SATHEE: Laws Of Motion Question 324

Laws Of Motion Question 324

Question: The force required to just move a body up the inclined plane is double the force required to just prevent the body from sliding down the plane. The coefficient of friction is $μ$ . The inclination $θ$ of the plane is

Options:

A) $\tan^{- 1} μ$

B) $\tan^{- 1} (μ / 2)$

C) $\tan^{- 1} 2 μ$

D) $\tan^{- 1} 3 μ$

Show Answer

Answer:

Correct Answer: D

Solution:

[d] In case [a] In case [b] $m g \sin θ = F_{1} - μ N$

$N=mg sin θ ∴ mg sin θ + μ mg cos θ = F_{1}$

$In second case (b)$

$μ N + F_{2} = mg sin θ$

$\Rightarrow μ mg cos θ - F_{2} = m g \sin θ$

$or F_{2} = m g \sin θ - μ m g \cos θ but F_{1} = 2 F_{2}$

$therefore mg sin θ + μ m g \cos θ$

$= 2 (m g \sin θ - μ m g \cos θ)$

$m g \sin θ = 3 μ m g \cos θ$

$or \tan θ = 3 μ or θ =ta n^{- 1} (3 μ)$

[d]

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Laws Of Motion Question 324

Question: The force required to just move a body up the inclined plane is double the force required to just prevent the body from sliding down the plane. The coefficient of friction isμ . The inclination θ of the plane is

Options:

Answer:

Solution:

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Question: The force required to just move a body up the inclined plane is double the force required to just prevent the body from sliding down the plane. The coefficient of friction is $μ$ . The inclination $θ$ of the plane is