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Friction on inclined plane of angle more than angle of repose

Question

A box of mass 8 Kg is placed on a rough inclined plane of inclination θ . Its downward motion can be prevented by applying an upward pull F parallel to the inclined plane. And it can be made to slide upwards by applying a force 2F parallel to the inclined plane. The coefficient of friction between the box and inclined plane is

Moderate

A
$\frac{1}{3} \tan θ$
B
$3 \tan θ$
C
$\frac{1}{2} \tan θ$
D
$2 \tan θ$

Solution

$f_{l} = μ N = μ m g \cos θ$
$F = m g \sin θ - f_{l} = m g \sin θ - μ m g \cos θ . . . . . . . . . . . . . . . (1)$
$2 F = m g \sin θ + f_{l} = m g \sin θ + μ m g \cos θ . . . . . . . . . . (2)$
$2 (m g \sin θ - μ m g \cos θ) = m g \sin θ + μ m g \cos θ$
$μ = \frac{1}{3} \tan θ$

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