Question Number 109201 by aurpeyz last updated on 21/Aug/20
Commented by aurpeyz last updated on 21/Aug/20
$${pls}\:{help} \\ $$
Answered by Dwaipayan Shikari last updated on 22/Aug/20
$${Accelaration}=\frac{{F}}{{M}_{{sys}} }=\frac{\mathrm{6}}{\mathrm{13}}\:\frac{{m}}{{s}^{\mathrm{2}} } \\ $$$${T}_{\mathrm{1}} ={F}−{ma}=\mathrm{6}−\mathrm{6}.\frac{\mathrm{6}}{\mathrm{13}}=\mathrm{3}.\mathrm{2}{N} \\ $$$${T}_{\mathrm{2}} ={T}_{\mathrm{1}} −{m}_{\mathrm{0}} {a}=\mathrm{3}.\mathrm{2}−\mathrm{4}.\frac{\mathrm{6}}{\mathrm{13}}=\mathrm{1}.\mathrm{36}{N} \\ $$
Commented by aurpeyz last updated on 02/Sep/20
$${is}\:{line}\:\mathrm{3}\:{not}\:{supposed}\:{to}\:{be}\:\mathrm{T}_{\mathrm{2}} =\mathrm{T}_{\mathrm{1}} −{m}_{\mathrm{2}} {a}\:?? \\ $$