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(D) In the frame of the trolley,a pseudo force $F_p = ma_0$ acts on block $m$ in the negative $x$-direction and $F_p = Ma_0$ acts on block $M$ in the

Vedclass · Accepted Answer

All of the above.

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(D) In the frame of the trolley,a pseudo force $F_p = ma_0$ acts on block $m$ in the negative $x$-direction and $F_p = Ma_0$ acts on block $M$ in the negative $x$-direction.For block $m$: The forces are tension $T$ upwards,gravity $mg$ downwards,and pseudo force $ma_0$ horizontally. Since it is constrained to move vertically,the vertical equation is $T - mg = ma_y$. If $a_y = 0$,then $T = mg$.For block $M$: The forces are tension $T$ horizontally,friction $f$ horizontally,and normal force $N = Mg$ vertically. The pseudo force $Ma_0$ acts in the negative $x$-direction. The equation of motion is $T - f - Ma_0 = Ma_x$. If $a_x = 0$,then $T = f + Ma_0$.$1$. If $a_0$ is such that the pseudo force $Ma_0$ balances the tension $T$,$T$ can be zero.$2$. If $a_0 = 0$,then $T = mg$ is possible if the system is in equilibrium.$3$. Since $f \leq \mu Mg$,if $T  \mu Mg$ a necessary condition for motion or equilibrium in certain states.Thus,all statements are correct.

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