Find minimum value of F such that m 2 starts its motion on the groundA- -02 m 1- m 2 g -2B- -02 m 1- m 2 gC- -0 m 1- m 2 g -2D- -0 m 1- m 2 g

The correct option is A μ_0(2m_1+m_2)g/2 Fx μ_0 m_1 gx nbsp;1/2 kx^2=0 X = μ_0 m_2 g/k nbsp; ...

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Acceleration in 2D

Find minimum ...

Question

Find minimum value of F such that $m_{2}$ starts its motion on the ground

$\frac{μ_{0} (2 m_{1} + m_{2}) g}{2}$

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$μ_{0} (2 m_{1} + m_{2}) g$

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$μ_{0} (m_{1} + m_{2}) g$

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$\frac{μ_{0} (m_{1} + m_{2}) g}{2}$

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F = \frac{m_{2} g}{2}

m_{1}

m_{1}

^{'} M^{'}

^{'} R^{'}

m_{1}

m_{2}

a

\begin{matrix} Column I & Column II (a) & Acceleration of m_{1} in elevator frame & (p) & \frac{1}{2} (\frac{(m_{2} - m_{1}) g + 2 m_{2} a}{m_{1} + m_{2}}) t^{2} (b) & Acceleration of m_{1} in ground frame & (q) & \frac{1}{2} [\frac{m_{2} - m_{1}}{(m_{1} + m_{2})} (g + a)] t^{2} (c) & Distance covered of m_{1} in elevator frame & (r) & (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) g + (\frac{2 m_{2}}{m_{1} + m_{2}}) a (d) & Distance covered of m_{1} in ground frame & (s) & \frac{(m_{2} - m_{1})}{(m_{1} + m_{2})} (g + a) (t) & None of the above \end{matrix}

m_{1}, m_{2}

m_{3}

A

F

m_{1}

θ

m_{2}

m_{1}

\begin{matrix} Column I & Column II i. & F & a. & (m_{1} + m_{2}) g ii. & Force acting on m_{2} is & b. & m_{2} g sec θ iii. & Tension in the string is & c. & m_{2} \frac{F}{m_{1} + m_{2}} iv. & Force acting on m_{1} by the wire is & d. & (m_{1} + m_{2}) g tan θ \end{matrix}

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