}.
\]
\[
10\sqrt3+F=50.
\]
\[
F=50-10\sqrt3.
\]
Using
\[
\sqrt3=1.732,
\]
we get
\[
F=50-17.32.
\]
\[
F=32.68\ \text{N}.
\]
\[
F\approx 32.7\ \text{N}.
\]
Step 6: Final conclusion.
Therefore, the minimum force required is
\[
\boxed{32.7\ \text{N}}
\]
Hence, the correct option is
\[
\boxed{(3)}
\]
