Three blocks of masses m1, m2 and m3 are connected as shown in figure. All the surfaces are frictionless and the string and the pulleys are light. Find the acceleration of m1

hi ‚ÄčAtul Tripathi ,

Please find below the solution to the asked query:


as we can see in the diagram:

let us say the tension in the string attached to m1 be T1 and the tension in the string attached to m2 and m3 be T2

Let us say the acceleration of m1 be a1 and the acceleration of m2 be a2 while the acceleration of m3 be a3.


For m1 mass:writing the equation of motion:T1 = m1a1   ...equation 1for m2 mass:T2 - m2g = m2a2   ...equation 2for m3 mass:m3g -T2 = m3a3    ...equation 3as we know the pulley is massless so T1 = 2T2   ...equation 4while if we assume our reference frame to be the hanging pulley which is accelerated by a1 downwardswe can say a2 + a1 = a3 - a1 (as in its frame both the masses will have the same acceleration)so 2a1 =  a3 -a2   ...equation 5substituting  equation 4and equation 5  in equation 1we get :2T2 = m1(a3 - a2)2or, (a3 - a2) =-(a2- a3)= 4T2m1  .....equation 6from equation 2 and equation 3a2 = T2 - m2g m2 and a3 = m3g -T2 m3so -a2+ a3=- T2 - m2g m2+ m3g -T2 m3using equation 64T2m1 =- T2 - m2g m2+m3g -T2 m34T2m1 = -m3(T2 - m2g ) +m2(m3g -T2 )m2m34T2m1 = -m3T2 +m3m2g +m3m2g -m2T2 m2m34T2m2m3 = -m1m3T2 +2m3m2m1g -m1m2T2T2(4m2m3 +m1m3+m1m2) = +2m3m2m1gT2(+4m2m3 +m1m3+m1m2)  =2m3m2m1gT2 = 2m3m2m1g(m1m3+m1m2+4m2m3 )  so as a1 = 2T2m1 = 4m3m2g(m1m3+m1m2+4m2m3 )  Answer


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