Three blocks of masses m1,m2 and m3 are connected as shown in the figure. All the surfaces are frictionless and the string and the pulleys are light. Find the acceleration of m1
Since the pulley is light string A will have twice the tension as compared to B
Assuming string B has tension T
string A has tension 2T
m3 moves down with acceleration a with respect to pulley
m2 moves down with acceleration -a with respect to pulley
m1 moves with acceleration a’
2T = m1a’
m3 moves with acceleration a+a’
m2 moves with acceleration a’-a
T= m3g – (a+a’)m3
T=m2(a’-a) + m2g
m1a’/2 = m3g – (a+a’)m3
=> 2m3a’ + m1a’/2 = m3g – m3a – (1)
m1a/2 = m2g + m2(a’-a)
=>2m2a’ + m1a’/2 = m2g + m2a – (2)
To eliminate a multiply by m2 and m3
m2(2m3a’ + m1a’)/2 = m3m2g – m3m2a
+
m3(2m2a’ + m1a’)/2 = m3m2g + m3m2a
=> m2(2m3a’ + m1a’) + m3(2m2a’ + m1a’) / 2 = 2m3m2g
=>a’ ( m3(2m2+m2) + m2(2m3+m1)) = 4gm2m3
=> a’ = 4gm2m3/m1m3 + m1m2 + m2m3