Question-225980

Question Number 225980 by fantastic2 last updated on 17/Nov/25

Commented by mr W last updated on 18/Nov/25

α_(max) =sin^(−1) ((mμ)/((m+M)(√(1+μ^2 ))))

$$\alpha_{{max}} =\mathrm{sin}^{−\mathrm{1}} \frac{{m}\mu}{\left({m}+{M}\right)\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }} \\ $$

Commented by fantastic2 last updated on 18/Nov/25

$${right}\:{sir}!\: \\ $$

Commented by mr W last updated on 18/Nov/25

$${have}\:{you}\:{solved}\:{it}? \\ $$

Commented by fantastic2 last updated on 18/Nov/25

$${yes}\:{sir} \\ $$

Answered by fantastic2 last updated on 18/Nov/25

Commented by fantastic2 last updated on 18/Nov/25

as the inclination increases the cylender tends more to move so the block has to go further from the touching point(inclination−cylender) at α_(max) μ=tan γ moment of force formed by the cylender to move it clockwise F_c =Mg×d=MgRsin α_(max) moment of force formed by the block to move the cylender anticlockwise F_b =mg×d=mg(Rsin γ−Rsin α_(max) ) the normal force and friction force both are acting about the point the moment of force is considered.so the have no effect in net moment of force so MgRsin α_(max) =mg(Rsin γ−Rsin α_(max) ) sin α_(max) =((msin θ)/(M+m)) sin α_(max) =((mμ)/((M+m)(√(1+μ^2 ))))[∵tan γ=μ] α_(max) =sin^(−1) (((mμ)/((M+m)(√(1+μ^2 )))))

$${as}\:{the}\:{inclination}\:{increases}\: \\ $$$${the}\:{cylender}\:{tends}\:{more}\:{to}\:{move} \\ $$$${so}\:{the}\:{block}\:{has}\:{to}\:{go}\:{further} \\ $$$${from}\:{the}\:{touching}\:{point}\left({inclination}−{cylender}\right) \\ $$$${at}\:\alpha_{{max}} \\ $$$$\mu=\mathrm{tan}\:\gamma \\ $$$${moment}\:{of}\:{force}\:{formed}\:{by}\:{the}\:{cylender}\:{to} \\ $$$${move}\:{it}\:{clockwise} \\ $$$${F}_{{c}} ={Mg}×{d}={MgR}\mathrm{sin}\:\alpha_{{max}} \\ $$$${moment}\:{of}\:{force}\:{formed}\:{by}\:{the}\:{block}\:{to} \\ $$$${move}\:{the}\:{cylender}\:{anticlockwise} \\ $$$${F}_{{b}} ={mg}×{d}={mg}\left({R}\mathrm{sin}\:\gamma−{R}\mathrm{sin}\:\alpha_{{max}} \right) \\ $$$${the}\:{normal}\:{force}\:{and}\:{friction}\: \\ $$$${force}\:{both}\:{are}\:{acting}\:{about} \\ $$$${the}\:{point}\:{the}\:{moment}\:{of}\:{force} \\ $$$${is}\:{considered}.{so}\:{the}\:{have}\:{no}\: \\ $$$${effect}\:{in}\:{net}\:{moment}\:{of}\:{force} \\ $$$${so} \\ $$$${MgR}\mathrm{sin}\:\alpha_{{max}} ={mg}\left({R}\mathrm{sin}\:\gamma−{R}\mathrm{sin}\:\alpha_{{max}} \right) \\ $$$$\mathrm{sin}\:\alpha_{{max}} =\frac{{m}\mathrm{sin}\:\theta}{{M}+{m}} \\ $$$$\mathrm{sin}\:\alpha_{{max}} =\frac{{m}\mu}{\left({M}+{m}\right)\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}\left[\because\mathrm{tan}\:\gamma=\mu\right] \\ $$$$\alpha_{{max}} =\mathrm{sin}^{−\mathrm{1}} \left(\frac{{m}\mu}{\left({M}+{m}\right)\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}\right) \\ $$

Commented by mr W last updated on 18/Nov/25

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Answered by mr W last updated on 18/Nov/25

Commented by mr W last updated on 19/Nov/25

C is common COM from m and M. tan φ=μ ⇒sin φ=(μ/( (√(1+μ^2 )))) ((AC)/(AB))=(m/(M+m))=((AC)/(AD))=((sin α)/(sin φ)) ⇒sin α=((m sin φ)/(M+m))=((mμ)/((M+m)(√(1+μ^2 )))) ⇒α_(max) =sin^(−1) ((mμ)/((M+m)(√(1+μ^2 ))))

$${C}\:{is}\:{common}\:{COM}\:{from}\:{m}\:{and}\:{M}. \\ $$$$\mathrm{tan}\:\phi=\mu\:\Rightarrow\mathrm{sin}\:\phi=\frac{\mu}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }} \\ $$$$\frac{{AC}}{{AB}}=\frac{{m}}{{M}+{m}}=\frac{{AC}}{{AD}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\phi} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{{m}\:\mathrm{sin}\:\phi}{{M}+{m}}=\frac{{m}\mu}{\left({M}+{m}\right)\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }} \\ $$$$\Rightarrow\alpha_{{max}} =\mathrm{sin}^{−\mathrm{1}} \frac{{m}\mu}{\left({M}+{m}\right)\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }} \\ $$

Commented by fantastic2 last updated on 18/Nov/25

$$\underset{\overset{\frown} {\smile}} {\overset{\frown} {\bullet}\mid\:\bullet}\:\left[{rate}\:{my}\:{creativity}\right] \\ $$

Commented by ajfour last updated on 19/Nov/25

MgRsin α=mg(R−c)sin φ μmgcos φ=mgsin φ tan φ=μ so sin φ=(μ/( (√(1+μ^2 )))) MgRsin α=(μ/( (√(1+μ^2 ))))((M/(M+m)))mgR sin α=((μm)/((M+m)(√(1+μ^2 ))))

$${MgR}\mathrm{sin}\:\alpha={mg}\left({R}−{c}\right)\mathrm{sin}\:\phi \\ $$$$\mu{mg}\mathrm{cos}\:\phi={mg}\mathrm{sin}\:\phi \\ $$$$\mathrm{tan}\:\phi=\mu \\ $$$${so}\:\:\:\:\mathrm{sin}\:\phi=\frac{\mu}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }} \\ $$$${MgR}\mathrm{sin}\:\alpha=\frac{\mu}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}\left(\frac{{M}}{{M}+{m}}\right){mgR} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mu{m}}{\left({M}+{m}\right)\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }} \\ $$$$ \\ $$

Commented by fantastic2 last updated on 19/Nov/25

$${wow}\:{sir} \\ $$

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