The minimum force required to start pushing a body up a rough (frictional coefficient $\mu $ ) inclined plane is ${F_1}$ while the minimum force needed to prevent it from sliding down is ${F_2}$ . If the inclined plane makes an angle of $\theta $ with the horizontal such that $\tan \theta = 2\mu $ then find the ratio $\dfrac{{{F_1}}}{{{F_2}}}$ is:

Hint: The force is resolved in two components: horizontal component and vertical component. These are dependent on the sine and cosine of angle. In the above question the vertical and the horizontal components are added in push force. The vertical and the horizontal components are subtracted in pull force.


Complete step by step solution:
According to the question it is given that the minimum force required to start pushing a body up a rough inclined plane is ${F_1}$ and the minimum force needed to prevent it from sliding down is ${F_2}$.

The frictional coefficient is $\mu $ and the inclined plane is making an angle of $\theta $ with the horizontal having the condition as shown below.
$\tan \theta = 2\mu $

The sliding of the body developed the force of $mg\sin \theta $. The force $\mu mg\cos \theta $ is also added to slide the body.

So, the force required for pushing the object in the inclined plane is written as,
$
   \Rightarrow {F_1} = W\sin \theta + \mu W\cos \theta \\
   \Rightarrow {F_1} = mg\left( {\sin \theta + \mu \cos \theta } \right) \\
 $ (1)

So, the force needed to prevent it from sliding in the inclined plane is written as,
$
   \Rightarrow {F_2} = W\sin \theta - \mu W\cos \theta \\
   \Rightarrow {F_2} = mg\left( {\sin \theta - \mu \cos \theta } \right) \\
 $ (2)

Take the ratio of both the forces as mentioned in the equation (1) and (2).

$
   \Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{mg\left( {\sin \theta + \mu \cos \theta } \right)}}{{mg\left( {\sin \theta - \mu \cos \theta } \right)}} \\
   \Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{\left( {\tan \theta + \mu } \right)}}{{\left( {\tan \theta - \mu } \right)}} \\
 $

Substitute $2\mu $ for $\tan \theta $ in the above expression and simplify the equation as shown below.
$
   \Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{\left( {2\mu + \mu } \right)}}{{\left( {2\mu - \mu } \right)}} \\
   \Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = 3 \\
 $
Therefore, the value of the ratio of the minimum force required to start pushing a body up a rough inclined plane and the minimum force needed to prevent it from sliding down $\left( {\dfrac{{{F_1}}}{{{F_2}}}} \right)$ is $3$.

Hence, the correct answer is 3.

Note: For pushing the object both the components of the forces. For preventing the object from sliding the component of the forces subtracted. So, the ratio of the force is given as 3.