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The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. if coefficient of friction between the body and inclined plane is $\frac{1}{2 \sqrt{3}}$ ,the angle of inclined plane is

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60 °

45 °

30 °

15 °

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detailed solution

Correct option is C

$F_{up} = 3 F_{down}$

$mg (\sin θ + μ_{k} \cos θ) = 3 mg (\sin θ - μ_{k} \cos θ)$ $4 μ_{k} \cos θ = 2 \sin θ$

$4 μ_{k} = 2 \tan θ$

$2. \frac{1}{2 \sqrt{3}} = \tan θ$

$\tan θ = \frac{1}{\sqrt{3}}$

$θ = 30 °$

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The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. if coefficient of friction between the body and inclined plane is 123 ,the angle of inclined plane is

Unlock the full solution & master the concept.

Ready to Test Your Skills?

free study material

The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. if coefficient of friction between the body and inclined plane is $\frac{1}{2 \sqrt{3}}$ ,the angle of inclined plane is