A constant force F=20N acts on a block of mass 2 kg which is connected to two blocks of masses m_(1) = 1.0 kgand m_(2) = 2 kg as shown in Calculate the accelerations produced in ball the three blocks Assume pulleys are frictionless and weightless .

A constant force F=20N acts on a block of mass 2 kg which is connected

1.	A constant force F=20N acts on a block of mass 2 kg which is connected to two blocks of masses m_(1) = 1.0 kgand m_(2) = 2 kg as shown in Calculate the accelerations produced in ball the three blocks Assume pulleys are frictionless and weightless .
Answer» Solution :The free body diagrams of three blocks of masses `M,m_(1)` and `m_(2)` are shown in Eduations of MOTION of three blocks are `Ma = F - T`…(i) `m_(1) a_(1) = 2 T - m_(1) g` …(ii) `m_(2)a_(2) = m_(2)g - T `…(iii) Now if mass `M` MOVES to the LEFT through a distance X and mass `m_(2)` moves downwards through the same distance x then the distance travelled by mass `m_(1)` is 2x upwards Therefore sum ofthe accelerations of M and `m_(2)` is double the acceleration of `m_(1)` `a + a_(2) = 2 a_(1)` From (i) `T = F - Ma = 20 - 2 a ` ...(iv) ltbr) From (iii) `T = m_(2)g - m_(2) a_(2) = 2xx 9.8 - 2 a_(2)` ...(v) From (ii), `2 T = m_(1) a_(1) + m_(1) g = 1 a_(1) + 9.8 ` ...(vi) Add (v) and (vi) `2 T = (20 -2a) + (19.6 -2 a_(2)) = 39.6 -2 (a+ a_(2)) = 39.6-4 a_(1)`...(VII) Using (vii),` a_(1) + 9.8 = 39.6 - 4 a_(1) ` ` a_(1) = 39.6 - 9.8 = 29.8 ` `a_(1) = (29.8)/(5) = 5.96 m//s^(2)` From (vii) ` 2 T = 5.96 + 9.8 = 15.76 ` ` T = (15.76)/(2) ` From (iv) ` 2 a_(2) = 19.6 - T = 19.6 - 7.88 = 11.78 ` `a_(2) = (11.78)/(2)` From (iv) ` a =2 a_(1) - a_(2) = 2 xx 5.96 - 5.89 = 11.92 - 5.89 = 6.03 m//s^(2)` .