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Discussion Response Post to: hc verma pg no 81(newton's laws of motion) no 31

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sipnali.... (152)

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For block of mass M, T + Ma – Mg = 0 ...........(i)

For pulley B, T₂= 2T

For block of mass 2M, 2T = 2M(a/2)

or, Ma = 2T

T = Ma /2.

put T = Ma/2 in eq. (i),

Ma/2 + ma = Mg. (because T = Ma/2)

3 Ma = 2 Mg

a = 2g/3

a) acceleration of mass M is 2g/3.

b) Tension T = Ma/2 = M/2 =2g/3 =Mg/3

c) resultant of tensions = force exerted by the clamp on the pulley

R = root( T² + T² ) = T(root2) = (root2)Mg/3

= T/T = 1

= 45

So, it is (root2)Mg/3 at an angle of 45° with horizontal

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