Question Number 224358 by fantastic last updated on 05/Sep/25
Commented by fantastic last updated on 05/Sep/25
Answered by mr W last updated on 05/Sep/25
$${tension}\:{between}\:{m}_{\mathrm{2}} \:{and}\:{m}_{\mathrm{3}} : \\ $$$${T}_{\mathrm{2}} =\mu{m}_{\mathrm{3}} {g}=\mathrm{0}.\mathrm{25}×\mathrm{4}×\mathrm{9}.\mathrm{81}=\mathrm{9}.\mathrm{81}\:{N} \\ $$$${tension}\:{between}\:{m}_{\mathrm{1}} \:{and}\:{m}_{\mathrm{2}} : \\ $$$${T}_{\mathrm{1}} ={m}_{\mathrm{1}} {g}={T}_{\mathrm{2}} +{m}_{\mathrm{2}} {g}\left(\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow{m}_{\mathrm{1}} =\mu{m}_{\mathrm{3}} +{m}_{\mathrm{2}} \left(\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}.\mathrm{25}×\mathrm{4}+\mathrm{4}×\left(\mathrm{0}.\mathrm{6}+\mathrm{0}.\mathrm{25}×\mathrm{0}.\mathrm{8}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}.\mathrm{2}\:{kg} \\ $$
Commented by fantastic last updated on 05/Sep/25
$${Thanks}\:{sir}. \\ $$$${Your}\:{approach}\:{is}\:{always}\:{great}\:\:\:\: \\ $$
Answered by fantastic last updated on 05/Sep/25
Commented by fantastic last updated on 05/Sep/25
$${m}_{\mathrm{1}} {g}−{T}_{{m}_{\mathrm{1}} } ={ma}=\mathrm{0} \\ $$$${T}_{{m}_{\mathrm{1}} } ={m}_{\mathrm{1}} {g} \\ $$$${m}_{\mathrm{1}} {g}={m}_{\mathrm{2}} {g}\mathrm{sin}\:\theta+\mu{m}_{\mathrm{2}} {g}\mathrm{cos}\:\theta+\mu{gm}_{\mathrm{3}} \\ $$$${m}_{\mathrm{1}} ={m}_{\mathrm{2}} \left(\mathrm{sin}\:\theta+\mu\mathrm{cos}\:\theta\right)+\mu{m}_{\mathrm{3}} \\ $$$$=\mathrm{4}\left(\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{4}}{\mathrm{5}}\right)+\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{4}=\frac{\mathrm{16}}{\mathrm{5}}+\mathrm{1}=\frac{\mathrm{21}}{\mathrm{5}}=\mathrm{4}.\mathrm{2}{kg} \\ $$$${T}−\mu{m}_{\mathrm{3}} {g}={m}_{\mathrm{3}} {a} \\ $$$${T}={m}_{\mathrm{3}} \left({a}+\mu{g}\right)=\mathrm{4}×\left(\mathrm{0}+\frac{\mathrm{9}.\mathrm{8}}{\mathrm{4}}\right)=\mathrm{9}.\mathrm{8}{N} \\ $$