$$mg\sin \theta = ma\,\,\,\,\therefore a = g\sin \theta $$
where $$a$$ is along the inclined plane
$$\therefore $$ vertical component of acceleration is $$g\,{\sin ^2}\theta $$
$$\therefore $$ Telative vertical acceleration of $$A$$ with respect to $$B$$ is $$g\left( {{{\sin }^2}60 - {{\sin }^2}30} \right] = \frac{g}{2} = 4.9\,m/{s^2}$$       in vertical direction

Let $$T$$ be the tension in the branch of a tree when monkey is descending with acceleration $$a.$$ Then $$mg - T = ma;$$    and $$T = 75\% $$  of weight of monkey
$$\eqalign{ & \therefore ma = mg - \left( {\frac{{75}}{{100}}} \right)mg = \left( {\frac{1}{4}} \right)mg \cr & {\text{or}}\,\,a = \frac{g}{4}. \cr} $$

$$\eqalign{ & F = \left( {m + m + m} \right) \times a\,\,\,\,\,\, \cr & \therefore a = \frac{{10.2}}{6}m/{s^2} \cr & \therefore {T_2} = ma = 2 \times \frac{{10.2}}{6} = 3.4N \cr} $$

For the bag accelerating down
$$\eqalign{ & mg - T = ma \cr & \therefore T = m\left( {g - a} \right) = \frac{{49}}{{10}}\left( {10 - 5} \right) \cr & = 24.5\,N \cr} $$
$$ \approx 24\,N$$

$$\eqalign{ & N = ma\,\sin \theta + mg\cos \theta \,......\left( 1 \right) \cr & {\text{also}}\,mg\sin \theta = ma\,\cos \theta \,......\left( 2 \right) \cr} $$

$$\eqalign{ & {\text{from}}\,\left( 2 \right)a = g\,\tan \theta \cr & \therefore N = mg\frac{{{{\sin }^2}\theta }}{{\cos \theta }} + mg\cos \theta , \cr & {\text{or}}\,\,N = \frac{{mg}}{{\cos \theta }} \cr} $$

Breaking all the forces in $$x - y$$  axis.
Total force along $$\left( { + x} \right)$$  axis $$ = \left( {1\cos {{60}^ \circ } + 2\sin {{30}^ \circ }} \right)$$
along $$\left( { - x} \right)$$  axis $$ = \left( {4\sin {{30}^ \circ }} \right)$$
along $$\left( { + y} \right)$$  axis $$ = \left( {4\cos {{30}^ \circ } + 1\sin {{60}^ \circ }} \right)$$
along $$\left( { - y} \right)$$  axis $$ = \left( {2\cos {{30}^ \circ }} \right)$$
⇒ Net force along $$x$$-axis
$$\eqalign{ & = - \left( {1\cos {{60}^ \circ } + 2\sin {{30}^ \circ }} \right) + 4\sin {30^ \circ } \cr & \Rightarrow - \left( {\frac{1}{2} + 2 \times \frac{1}{2}} \right) + 4 \times \frac{1}{2} \cr & \Rightarrow \frac{{ - 3}}{2} + 2 = + \frac{1}{2} \cr} $$
Net force along $$y$$-axis
$$\eqalign{ & = 4\cos {30^ \circ } + 1\sin {60^ \circ } - 2\cos {30^ \circ } \cr & \Rightarrow 4 \times \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{2} - 2 \times \frac{{\sqrt 3 }}{2} \cr & = \frac{{5\sqrt 3 }}{2} - \frac{{2\sqrt 3 }}{2} \cr & = \frac{{3\sqrt 3 }}{2} \cr} $$
To have, resultant only in y-axis we must have $$\frac{1}{2}N$$  force towards $$+x$$ -axis, so that it can compensate the net force of $$- x$$ axis.

As shown in the figure, the three forces are represented by the sides of a triangle taken in the same order.
Therefore the resultant force is zero. $${{\vec F}_{net}} = m\vec a.$$
Therefore acceleration is also zero ie velocity remains unchanged.

$$\eqalign{ & {T_1} = m{\omega ^2}\left( {3\ell } \right) \cr & {\text{and}}\,{T_2} - {T_1} = m{\omega ^2}\ell \cr & {\text{or}}\,\,{T_2} = {T_1} + m{\omega ^2}\ell = m{\omega ^2}\left( {3\ell } \right) + m{\omega ^2}\ell \cr & = 4m{\omega ^2}\ell \cr & \therefore \frac{{{T_2}}}{{{T_1}}} = \frac{4}{3}. \cr} $$

Taking the rope and the block as a system

we get $$P = \left( {m + M} \right)a\,\,\,\,\,\therefore a = \frac{P}{{m + M}}$$
Taking the block as a system, we get $$T = Ma$$
$$\therefore T = \frac{{MP}}{{m + M}}$$

$$a = \frac{F}{{{m_1} + {m_2}}}$$
So the acceleration is same whether the force is applied on $${m_1}$$ or $${m_2}.$$