Question Number 32440 by Tinkutara last updated on 25/Mar/18
Answered by ajfour last updated on 25/Mar/18
$$\mathrm{5}{g}−{T}=\mathrm{5}\left(\mathrm{2}{a}\right) \\ $$$$\mathrm{2}{T}−\mathrm{5}{g}=\mathrm{5}{a} \\ $$$$\Rightarrow\:\:\mathrm{5}{g}−{T}=\mathrm{4}{T}−\mathrm{10}{g} \\ $$$$\Rightarrow\:\:\:{T}=\mathrm{30}{N}\:. \\ $$
Commented by Tinkutara last updated on 25/Mar/18
But why the system will move? How string will be released because 50 kg is at rest?
Commented by ajfour last updated on 25/Mar/18
$${only}\:{the}\:\mathrm{5}{kg}\:{blocks}\:{shall}\:{move}. \\ $$$${the}\:{tension}\:{is}\:{obtained}\:{for}\:{this} \\ $$$${little}\:{duration}\:{until}\:{the}\:{middle} \\ $$$$\mathrm{5}{kg}\:{block}\:{comes}\:{almost}\:{at}\:{the}\:{level}\:{of} \\ $$$${pulleys}. \\ $$
Commented by Tinkutara last updated on 25/Mar/18
But if 50 kg is at rest how will rope be released because it should be of fixed length?
Commented by ajfour last updated on 25/Mar/18
still there is some scope for
middle 5kg block to rise and
the other 5kg block to descend.
Commented by Tinkutara last updated on 29/Mar/18
Thank you very much Sir! I got the answer. ��������