Question Number 29198 by ajfour last updated on 05/Feb/18
Commented by ajfour last updated on 05/Feb/18
$${uniform}\:{rod}. \\ $$
Answered by mrW2 last updated on 05/Feb/18
Commented by mrW2 last updated on 05/Feb/18
![N_1 =f_2 N_2 =mg−f_1 N_2 l cos θ−mg(l/2)cos θ−N_1 l sin θ=0 case 1: f_1 =μ_1 N_1 , f_2 ≤μ_2 N_2 =μ_2 (mg−μ_1 f_2 ) f_2 ≤((μ_2 mg)/(1+μ_1 μ_2 )) (mg−μ_1 f_2 )l cos θ−mg(l/2)cos θ−f_2 l sin θ=0 f_2 (μ_1 +tan θ)=((mg)/2) ⇒f_2 =((mg)/(2(μ_1 +tan θ)))≤((μ_2 mg)/(1+μ_1 μ_2 )) ⇒(1/(μ_1 +tan θ))≤((2μ_2 )/(1+μ_1 μ_2 )) ⇒tan θ≥((1−μ_1 μ_2 )/(2μ_2 )) ⇒θ≥tan^(−1) (((1−μ_1 μ_2 )/(2μ_2 ))) case 2: f_2 =μ_2 N_2 , f_1 ≤μ_1 N_1 =μ_1 μ_2 (mg−f_1 ) ⇒f_1 ≤((μ_1 μ_2 mg)/(1+μ_1 μ_2 )) (mg−f_1 )l cos θ−mg(l/2)cos θ−μ_2 (mg−f_1 )l sin θ=0 −f_1 cos θ+((mg)/2)cos θ−μ_2 mgsin θ+μ_2 f_1 sin θ=0 (1−μ_2 tan θ)f_1 =((mg)/2)(1−2μ_2 tan θ) ⇒f_1 =((mg(1−2μ_2 tan θ))/(2(1−μ_2 tan θ))) ⇒f_1 =((mg(−(1/2)+1−μ_2 tan θ))/(1−μ_2 tan θ)) ⇒f_1 =mg[1−(1/(1−μ_2 tan θ))]≤((mgμ_1 μ_2 )/(1+μ_1 μ_2 )) (1/(1+μ_1 μ_2 ))≤(1/(1−μ_2 tan θ)) tan θ≥−μ_1 ⇒always true ⇒the rod slips when θ≤tan^(−1) (((1−μ_1 μ_2 )/(2μ_2 )))](https://www.tinkutara.com/question/Q29244.png)
$${N}_{\mathrm{1}} ={f}_{\mathrm{2}} \\ $$$${N}_{\mathrm{2}} ={mg}−{f}_{\mathrm{1}} \\ $$$${N}_{\mathrm{2}} {l}\:\mathrm{cos}\:\theta−{mg}\frac{{l}}{\mathrm{2}}\mathrm{cos}\:\theta−{N}_{\mathrm{1}} {l}\:\mathrm{sin}\:\theta=\mathrm{0} \\ $$$$ \\ $$$${case}\:\mathrm{1}:\:{f}_{\mathrm{1}} =\mu_{\mathrm{1}} {N}_{\mathrm{1}} ,\:{f}_{\mathrm{2}} \leqslant\mu_{\mathrm{2}} {N}_{\mathrm{2}} =\mu_{\mathrm{2}} \left({mg}−\mu_{\mathrm{1}} {f}_{\mathrm{2}} \right) \\ $$$${f}_{\mathrm{2}} \leqslant\frac{\mu_{\mathrm{2}} {mg}}{\mathrm{1}+\mu_{\mathrm{1}} \mu_{\mathrm{2}} } \\ $$$$\left({mg}−\mu_{\mathrm{1}} {f}_{\mathrm{2}} \right){l}\:\mathrm{cos}\:\theta−{mg}\frac{{l}}{\mathrm{2}}\mathrm{cos}\:\theta−{f}_{\mathrm{2}} {l}\:\mathrm{sin}\:\theta=\mathrm{0} \\ $$$${f}_{\mathrm{2}} \left(\mu_{\mathrm{1}} +\mathrm{tan}\:\theta\right)=\frac{{mg}}{\mathrm{2}} \\ $$$$\Rightarrow{f}_{\mathrm{2}} =\frac{{mg}}{\mathrm{2}\left(\mu_{\mathrm{1}} +\mathrm{tan}\:\theta\right)}\leqslant\frac{\mu_{\mathrm{2}} {mg}}{\mathrm{1}+\mu_{\mathrm{1}} \mu_{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mu_{\mathrm{1}} +\mathrm{tan}\:\theta}\leqslant\frac{\mathrm{2}\mu_{\mathrm{2}} }{\mathrm{1}+\mu_{\mathrm{1}} \mu_{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{tan}\:\theta\geqslant\frac{\mathrm{1}−\mu_{\mathrm{1}} \mu_{\mathrm{2}} }{\mathrm{2}\mu_{\mathrm{2}} } \\ $$$$\Rightarrow\theta\geqslant\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\mu_{\mathrm{1}} \mu_{\mathrm{2}} }{\mathrm{2}\mu_{\mathrm{2}} }\right) \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:{f}_{\mathrm{2}} =\mu_{\mathrm{2}} {N}_{\mathrm{2}} ,\:{f}_{\mathrm{1}} \leqslant\mu_{\mathrm{1}} {N}_{\mathrm{1}} =\mu_{\mathrm{1}} \mu_{\mathrm{2}} \left({mg}−{f}_{\mathrm{1}} \right) \\ $$$$\Rightarrow{f}_{\mathrm{1}} \leqslant\frac{\mu_{\mathrm{1}} \mu_{\mathrm{2}} {mg}}{\mathrm{1}+\mu_{\mathrm{1}} \mu_{\mathrm{2}} } \\ $$$$\left({mg}−{f}_{\mathrm{1}} \right){l}\:\mathrm{cos}\:\theta−{mg}\frac{{l}}{\mathrm{2}}\mathrm{cos}\:\theta−\mu_{\mathrm{2}} \left({mg}−{f}_{\mathrm{1}} \right){l}\:\mathrm{sin}\:\theta=\mathrm{0} \\ $$$$−{f}_{\mathrm{1}} \mathrm{cos}\:\theta+\frac{{mg}}{\mathrm{2}}\mathrm{cos}\:\theta−\mu_{\mathrm{2}} {mg}\mathrm{sin}\:\theta+\mu_{\mathrm{2}} {f}_{\mathrm{1}} \mathrm{sin}\:\theta=\mathrm{0} \\ $$$$\left(\mathrm{1}−\mu_{\mathrm{2}} \mathrm{tan}\:\theta\right){f}_{\mathrm{1}} =\frac{{mg}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2}\mu_{\mathrm{2}} \mathrm{tan}\:\theta\right) \\ $$$$\Rightarrow{f}_{\mathrm{1}} =\frac{{mg}\left(\mathrm{1}−\mathrm{2}\mu_{\mathrm{2}} \mathrm{tan}\:\theta\right)}{\mathrm{2}\left(\mathrm{1}−\mu_{\mathrm{2}} \mathrm{tan}\:\theta\right)} \\ $$$$\Rightarrow{f}_{\mathrm{1}} =\frac{{mg}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}−\mu_{\mathrm{2}} \mathrm{tan}\:\theta\right)}{\mathrm{1}−\mu_{\mathrm{2}} \mathrm{tan}\:\theta} \\ $$$$\Rightarrow{f}_{\mathrm{1}} ={mg}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−\mu_{\mathrm{2}} \mathrm{tan}\:\theta}\right]\leqslant\frac{{mg}\mu_{\mathrm{1}} \mu_{\mathrm{2}} }{\mathrm{1}+\mu_{\mathrm{1}} \mu_{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mu_{\mathrm{1}} \mu_{\mathrm{2}} }\leqslant\frac{\mathrm{1}}{\mathrm{1}−\mu_{\mathrm{2}} \mathrm{tan}\:\theta} \\ $$$$\mathrm{tan}\:\theta\geqslant−\mu_{\mathrm{1}} \:\Rightarrow{always}\:{true} \\ $$$$ \\ $$$$\Rightarrow{the}\:{rod}\:{slips}\:{when}\:\theta\leqslant\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\mu_{\mathrm{1}} \mu_{\mathrm{2}} }{\mathrm{2}\mu_{\mathrm{2}} }\right) \\ $$
Commented by ajfour last updated on 05/Feb/18
$${Great}!\:{you}\:{have}\:{been}\:{too}\:{careful} \\ $$$${Sir},\:{i}\:{simply}\:{took}\:{friction}\:{at} \\ $$$${each}\:{surface}\:{at}\:{its}\:{maximum}, \\ $$$${shall}\:{try}\:{to}\:{understand}\:{your} \\ $$$${carefulness}\:{Sir}. \\ $$
Commented by mrW2 last updated on 05/Feb/18
$${Your}\:{right}.\:{There}\:{is}\:{no}\:{need}\:{to}\:{be} \\ $$$${such}\:{careful}.\:{If}\:{one}\:{end}\:{slips},\:{the} \\ $$$${other}\:{end}\:{slips}\:{too}.\:{We}\:{can}\:{put}\:{max}. \\ $$$${friction}\:{on}\:{both}\:{ends}\:{at}\:{the}\:{same}\:{time}. \\ $$