A cubical block rests on a plane of μ=√3. The angle through which the plane be inclined to the horizontal so that the block just slides down will be

Question

A cubical block rests on a plane of μ=√3. The angle through which the plane be inclined to the horizontal so that the block just slides down will be (A) 30° (B) 45° (C) 60° (D) 75°

Tardigrade Admin · Accepted Answer

fs=m g sin θ μ m g cos θ=m g sin θ tan θ=μ=√3 θ=60°

Q. A cubical block rests on a plane of $μ = 3$ . The angle through which the plane be inclined to the horizontal so that the block just slides down will be

$3 0^{\circ}$

$4 5^{\circ}$

$6 0^{\circ}$

$7 5^{\circ}$

$f_{s} = m g sin θ$
$μ m g cos θ = m g sin θ$
$tan θ = μ = 3$
$θ = 6 0^{\circ}$

Q. A cubical block rests on a plane of μ=3​. The angle through which the plane be inclined to the horizontal so that the block just slides down will be

30∘

45∘

60∘

75∘