Question

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

Answer 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

Answer 2

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