Answer:
The tension in the cable [tex]T_3[/tex] = 993.5 N
The tension in the cable [tex]T_2 =[/tex] 496.75 N
The tension in the cable [tex]T_1[/tex] = 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:
[tex]L_1 = s_B + 2 s_A ---------- equation \ (1)[/tex]
where;
[tex]s_B[/tex] = distance from the fixed point to point B
[tex]s_A[/tex] = distance from the fixed point to pulley A
From the cable 2 as well.we obtain its length
[tex]L_2 = ( s_W - s_A) + s_W ------- equation \ (2)[/tex]
where :
[tex]s_W =[/tex] 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 [tex]L_1[/tex] is constant ; Then:
[tex]\frac{dL_1}{dt} = \frac{ds}{dt}+ 2\frac{ds_A}{dt} ---------- euqation \ (3)[/tex]
[tex]0 = v_B +2 v_A[/tex]
where;
[tex]v_B =[/tex] velocity at point B
[tex]v_A[/tex] = 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 [tex]L_2[/tex] is constant ; Then:
[tex]L_2 = (s_W - s_A) +s_W[/tex]
[tex]\frac{dL_2}{dt}=2\frac{ds_W}{dt}-\frac{ds_A}{dt} ----------- equation \ (4)[/tex]
[tex]0 = 2v_W -v_A[/tex]
where;
[tex]v_W =[/tex] the velocity of the weight
Let differentiate equation (3) in order to deduce a relation between accelerations A and B with respect to time
[tex]\frac{dv_A}{dt} + 2 \frac{dv_A}{dt } = 0[/tex]
[tex]a_B +2a_A = 0 --------- equation \ (5)[/tex]
[tex]a_A = - \frac{1}{2}a_B[/tex]
where;
[tex]a_A[/tex] = acceleration at A
[tex]a_B=[/tex] acceleration at B
Replacing 0.5 m/s ² for [tex]a_B[/tex] in equation (5); then
[tex]a_A = - \frac{1}{2}*0.5[/tex]
[tex]a_A = - 0.25 \ m/s^2[/tex]
Let differentiate equation (4) in order to deduce a relation between W and A with respect to time
[tex]2\frac{dv__W}{dt}- \frac{dv_A}{dt} = 0[/tex]
[tex]2a__W} -a_A = 0 ----------- equation \ (6)[/tex]
[tex]a__W }= \frac{1}{2}a_A[/tex]
where;
[tex]a_W[/tex] = acceleration of weight W
Replacing - 0.25 m/s² for [tex]a_A[/tex]
[tex]a__W }= \frac{1}{2}*(-0.25)[/tex]
[tex]a__W }= -0.125 \ m/s^2[/tex]
From the second diagram, if we consider the equilibrium of forces acting on the cylinder in y-direction; we have:
[tex]\sum F_y = ma_y[/tex]
[tex]mg - T_3 = ma_w[/tex]
where;
m= mass of the cylinder = 100 kg
[tex]T_3[/tex] = tension in the string = ???
g = acceleration due to gravity = 9.81 m/s²
[tex]a_w[/tex] = acceleration of the cylinder = [tex]- 0.125 \ m/s^2[/tex]
Plugging all values into above equation; we have
(100 × 9.81) - [tex]T_3[/tex] = 100(-0.125)
[tex]T_3[/tex] = 993.5 N
∴ The tension in the cable [tex]T_3[/tex] = 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:
[tex]\sum F _y = 0 \\ \\2T_2 -T_3 = 0 \\ \\ T_2 = \frac{T_3}{2}[/tex]
where ;
[tex]T_2[/tex] = tension in cable 2
Replacing 993.5 N for [tex]T_3[/tex] ; we have
[tex]T_2 = \frac{993.5 \ N}{2}[/tex]
[tex]T_2 = 496.75 \ N[/tex]
∴ The tension in the cable [tex]T_2 = 496.75 \ N[/tex]
From the fourth diagram, if we consider the equilibrium of forces acting on the cylinder in y-direction on the pulley A ; we have
[tex]\sum F _y = 0 \\ \\2T_1 -T_2 = 0 \\ \\ T_1 = \frac{T_2}{2}[/tex]
where;
[tex]T_1[/tex] = tension in cable 1
Replacing 496.75 N for [tex]T_2[/tex] in the above equation; we have:
[tex]T_1 = \frac{496.75}{2}[/tex]
[tex]T_1[/tex] = 248.375 N
∴ The tension in the cable [tex]T_1[/tex] = 248.375 N
