Question Number 18379 by virus last updated on 19/Jul/17
Answered by mrW1 last updated on 19/Jul/17
$$\mathrm{T}=\mathrm{tension}\:\mathrm{in}\:\mathrm{rope} \\ $$$$\mathrm{4T}=\mathrm{m}_{\mathrm{A}} \mathrm{a}_{\mathrm{A}} \:\:\:...\left(\mathrm{i}\right) \\ $$$$\mathrm{F}−\mathrm{5T}=\mathrm{m}_{\mathrm{B}} \mathrm{a}_{\mathrm{B}} \:\:\:...\left(\mathrm{ii}\right) \\ $$$$\mathrm{5a}_{\mathrm{B}} =\mathrm{4a}_{\mathrm{A}} \:\:\:\:...\left(\mathrm{iii}\right) \\ $$$$\Rightarrow\frac{\mathrm{F}−\mathrm{5T}}{\mathrm{4T}}=\frac{\mathrm{m}_{\mathrm{B}} }{\mathrm{m}_{\mathrm{A}} }×\frac{\mathrm{a}_{\mathrm{B}} }{\mathrm{a}_{\mathrm{A}} }=\frac{\mathrm{3M}}{\mathrm{M}}×\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\mathrm{5F}−\mathrm{25T}=\mathrm{48T} \\ $$$$\mathrm{T}=\frac{\mathrm{5}}{\mathrm{73}}\mathrm{F} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{A}} =\frac{\mathrm{4T}}{\mathrm{m}_{\mathrm{A}} }=\frac{\mathrm{20F}}{\mathrm{73M}} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{B}} =\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{20F}}{\mathrm{73M}}=\frac{\mathrm{16F}}{\mathrm{73M}} \\ $$
Commented by virus last updated on 21/Jul/17
$${yes}\:{sir}\:{answer}\:{is}\:{correct}\:{and}\:{thank}\:{you} \\ $$