For the system (in equilibrium) shown in fig ,the acceleration of the mass m4 immediately after the lower thread x is cut will be (assume that the threads are weightless and inextensible,the spring are weightless ,the mass of pulley is negligible and there is no friction).
Asked by m.nilu
24th July 2018,
11:40 AM
Answered by Expert
Answer:
For equilibrium of system of loads, we have (m1 + m2 ) > (m3 + m4 )
Force T1 in the left spring is given by, T1 = m2×g....................(1)
Let T2 be the force in right spring. For equilibrium, we have, m3×g + T2 = T0 ................(2)
where T0 is the tension in the string around pulley.
for the load m1 , we have, m1×g + T1 = T0 ..................(3)
by substituting for T1 from eqn.(1), we rewrite eqn. (3) as m1×g + m2×g = T0 .......................(4)
from (2) and (4), we write T2 = (m1 + m2 - m3)×g .......................(5)
After cutting the lower thread, the equations of motion for the loads are
m1×g + T1 - T0 = m1×a1 ..................................(6)
m2×g - T1 = m2×a2 ............................(7)
T2 + m3×g - T0 = m3×a3 ..........................(8)
T2 + m4×g = m4×a4 .........................(9)
Solving above equations, we get, a1 = a2 = a3 = 0
and 
Answered by Expert
9th August 2018,
4:15 PM
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