Two bodies of masses m1 & m2 are connected by a light string which passes over a frictionless, massless pulley. If the pulley is moving upward with uniform acceleration g/2, then tension in the string will be
A) 3m1m2g/m1+m2 B)m1+m2 g/4m1m2 C)2m1m2 g/m1+m2 D)m1m2 g/m1+m2

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2014-12-18T23:53:54+05:30

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Let   m2 > m1.

Consider the masses m1 and m2, with forces Tension T in the string upwards and their weights downwards, acting on them.  Both of these masses are moving with same acceleration a up (m1) or down (m2 as m2>m1), as the string is tight and has a uniform tension all along its length.    This explanation is valid in the frame of reference of moving pulley with an acceleration g/2 upwards.

In the frame of reference of pulley, with an acceleration g/2 upwards, the two masses are having acceleration a and -a.  To balance forces in this non-inertial frame, we need to add a fictitious force of  m1 g/2 downwards on m1 and -m2 g/2 downwards on m2.  Then we write free body diagrams and apply Newtons' laws.

           T - (m1 g +m1 g/2) = m1 a      =>    T - 3 m1*g/2 = m1 a
           (m2 g +m2 g/2) - T = m2 a      =>    3 m2*g/2 - T = m2 a

   Add the two equations to get

                   a = (3g/2) * (m2 - m1) / (m1 + m2)
    then,
                   T = 3m1 g/2 + m1 a  =  3 m1 m2  g  / (m1 + m2)

     Option A is correct.

In the inertial frame wrt  a stationary person,

   acceleration of the mass m1 = a1 = a + g / 2    upwards
               a1 = g * (2 m2 - m1) / (m1 + m2)      upwards

   acceleration of mass m2 = a2 = a - g / 2        downwards
               a2 =  g * (m2 - 2 m1) / (m1 + m2)    downwards

You can verify,  by putting, m1=m2,  then a = 0 in the frame  of pulley, as both weights are same.  But in the inertial frame, a1 = g/2,  and
          a2 = -g/2 downwards or,  g/2 upwards.


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