Answer
See the detailed answer below.
Work Step by Step
a)
Since the strings and the ropes are massless and since the pullies are frictionless, the tension in the same wire is constant along it.
So that the tension in the wire of pulley B is constant along this wire and so do the wire of pulley A.
We name the tension in wire A as $T_A$ and the tension in wire B as $T_B$. We can assume that these wires are non-stretchable which means that their length is constant.
This means that the length of wire B is given in the given figure.
the length right side of wire B is given by $y_B-y_A$ and the length of the left side of the wire is given by $y_B-y_3$.
Thus, the whole length of wire B is given by
$$L_B=(y_B-y_A)+(y_B-y_3)$$
$$L_B=2y_B-y_A-y_3 \tag K$$
By the same approach, the length of wire A is given by
$$L_A=(y_A-y_1)+(y_A-y_2)$$
$$L_A= 2y_A-y_1 -y_2 $$
Solving (K) for $y_A$ and plugging it into the previous equation;
$$L_A= 2\left[2y_B-L_B-y_3\right]-y_1 -y_2 $$
$$L_A= 4y_B-2L_B-2y_3 -y_1 -y_2 $$
Thus,
$$\overbrace{L_A+2L_B}^{\rm Constant}= 4y_B-2y_3 -y_1 -y_2 $$
The left side is constant since the length of the two strings is constant.
$$-2y_3 -y_1 -y_2=\overbrace{{\rm constant}-4y_B}^{\rm Constant_2} $$
The right side now is a new constant since we add to it another constant value which is the height of pulley B, as we see from the given figure, it remains stationary in the vertical direction.
$$-2y_3 -y_1 -y_2= {\rm Constant_2} $$
$$ 2y_3 +y_2+y_1 = {\rm Constant_3} $$
So,
$$ \dfrac{d}{dt}\left[2y_3 +y_1 +y_2\right]= \dfrac{d}{dt} {\rm Constant_3} =0$$
$$ \dfrac{2dy_3 }{dt}+ \dfrac{dy_2}{dt}+ \dfrac{dy_1 }{dt}=0$$
$$ 2v_{y3}+ v_{y2}+ v_{y1}=0$$
$$ \dfrac{d}{dt}\left[2v_{y3}+ v_{y2}+ v_{y1}\right]= \dfrac{d}{dt}0 =0$$
$$ \boxed{2a_{y3}+ a_{y2}+a_{y1}=0}$$
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b)
We need to apply Newton's second law on each object.
$$\sum F_{y3}=T_B-m_3g=m_3a_{y3}$$
Thus,
$$T_B-m_3g=m_3a_{y3}\tag 1$$
$$\sum F_{y,pulley_A}=T_B -2T_A=\overbrace{m_{pulley_A}}^{0}a_{y}$$
Hence,
$$T_B=2T_A\tag 2$$
$$\sum F_{y2}=T_A-m_2g=m_2a_{y2}$$
$$T_A-m_2g=m_2a_{y2}\tag 3$$
$$\sum F_{y1}=T_A-m_1g=m_1a_{y1}$$
$$T_A-m_1g=m_1a_{y1}\tag 4$$
Divide both sides of (1) on $m_3$, (3) on $m_2$, and (4) on $m_1$, and then adding the resulting equations.
$$\dfrac{T_B}{m_3}- g= a_{y3}$$
Plugging from (2);
$$\dfrac{2T_A}{m_3}- g= a_{y3}$$
Mutliplying both sides by 2;
$$\dfrac{4T_A}{m_3}- 2g= 2a_{y3}\tag 5$$
$$\dfrac{T_A}{m_2}- g= a_{y2} \tag 6$$
$$ \dfrac{T_A}{m_1 }-g=a_{y1}\tag 7$$
Adding these last 3 formulas
$$\dfrac{4T_A}{m_3}- 2g+\dfrac{T_A}{m_2}- g+ \dfrac{T_A}{m_1 }-g=\overbrace{2a_{y3}+ a_{y2}+a_{y1}}^{0}$$
Note the boxed formula above;
$$\dfrac{4T_A}{m_3}+\dfrac{T_A}{m_2}+\dfrac{T_A}{m_1 }-4g=0$$
Solving for $T_A$;
$$\dfrac{4T_A}{m_3}+\dfrac{T_A}{m_2}+\dfrac{T_A}{m_1 }=4g$$
$$T_A\left[\dfrac{4}{m_3}+\dfrac{1}{m_2}+\dfrac{1}{m_1 }\right]=4g$$
$$\boxed{T_A=\dfrac{4g}{\left[\dfrac{4}{m_3}+\dfrac{1}{m_2}+\dfrac{1}{m_1 }\right]}}$$
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c) Plugging the given data into the previous boxed formula above;
$$ T_A=\dfrac{4\times 9.8}{\left[\dfrac{4}{4}+\dfrac{1}{1.5}+\dfrac{1}{2.5}\right]} =\bf 18.97\;\rm N$$
From (5);
$$\dfrac{2T_A}{m_3}- g= a_{y3} $$
$$ a_{y3}=\dfrac{2\times 18.97}{4}- 9.8=\color{red}{\bf -0.315}\;\rm m/s^2$$
From (6);
$$ a_{y2}=\dfrac{T_A}{m_2}- g =\dfrac{ 18.97}{1.5}- 9.8=\color{red}{\bf 2.85}\;\rm m/s^2$$
From (7);
$$ a_{y1}=\dfrac{T_A}{m_1 }-g =\dfrac{ 18.97}{2.5}- 9.8=\color{red}{\bf -2.21}\;\rm m/s^2$$
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d) The author asks about why $m_3$ is accelerating while its mass is equal to the mass of $m_1+m_2$.
$m_3$ will be in equilibrium if the tension exerting on it is just equal to its weight.
$$T_B=2T_A=2\times 18.97\approx 38\;\rm N$$
while its weight is $$m_3g=4\times 9.8\approx 39 \;\rm N$$
so that it must accelerate.
