SATHEE: Laws Of Motion Question 51

Laws Of Motion Question 51

Question: A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined at an angle $θ$ to the vertical. The block will remain in equilibrium, if the coefficient of friction between it and the surface is [Haryana CEE 1996]

Options:

A) $\frac{(P + Q \sin θ)}{(m g + Q \cos θ)}$

B) $\frac{(P \cos θ + Q)}{(m g - Q \sin θ)}$

C) $\frac{(P + Q \cos θ)}{(m g + Q \sin θ)}$

D) $\frac{(P \sin θ - Q)}{(m g - Q \cos θ)}$

Show Answer

Answer:

Correct Answer: A

Solution:

By drawing the free body diagram of the block for critical condition $F = μ R$ therefore $P + Q \sin θ$

$= μ (m g + Q \cos θ)$

$∴$ $μ = \frac{P + Q \sin θ}{m g + Q \cos θ}$

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Laws Of Motion Question 51

Question: A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined at an angle θ to the vertical. The block will remain in equilibrium, if the coefficient of friction between it and the surface is [Haryana CEE 1996]

Options:

Answer:

Solution:

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Question: A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined at an angle $θ$ to the vertical. The block will remain in equilibrium, if the coefficient of friction between it and the surface is [Haryana CEE 1996]