The tension in the cable
= 993.5 N
The tension in the cable
 496.75  N
The tension in the cable
= 248.375 N
Explanation:
The diagram attached below depicts the full understanding of what the question is all about.
Now, obtaining the length of cable 1 from the diagram; we have:
![L_1 = s_B + 2 s_A ---------- equation \ (1)](/tpl/images/0591/2653/da44b.png)
where;
= distance from the fixed point to point B
= distance from the fixed point to pulley A
From the cable 2 as well.we obtain its length
![L_2 = ( s_W - s_A) + s_W ------- equation \ (2)](/tpl/images/0591/2653/76a98.png)
where :
distance from the fixed point to the weight attached to the pulley
Let differentiate equation (1) in order to deduce a relation between the velocities of A and B with respect to time ;
Since
is constant ; Then:
![\frac{dL_1}{dt} = \frac{ds}{dt}+ 2\frac{ds_A}{dt} ---------- euqation \ (3)](/tpl/images/0591/2653/c2143.png)
![0 = v_B +2 v_A](/tpl/images/0591/2653/866c9.png)
where;
velocity at point B
= velocity at pulley A
Let differentiate equation (2) as well in order to deduce a relation between the velocities of W and A with respect to time :
Since
is constant ; Then:
![L_2 = (s_W - s_A) +s_W](/tpl/images/0591/2653/03970.png)
![\frac{dL_2}{dt}=2\frac{ds_W}{dt}-\frac{ds_A}{dt} ----------- equation \ (4)](/tpl/images/0591/2653/4b3f5.png)
![0 = 2v_W -v_A](/tpl/images/0591/2653/886ca.png)
where;
the velocity of the weight
Let differentiate equation (3) in order to deduce a relation between accelerations A and B with respect to time Â
![\frac{dv_A}{dt} + 2 \frac{dv_A}{dt } = 0](/tpl/images/0591/2653/aeece.png)
![a_B +2a_A = 0 --------- equation \ (5)](/tpl/images/0591/2653/35334.png)
![a_A = - \frac{1}{2}a_B](/tpl/images/0591/2653/0130e.png)
where;
= acceleration at A
acceleration at B
Replacing 0.5 m/s ² for
in equation (5); then
![a_A = - \frac{1}{2}*0.5](/tpl/images/0591/2653/83d77.png)
![a_A = - 0.25 \ m/s^2](/tpl/images/0591/2653/13486.png)
Let differentiate equation (4) in order to deduce a relation between W and A with respect to time
![2\frac{dv__W}{dt}- \frac{dv_A}{dt} = 0](/tpl/images/0591/2653/6c2d8.png)
![2a__W} -a_A = 0 ----------- equation \ (6)](/tpl/images/0591/2653/f4c47.png)
![a__W }= \frac{1}{2}a_A](/tpl/images/0591/2653/fa92d.png)
where;
= acceleration of weight W
Replacing - 0.25 m/s² for ![a_A](/tpl/images/0591/2653/c657a.png)
![a__W }= \frac{1}{2}*(-0.25)](/tpl/images/0591/2653/cec67.png)
![a__W }= -0.125 \ m/s^2](/tpl/images/0591/2653/343e9.png)
From the second diagram, if we consider the equilibrium of forces acting on the cylinder in y-direction; we have:
![\sum F_y = ma_y](/tpl/images/0591/2653/352b1.png)
![mg - T_3 = ma_w](/tpl/images/0591/2653/55e84.png)
where;
m= mass of the cylinder = 100 kg
= tension in the string = ???
g = acceleration due to gravity = 9.81 m/s²
= acceleration of the cylinder = ![- 0.125 \ m/s^2](/tpl/images/0591/2653/5577b.png)
Plugging all values into above equation; we have
(100 × 9.81) -
= 100(-0.125)
= 993.5 N
∴ The tension in the cable
= 993.5 N
From the third diagram, if we consider the equilibrium of forces acting on the cylinder in y-direction on the pulley ; we have:
![\sum F _y = 0 \\ \\2T_2 -T_3 = 0 \\ \\ T_2 = \frac{T_3}{2}](/tpl/images/0591/2653/f7035.png)
where ;
= tension in cable 2
Replacing 993.5 N for
; we have
![T_2 = \frac{993.5 \ N}{2}](/tpl/images/0591/2653/0fbbf.png)
![T_2 = 496.75 \ N](/tpl/images/0591/2653/97f2e.png)
∴ The tension in the cable ![T_2 = 496.75 \ N](/tpl/images/0591/2653/97f2e.png)
From the fourth  diagram, if we consider the equilibrium of forces acting on the cylinder in y-direction on the pulley A ; we have
![\sum F _y = 0 \\ \\2T_1 -T_2 = 0 \\ \\ T_1 = \frac{T_2}{2}](/tpl/images/0591/2653/caa64.png)
where;
= tension in  cable 1
Replacing 496.75 N for
in the above equation; we have:
![T_1 = \frac{496.75}{2}](/tpl/images/0591/2653/cb09b.png)
= 248.375 N
∴ The tension in the cable
= 248.375 N
![3/20 The winch takes in cable at the rate of 200 mm/s, and this rate is momentarily increasing at 50](/tpl/images/0591/2653/9e605.jpg)
![3/20 The winch takes in cable at the rate of 200 mm/s, and this rate is momentarily increasing at 50](/tpl/images/0591/2653/c64e8.jpg)
![3/20 The winch takes in cable at the rate of 200 mm/s, and this rate is momentarily increasing at 50](/tpl/images/0591/2653/8bc85.jpg)
![3/20 The winch takes in cable at the rate of 200 mm/s, and this rate is momentarily increasing at 50](/tpl/images/0591/2653/41fc1.jpg)