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Two masses of 5kg and 3kg are suspended ...

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))`.

Text Solution

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Given `m_(1) = 5` kg , `m_(2) = 3` kg
g = `9.8 m//s^(2)` and a = 2 m/ `s^(2)`

For the upper block
`T_(1) - T_(2) - 5 g = 5a `
`implies " " T_(1) - T_(2) = 5(g +a) " " .....(i) `
For the lower block `" " T_(2) - 3g = 3a`
`implies " " T_(2) = 3(g+a) = 3(9.8 + 2) = 35.4 N`
From Eq. (i) ` " " T_(1) = T_(2) + 5(g +a)`
= `35.4 + 5(9.8 + 2) = 94.4 N`
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