Question Number 224302 by fantastic last updated on 31/Aug/25
Commented by fantastic last updated on 31/Aug/25
$${friction}\:{coifficicient}=\mu \\ $$$${F}\:\:{force}\:{is}\:{working}\:{on}\:{block}\:{horizontally} \\ $$$${the}\:{block}\:{is}\:{going}\:{upward}\:{with}\: \\ $$$${constant}\:{velocity}. \\ $$$${F}=? \\ $$
Answered by mr W last updated on 01/Sep/25
$$\mathrm{tan}\:\phi=\mu \\ $$$${F}={mg}\:\mathrm{tan}\:\left(\theta+\phi\right) \\ $$$$\:\:\:\:\:=\frac{{mg}\left(\mathrm{tan}\:\theta+\mathrm{tan}\:\phi\right)}{\mathrm{1}−\mathrm{tan}\:\theta\:\mathrm{tan}\:\phi}=\frac{{mg}\left(\mu+\mathrm{tan}\:\theta\right)}{\mathrm{1}−\mu\:\mathrm{tan}\:\theta} \\ $$
Commented by mr W last updated on 01/Sep/25
Commented by mr W last updated on 01/Sep/25
Commented by mr W last updated on 01/Sep/25
$${we}\:{see}\:{the}\:{smallest}\:{force}\:{we}\:{need} \\ $$$${to}\:{make}\:{the}\:{block}\:{to}\:{move}\:{upwards} \\ $$$${is}\:{F}_{{min}} ={mg}\:\mathrm{sin}\:\left(\theta+\phi\right),\:{when}\:{F}\:{is}\: \\ $$$${not}\:{parallel}\:{to}\:{the}\:{inclined}\:{plane},\: \\ $$$${but}\:{at}\:{an}\:{angle}\:\phi. \\ $$