Question

The force required just to move a body up an inclined plane is double the force required just to prevent the body sliding down. If the coefficient of friction is 0.25, then the angle of inclination of the plane is

• A. ${\left(tan\right)}^{-1}\left(\frac{3}{4}\right)$
• B. ${\left(tan\right)}^{-1}\left(\frac{1}{4}\right)$
• C. 45°
• D. 30°

Answer 1

Answer: (A)

case (i)

$F_1=Mgsin\theta +f_r$
$=Mg\left(sin\theta +\mu cos\theta \right)$ ... (i)
Case (ii)

$F_2+f_r=Mgsin\theta$
$F_2=Mgcos\theta -\mu Mgcos\theta$
$=Mg\left(sin\theta -\mu cos\theta \right)$ ... (ii)
$F_1=2F_2\left(Given\right)$
From Eqs. (i) and (ii)
$Mg\left(sin\theta +\mu cos\theta \right)=2Mg\left(sin\theta -\mu cos\theta \right)$
$⟹sin\theta +\frac{1}{4}cos\theta =2\left(sin\theta -\frac{1}{4}cos\theta \right)$
$⟹cos\theta \left(\frac{1}{4}+\frac{1}{2}\right)=sin\theta$
$⟹\frac{3}{4}cos\theta =sin\theta$
$⟹tan\theta =\frac{3}{4}$
$⟹tan\theta =\frac{3}{4}$
$⟹\theta ={\left(tan\right)}^{-1}\left(\frac{3}{4}\right)$