Question

The force required to just move a body up the inclined plane is double the force required to just prevent the body from sliding down the plane. The coefficient of friction is $\mu .$ If $\theta$ is the angle of inclination of the plane, then $tan\theta$ is equal to

• A. $\mu$
• B. $3\mu$
• C. $2\mu$
• D. $0.5\mu$

Answer 1

The correct option is B $3\mu$

Let force required to prevent sliding $=$ F
so force required to pull up $=$ 2F
Friction force acting on body $=$ f $=$ $\mu$mgcos$\theta$
In case of pulling up friction will act down the plane but in case of just prevent body from sliding friction will act up along the plane.
case i : pulling up
Balancing the force on body along the plane.
2F $=$ f +mgsin$\theta$
2F $=$ $\mu$mgcos$\theta$ + mgsin$\theta$ -- (1)
Case ii : preventing from sliding
Balancing the force along the plane.
F + f $=$ mgsin$\theta$
F + $\mu$Cos$\theta$ $=$ mgsin$\theta$ --(2)
putting value of F from equation (1) in (2)
$\frac{mgsin\theta +\mu mgcos\theta }{2}+\mu mgcos\theta =mgsin\theta$

$mgsin\theta +\mu mgcos\theta +2\mu mgcos\theta =2mgsin\theta$
$3\mu cos\theta =sin\theta$
$3\mu cos\theta =sin\theta$
$\Rightarrow tan\theta =3\mu$
B, is correct answer.