A block is moving on an inclined plane making an angle 45° with the horizontal and the coefficient of friction is μ The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N=10 μ, then N is

Question

A block is moving on an inclined plane making an angle 45° with the horizontal and the coefficient of friction is μ The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N=10 μ, then N is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

F1 - mg sinθ + μ mg cosθ F2=mg sin θ- μ mg cos θ Given that F1=3F2 or (sin 45° +μ cos 45°) = 3 (sin 45° - μ cos 45°) On solving, we get=0.5 ∴ N=10 μ =5

Q. A block is moving on an inclined plane making an angle $4 5^{\circ}$ with the horizontal and the coefficient of friction is $μ$ The force required to just push it up the inclined plane is $3$ times the force required to just prevent it from sliding down. If we define $N = 10 μ,$ then $N$ is

$F_{1} - m g sin θ + μ m g cos θ$
$F_{2} = m g sin θ - μ m g cos θ$
Given that $F_{1} = 3 F_{2}$ or $(s in 4 5^{\circ} + μ cos 4 5^{\circ})$
$= 3 (s in 4 5^{\circ} - μ cos 4 5^{\circ})$
On solving, we get $= 0.5 ∴ N = 10 μ = 5$

Q. A block is moving on an inclined plane making an angle 45∘ with the horizontal and the coefficient of friction is μ The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N=10μ,then N is

F1​−mgsinθ+μmgcosθ F2​=mgsinθ−μmgcosθ Given that F1​=3F2​ or (sin45∘+μcos45∘) =3(sin45∘−μcos45∘) On solving, we get=0.5∴N=10μ=5

Q. A block is moving on an inclined plane making an angle $4 5^{\circ}$ with the horizontal and the coefficient of friction is $μ$ The force required to just push it up the inclined plane is $3$ times the force required to just prevent it from sliding down. If we define $N = 10 μ,$ then $N$ is

$F_{1} - m g sin θ + μ m g cos θ$
$F_{2} = m g sin θ - μ m g cos θ$
Given that $F_{1} = 3 F_{2}$ or $(s in 4 5^{\circ} + μ cos 4 5^{\circ})$
$= 3 (s in 4 5^{\circ} - μ cos 4 5^{\circ})$
On solving, we get $= 0.5 ∴ N = 10 μ = 5$