Question Number 204405 by mathdave last updated on 16/Feb/24
Commented by mathdave last updated on 16/Feb/24
$${pls}\:{someone}\:{should}\:{me}\:{out}\: \\ $$
Answered by mr W last updated on 16/Feb/24
Commented by mr W last updated on 16/Feb/24
$${N}_{\mathrm{1}} +{T}\:\mathrm{sin}\:\theta={G}_{\mathrm{1}} \\ $$$${T}\:\mathrm{cos}\:\theta={f}_{\mathrm{1}} \:\:{with}\:{f}_{\mathrm{1}} =\mu{N}_{\mathrm{1}} \\ $$$${N}_{\mathrm{1}} +{G}_{\mathrm{2}} ={N}_{\mathrm{2}} \\ $$$${P}={f}_{\mathrm{1}} +{f}_{\mathrm{2}} \:\:{with}\:{f}_{\mathrm{2}} =\mu{N}_{\mathrm{2}} \\ $$$$ \\ $$$${N}_{\mathrm{1}} ={G}_{\mathrm{1}} −{T}\:\mathrm{sin}\:\theta={G}_{\mathrm{1}} −{f}_{\mathrm{1}} \:\mathrm{tan}\:\theta={G}_{\mathrm{1}} −\mu{N}_{\mathrm{1}} \:\mathrm{tan}\:\theta \\ $$$$\Rightarrow{N}_{\mathrm{1}} =\frac{{G}_{\mathrm{1}} }{\mathrm{1}+\mu\:\mathrm{tan}\:\theta} \\ $$$${N}_{\mathrm{2}} ={G}_{\mathrm{2}} +\frac{{G}_{\mathrm{1}} }{\mathrm{1}+\mu\:\mathrm{tan}\:\theta} \\ $$$${P}=\mu\left({G}_{\mathrm{2}} +\frac{\mathrm{2}{G}_{\mathrm{1}} }{\mathrm{1}+\mu\:\mathrm{tan}\:\theta}\right) \\ $$$$\:\:\:=\mathrm{0}.\mathrm{3}\left(\mathrm{2}+\frac{\mathrm{2}×\mathrm{1}}{\mathrm{1}+\mathrm{0}.\mathrm{3}×\mathrm{tan}\:\mathrm{30}°}\right)\approx\mathrm{1}.\mathrm{11}\:{KN} \\ $$
Commented by mathdave last updated on 16/Feb/24
$${thanks}\:{so}\:{much}\:{mr}\:{W} \\ $$