First slide

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 and it can be made to slide upwards by applying a force 2 F. The coefficient of friction between the box and the inclined plane is

Easy

A

$(t a n θ) / 3$
B

$3 t a n θ$
C

$(t a n θ) / 2$
D

$2 t a n θ$

Solution

During downward motion: $F = m g s i n θ - m g c o s θ$
During upward motion: $2 F = m g s i n θ + μ m g c o s θ$
Solving above two equations, we get $μ = (t a n θ) / 3$

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