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The pulley A and C are fixed while the pulley B is movable. A mass M_2 is attached to pulley B , while the string has masses M_1 and M_3 at the two ends. Find the acceleration of each mass . |
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Answer» SOLUTION :The force acting on masses ` M_1 , M_2 and M_3` are shown in figure . Let ` a_2 and a_3` be upward accelerationof masses ` M_1 and M_3 and a_2` the downward acceleration of mass `M_2` . For motion of mass `M_1 ` ` T-M_1 g = M_1 a_1 `....... (1) For motion of mass `M_3` `T-M_3g = M_3 a_3` ........ (2) For motion of mass `M_2` `M_2g - 2T= M_2 a_2` ......... (3) Let ` l_1 , l_2 , l_3` be the lengths of vertical portions of the STRING between pulley at any instant. `M_1` goes up through ` x , M_2` goes down through y and `M_3` goes up through z. Then ` l_1 + 2l_2 + l_3 = ( l_1 - x) + 2 (l_2 + y) + l_3 -z` `implies x + z = 2y implies (d^(2) x)/( dt^(2)) + (d^(2) z)/(dt^(2)) = 2(d^(2) y)/(dt^(2))` ` a_1 + a_3 = 2a_2 implies a_2 =(a_1 + a_3)/( 2) `.......... (4) From (1) , ` a_1 =(T)/(M_1) - g ` ....... (5)  From (2) ` , a_3 = (T)/(M_3)-g` ............ (6) substituting ` a_1 and a_3 ` is (4) , we get ` a_2 =(T)/(2) ((1)/(M_1) + (1)/(M_3))-g`......... (7) From (3) , `T=(M_2(g-a_2))/(2)` .......... (8) substituting this value in (7) , we get `a_2 =(M_2)/(4) (g-a_2)((1)/(M_1 ) + (1)/(M_3))-g or a_2 = (M_2 (g-a_2) (M_3+M_1))/(4M_1M_3)-g` ` or 4M_1M_3 a_2 _ M_2(M_1 + M_3) g -M_2(M_1 + M_3) a_2- 4M_1 M_3 g ` `or [4M_1M_3 + M_2(M_1 + M_3)]a_2 = (M_1 + M_3) M_2 g - 4M_1 M_3 g ` ` a_2 = ((M_1 + M_3) M_2 g - 4M_1 M_3g)/(4M_1 M_3 + M_2(M_1 + M_3))` `a_2=(M_1M_2 + M_2 M_3 - 4M_1 M_3)/(M_1 M_2 + M_2M_3 + 4M_1 M_3 )g`...... (9) substituting value of T from (8) in (5) ` a_1 =(M_2(g-a_2))/(2M_1)-g` Substituting value of`a_2` in above equation , we get ` a_1 =(-M_1 M-2 + 3M_2 M_3- 4M_1 M_3)/(M_1 M_2 + M_2M_3 + 4M_1 M_3)g` ........ (10) SIMILARLY from (6) and (9) `a_3=(3M_1M_2 - M_2 M_3 - 4M_1 M_3)/( M_1 M_2 + M_2 M_3 -4M_1 M_3)` ........ (11) |
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