Physics, 09.04.2020 03:31 toxsicity

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.

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Answer from: Maaarii

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

3/20 The winch takes in cable at the rate of 200 mm/s, and this rate is momentarily increasing at 50

Answer from: Quest

Ithink they do have the same properties just in less quantities i think

Answer from: Quest

i would say

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3/20 The winch takes in cable at the rate of 200 mm/s, and this rate is momentarily increasing at 50...