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Question-29198




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

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