The masses `m_(1) m_(2)` and `m_(3)` of the three bodies shown in fig . Are 5 , 2 and 3 kg respectively Calculate the valuse of tension `T_(1) T_(2)` and `T_(3)` when (i) the whole system is going upward with an acceleration of `2 m//s^(2)` (ii) the whole system is stationary ` (g=9.8 m//s^(2))` .
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Two masses of 5kg and 3kg are suspended with help of massless inextensible strings as shown in figure. Calculate T_(1) and T_(2) when whole system is going upwards with acceleration =2m//s^(2) (use g = 9.8 ms^(-2)) .

The masses M_(1), M_(2) and M_(3) are 5, 2 and 3 kg respectively. These have been joined using massless, inextensible pieces of strings as shown in the figure. If the whole system is going upward with an acceleration of 2ms^(-2) , then the value of tensions T_(1), T_(2) and T_(3) respectively are ["take "g=9.8ms^(-2)]

A lift of mass 100 kg is moving upwards with an acceleration of 1 m//s^(2) . The tension developed in the string, which is connected to lift is ( g=9.8m//s^(2) )

Two objects A and B of masses m_(A) and m_(B) are attached by strings as shown in fig. If they are given upward acceleration then the rato of tension T_(1) : T_(2) is-

Two masses m_(1) and m_(2) are accelerated uniformly on a frictionless surface as shown in figure given below. The ratio of the tensions T_(1)/T_(2) is:

In the figure shown, find acceleration of the system and tenslons T_(1),T_(2) and T_(3) (Take = 10 m//s^(2))

Three blocks m_(1), m_(2) , and m_(3) are arranged as shown in fig. if m_(2)=5 kg and m_(3)=8kg , at what value of m_(1) will 8kg mass be at rest?