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 $0.25$, the angle of inclination of the plane is ...... $^o$
$36.8$
$45$
$30$
$42.6$
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 $0.25$, the angle of inclination of the plane is ...... $^o$
Retardation in upward motion $ = g(\sin \theta + \mu \cos \theta )$
$\therefore $ Force required just to move up ${F_{up}} = mg(\sin \theta + \mu \cos \theta )$
Similarly for down ward motion a $ = g(\sin \theta - \mu \cos \theta )$
$\therefore $ Force required just to prevent the body sliding down ${F_{dn}} = mg(\sin \theta - \mu \cos \theta )$
According to problem ${F_{up}} = 2{F_{dn}}$
$⇒$ $mg(\sin \theta + \mu \cos \theta ) = 2mg(\sin \theta - \mu \cos \theta )$
$⇒$ $\sin \theta + \mu \;\cos \theta = 2\sin \theta - 2\mu \;\cos \theta $
$⇒$ $3\mu \cos \theta = \sin \theta $
$⇒$ $\tan \theta = 3\mu $
$⇒$ $\theta = {\tan ^{ - 1}}(3\mu ) = {\tan ^{ - 1}}(3 \times 0.25) = {\tan ^{ - 1}}(0.75)$$ = 36.8^\circ $
$\therefore $ Force required just to move up ${F_{up}} = mg(\sin \theta + \mu \cos \theta )$
Similarly for down ward motion a $ = g(\sin \theta - \mu \cos \theta )$
$\therefore $ Force required just to prevent the body sliding down ${F_{dn}} = mg(\sin \theta - \mu \cos \theta )$
According to problem ${F_{up}} = 2{F_{dn}}$
$⇒$ $mg(\sin \theta + \mu \cos \theta ) = 2mg(\sin \theta - \mu \cos \theta )$
$⇒$ $\sin \theta + \mu \;\cos \theta = 2\sin \theta - 2\mu \;\cos \theta $
$⇒$ $3\mu \cos \theta = \sin \theta $
$⇒$ $\tan \theta = 3\mu $
$⇒$ $\theta = {\tan ^{ - 1}}(3\mu ) = {\tan ^{ - 1}}(3 \times 0.25) = {\tan ^{ - 1}}(0.75)$$ = 36.8^\circ $
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