Question

3/20 The winch takes in cable at the rate of 200 mm/s, and this rate is momentarily increasing at 500 mm/s each second. Determine the tensions in the three cables. Neglect the weights of the pulleys.

Answer 1

Answer:

The tension in the cable $T_3$ = 993.5 N

The tension in the cable $T_2 =$  496.75  N

The tension in the cable $T_1$ = 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:

$L_1 = s_B + 2 s_A ---------- equation \ (1)$

where;

$s_B$ = distance from the fixed point to point B

$s_A$ = distance from the fixed point to pulley A

From the cable 2 as well.we obtain its length

$L_2 = ( s_W - s_A) + s_W ------- equation \ (2)$

where :

$s_W =$ 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 $L_1$ is constant ; Then:

$\frac{dL_1}{dt} = \frac{ds}{dt}+ 2\frac{ds_A}{dt} ---------- euqation \ (3)$

$0 = v_B +2 v_A$

where;

$v_B =$ velocity at point B

$v_A$ = 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 $L_2$ is constant ; Then:

$L_2 = (s_W - s_A) +s_W$

$\frac{dL_2}{dt}=2\frac{ds_W}{dt}-\frac{ds_A}{dt} ----------- equation \ (4)$

$0 = 2v_W -v_A$

where;

$v_W =$ the velocity of the weight

Let differentiate equation (3) in order to deduce a relation between accelerations A and B with respect to time  

$\frac{dv_A}{dt} + 2 \frac{dv_A}{dt } = 0$

$a_B +2a_A = 0 --------- equation \ (5)$

$a_A = - \frac{1}{2}a_B$

where;

$a_A$ = acceleration at A

$a_B=$ acceleration at B

Replacing 0.5 m/s ² for $a_B$ in equation (5); then

$a_A = - \frac{1}{2}*0.5$

$a_A = - 0.25 \ m/s^2$

Let differentiate equation (4) in order to deduce a relation between W and A with respect to time

$2\frac{dv__W}{dt}- \frac{dv_A}{dt} = 0$

$2a__W} -a_A = 0 ----------- equation \ (6)$

$a__W }= \frac{1}{2}a_A$

where;

$a_W$ = acceleration of weight W

Replacing - 0.25 m/s² for $a_A$

$a__W }= \frac{1}{2}*(-0.25)$

$a__W }= -0.125 \ m/s^2$

From the second diagram, if we consider the equilibrium of forces acting on the cylinder in y-direction; we have:

$\sum F_y = ma_y$

$mg - T_3 = ma_w$

where;

m= mass of the cylinder = 100 kg

$T_3$ = tension in the string = ???

g = acceleration due to gravity = 9.81 m/s²

$a_w$ = acceleration of the cylinder = $- 0.125 \ m/s^2$

Plugging all values into above equation; we have

(100 × 9.81) - $T_3$ = 100(-0.125)

$T_3$ = 993.5 N

∴ The tension in the cable $T_3$ = 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:

$\sum F _y = 0 \\ \\2T_2 -T_3 = 0 \\ \\ T_2 = \frac{T_3}{2}$

where ;

$T_2$ = tension in cable 2

Replacing 993.5 N for $T_3$ ; we have

$T_2 = \frac{993.5 \ N}{2}$

$T_2 = 496.75 \ N$

∴ The tension in the cable $T_2 = 496.75 \ N$

From the fourth  diagram, if we consider the equilibrium of forces acting on the cylinder in y-direction on the pulley A ; we have

$\sum F _y = 0 \\ \\2T_1 -T_2 = 0 \\ \\ T_1 = \frac{T_2}{2}$

where;

$T_1$ = tension in  cable 1

Replacing 496.75 N for $T_2$ in the above equation; we have:

$T_1 = \frac{496.75}{2}$

$T_1$ = 248.375 N

∴ The tension in the cable $T_1$ = 248.375 N