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Question: The force required just to move a body up an inclined plane is double the force required just to pre...

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

A

36.8°

B

45°

C

30°

D

42.6°

Answer

36.8°

Explanation

Solution

Retardation in upward motion

∴\thereforeForce required just to move up Fup=mg(sin⁡θ+μcos⁡θ)F _ { u p } = m g ( \sin \theta + \mu \cos \theta )

Similarly for down ward motion a =g(sin⁡θ−μcos⁡θ)= g ( \sin \theta - \mu \cos \theta )

∴\therefore Force required just to prevent the body sliding down

Fdn=mg(sin⁡θ−μcos⁡θ)F _ { d n } = m g ( \sin \theta - \mu \cos \theta )

According to problem Fup=2FdnF _ { u p } = 2 F _ { d n }

⇒ mg(sin⁡θ+μcos⁡θ)=2mg(sin⁡θ−μcos⁡θ)m g ( \sin \theta + \mu \cos \theta ) = 2 m g ( \sin \theta - \mu \cos \theta )

⇒ sin⁡θ+μcos⁡θ=2sin⁡θ−2μcos⁡θ\sin \theta + \mu \cos \theta = 2 \sin \theta - 2 \mu \cos \theta

⇒ 3μcos⁡θ=sin⁡θ3 \mu \cos \theta = \sin \theta ⇒ tan⁡θ=3μ\tan \theta = 3 \mu

⇒ θ=tan⁡−1(3μ)=tan⁡−1(3×0.25)=tan⁡−1(0.75)\theta = \tan ^ { - 1 } ( 3 \mu ) = \tan ^ { - 1 } ( 3 \times 0.25 ) = \tan ^ { - 1 } ( 0.75 ) =36.8∘= 36.8 ^ { \circ }