SATHEE: Laws Of Motion Question 389

Laws Of Motion Question 389

Question: If the coefficient of friction between A and B is $m_{2}$ , the maximum horizontal acceleration of the wedge A for which B will remain at rest with respect to the wedge is:

Options:

A) $m_{1} = m_{2}$

B) $m_{2}$

C) $m_{1} = m_{2}$

D) $m_{2}$

Show Answer

Answer:

Correct Answer: B

Solution:

FBD of block B w.r.t. wedge A, for maximum ‘a’ perpendicular to wedge:

$\sum f_{y^{'}} = (m g \cos θ + m a \sin θ - N) = 0$ and

$\sum f_{x^{'}} = m g \sin θ + μ N - m a \cos θ = 0$

(for maximum a)

$T^{'} = 2 \sec$

$m g \sin θ + μ (m g \cos θ + m a \sin θ) - m a \cos θ = 0$

$\Rightarrow a = \frac{(g \sin θ + μ g \cos θ)}{\cos θ - μ \sin θ}$

for $\frac{T_{1}}{T_{2}} = \sqrt{\frac{g + \frac{g}{4}}{g}} = \frac{\sqrt{5}}{2}$

$\Rightarrow a = g (\frac{\tan 45^{\circ} + μ}{\cot 45^{\circ} - μ}); a = g (\frac{1 + μ}{1 - μ})$

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Laws Of Motion Question 389

Question: If the coefficient of friction between A and B is m2 , the maximum horizontal acceleration of the wedge A for which B will remain at rest with respect to the wedge is:

Options:

Answer:

Solution:

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Question: If the coefficient of friction between A and B is $m_{2}$ , the maximum horizontal acceleration of the wedge A for which B will remain at rest with respect to the wedge is: