SATHEE: Laws Of Motion Question 145

Laws Of Motion Question 145

Question: The force required just to move a body up an inclined plane is double the force required just to prevent the body sliding down. If the coefficient of friction is 0.25, the angle of inclination of the plane is

Options:

A) $37^{\circ}$

B) $45^{\circ}$

C) $30^{\circ}$

D) $53^{\circ}$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] Retardation in upward motion $= g (s i n θ + μ c o s θ)$

$∴$ force required just to move up $F_{u p} = m g (\sin θ + μ \cos θ)$ Similarly for downward motion $a = g (\sin θ + μ \cos θ)$

$∴$ Force required just to prevent the body sliding down $F_{d n} = m g (\sin θ - μ \cos θ)$ According to problem $F_{u p} = 2 F_{d n}$

$\Rightarrow m g (s i n θ + μ c o s θ) = 2 m g (s i n θ - μ c o s θ)$

$\Rightarrow \sin θ + μ c o s θ = 2 s i n θ - 2 μ c o s θ$

$\Rightarrow 3 μ \cos θ = \sin θ \Rightarrow \tan θ = 3 μ$

$\Rightarrow θ = \tan^{- 1} (3 μ) = t a n^{- 1} (3 \times 0.25) = t a n^{- 1} (0.75) = 37^{\circ}$

Laws Of Motion Question 145

Question: The force required just to move a body up an inclined plane is double the force required just to prevent the body sliding down. If the coefficient of friction is 0.25, the angle of inclination of the plane is

Options:

Answer:

Solution:

Engineering Preparation

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NCERT Books & Solutions

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Syllabus

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Laws Of Motion Question 145

Question: The force required just to move a body up an inclined plane is double the force required just to prevent the body sliding down. If the coefficient of friction is 0.25, the angle of inclination of the plane is

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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