A cubical block rests on an inclined plane of coefficient of friction μ =(1/√3) . What should be the angle of inclination so that the block just slides down the inclined plane?

Question

A cubical block rests on an inclined plane of coefficient of friction μ =(1/√3) . What should be the angle of inclination so that the block just slides down the inclined plane? (A)  30°  (B)  60°  (C)  45°  (D)  90°

Tardigrade Admin · Accepted Answer

tan θ =μ tan θ =(1/√3) tan θ = tan 30° Angle of inclination, θ =30°

Q. A cubical block rests on an inclined plane of coefficient of friction $μ = \frac{1}{3}$ . What should be the angle of inclination so that the block just slides down the inclined plane?

$3 0^{\circ}$

$6 0^{\circ}$

$4 5^{\circ}$

$9 0^{\circ}$

$tan θ = μ$ $tan θ = \frac{1}{3}$
$tan θ = tan 3 0^{\circ}$
Angle of inclination, $θ = 3 0^{\circ}$

Q. A cubical block rests on an inclined plane of coefficient of friction μ=3​1​ . What should be the angle of inclination so that the block just slides down the inclined plane?

30∘

60∘

45∘

90∘