2 bodies of mass m1 and m2 are connected by a light string which passes over a frictionless massless pulley. If the pulley is moving upward with a uniform acceleration g/2, then tension in the string will be
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Answered by
14
hi buddy thank you for asking this question,
When the system is accelerating upward with acceleration (g/2), the net downward acceleration experienced by the masses is = g + g/2 = 3g/2
(the FBD diagram is in the photo)
Now,
M(3g/2) – T = Ma ……..(1)
T – m(3g/2) = ma ………(2)
Adding the above two equations we get,
(3g/2)(M – m) = a(M + m)
=> a = (3g/2)(M – m)/(M + m)
So,
(2) => T = m(3g/2) + ma
=> T = m(3g/2) + (m)(3g/2)(M – m)/(M + m)
=> T = 3mgM/(M + m)
This is the tension in the string.
hope it's helps pls give it brainliest and follow me please
When the system is accelerating upward with acceleration (g/2), the net downward acceleration experienced by the masses is = g + g/2 = 3g/2
(the FBD diagram is in the photo)
Now,
M(3g/2) – T = Ma ……..(1)
T – m(3g/2) = ma ………(2)
Adding the above two equations we get,
(3g/2)(M – m) = a(M + m)
=> a = (3g/2)(M – m)/(M + m)
So,
(2) => T = m(3g/2) + ma
=> T = m(3g/2) + (m)(3g/2)(M – m)/(M + m)
=> T = 3mgM/(M + m)
This is the tension in the string.
hope it's helps pls give it brainliest and follow me please
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ayushdey0970:
Brainliest
Answered by
1
hi buddy thank you for asking this question,
When the system is accelerating upward with acceleration (g/2), the net downward acceleration experienced by the masses is = g + g/2 = 3g/2
(the FBD diagram is in the photo)
Now,
M(3g/2) – T = Ma ……..(1)
T – m(3g/2) = ma ………(2)
Adding the above two equations we get,
(3g/2)(M – m) = a(M + m)
=> a = (3g/2)(M – m)/(M + m)
So,
(2) => T = m(3g/2) + ma
=> T = m(3g/2) + (m)(3g/2)(M – m)/(M + m)
=> T = 3mgM/(M + m)
This is the tension in the string.
hope it's helps pls give it brainliest and follow me please
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