The answer is Ta 4m1m2m3g4m1m2m3 m1m2m1m3 I want to know how

590 Find the tension in cord A in terms of the masses mi, m2, ms, and g. Neglect the masses of the pulleys

Solution

for upper pulley considering pulley A system is accelerating downward and m1 accelerating upward.

(m2 + m3)g - T = (m2 + m3)a

and T - m1g = m1a

adding both, m2g + m3g - m1g = (m1 + m2 + m3)a

a = (m2 + m3 - m1)g / (m1 + m2 + m3)

now come to pulleyA system.

As we sees from the frame of ppulley A then a psuedo force mass *a will act on blocks in upward direction.

suppose m2 is going down and m3 up then ,

applying Fnet = m a on m2.

m2g - Ta - Fpsuedo = m a1

m2g - Ta - m2((m2 + m3 - m1)g / (m1 + m2 + m3)) = m2 a1

multiplying by m3

m2m3g - m3 Ta - m2m3((m2 + m3 - m1)g / (m1 + m2 + m3) = m2 m3 a1   ...(i)

on m3:

Ta + m3 ((m2 + m3 - m1)g / (m1 + m2 + m3))   - m3g = m3 a1 ....(ii)

multiplying by -m2:

- m2 Ta - m2m3 (m2 + m3 - m1)g / (m1 + m2 + m3) + m2m3g = - m2m3a1

2m2m3g - 2m2m3((m2 + m3 - m1)g / (m1 + m2 + m3) - (m2 + m3) Ta = 0

(m2 + m3)(m1 + m2 + m3)Ta = (m1+m2+m3)(2m2m3g) - (m2 + m3 - m1)(2m2m3)g

(m2 + m3)(m1 + m2 + m3)Ta / 2g = m1m2m3 + m2^2 m3 + m2 m3^2 - m2^2m3 - m2m3^2 + m1m2m3

(m2 + m3)(m1 + m2 + m3)Ta / 2g = 2m1m2m3

Ta = (4m1m2m3)g / (m1m2 + m2^2 + m2m3 + m1m3 + m3m3 + m3^2)