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Question-22481

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Question Number 22481 by ajfour last updated on 19/Oct/17
Commented by ajfour last updated on 19/Oct/17
Q. 22479 (Solution)
$${Q}.\:\mathrm{22479}\:\left({Solution}\right) \\ $$
Answered by ajfour last updated on 19/Oct/17
When the ladder is about to slip: If μ_1 ≠0 , μ_2 ≠0 f_2 =μ_2 N_2 =N_1 ....(i) f_1 +N_2 =μ_1 N_1 +N_2 =mg ⇒ (μ_1 μ_2 +1)N_2 =mg N_2 =((mg)/(1+μ_1 μ_2 )) considering τorque about centre: N_2 ((l/2))cos θ=f_2 ((l/2))sin θ+ f_1 ((l/2))cos θ+N_1 ((l/2))sin θ ⇒ N_2 (1−μ_2 tan θ)=N_1 (μ_1 +tan θ) .....(iii) ⇒ If μ_1 =0 , μ_2 ≠0 then from (ii) we get N_2 =mg, and N_1 =μ_2 mg . Then from (iii) we get mg(1−μ_2 tan θ)=N_1 tan θ ⇒ mg−N_1 tan θ=N_1 tan θ or N_1 tan θ= ((mg)/2) . If μ_1 ≠0 , μ_2 =0 for the about to slip case f_2 =0 ⇒ N_1 =0 , θ=(π/2) N_2 =mg . Hence (3), (4) are correct .
$${When}\:{the}\:{ladder}\:{is}\:{about}\:{to}\:{slip}: \\ $$$${If}\:\:\:\:\:\mu_{\mathrm{1}} \neq\mathrm{0}\:,\:\mu_{\mathrm{2}} \neq\mathrm{0}\: \\ $$$${f}_{\mathrm{2}} =\mu_{\mathrm{2}} {N}_{\mathrm{2}} ={N}_{\mathrm{1}} \:\:\:….\left({i}\right) \\ $$$${f}_{\mathrm{1}} +{N}_{\mathrm{2}} =\mu_{\mathrm{1}} {N}_{\mathrm{1}} +{N}_{\mathrm{2}} ={mg}\:\:\: \\ $$$$\Rightarrow\:\:\:\left(\mu_{\mathrm{1}} \mu_{\mathrm{2}} +\mathrm{1}\right){N}_{\mathrm{2}} ={mg} \\ $$$$\:\:\:\:\:\:\:\:{N}_{\mathrm{2}} =\frac{{mg}}{\mathrm{1}+\mu_{\mathrm{1}} \mu_{\mathrm{2}} }\: \\ $$$${considering}\:\tau{orque}\:{about}\:{centre}: \\ $$$${N}_{\mathrm{2}} \left(\frac{{l}}{\mathrm{2}}\right)\mathrm{cos}\:\theta={f}_{\mathrm{2}} \left(\frac{{l}}{\mathrm{2}}\right)\mathrm{sin}\:\theta+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}_{\mathrm{1}} \left(\frac{{l}}{\mathrm{2}}\right)\mathrm{cos}\:\theta+{N}_{\mathrm{1}} \left(\frac{{l}}{\mathrm{2}}\right)\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:{N}_{\mathrm{2}} \left(\mathrm{1}−\mu_{\mathrm{2}} \mathrm{tan}\:\theta\right)={N}_{\mathrm{1}} \left(\mu_{\mathrm{1}} +\mathrm{tan}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\left({iii}\right) \\ $$$$\Rightarrow\:{If}\:\mu_{\mathrm{1}} =\mathrm{0}\:,\:\mu_{\mathrm{2}} \neq\mathrm{0}\:\:\:{then} \\ $$$${from}\:\left({ii}\right)\:{we}\:{get}\:{N}_{\mathrm{2}} ={mg},\:{and} \\ $$$${N}_{\mathrm{1}} =\mu_{\mathrm{2}} {mg}\:.\:{Then}\:{from}\:\left({iii}\right)\:{we}\:{get} \\ $$$$\:\:\:\:{mg}\left(\mathrm{1}−\mu_{\mathrm{2}} \mathrm{tan}\:\theta\right)={N}_{\mathrm{1}} \mathrm{tan}\:\theta \\ $$$$\Rightarrow\:\:\:{mg}−{N}_{\mathrm{1}} \mathrm{tan}\:\theta={N}_{\mathrm{1}} \mathrm{tan}\:\theta \\ $$$${or}\:\:\:\:\:\:{N}_{\mathrm{1}} \mathrm{tan}\:\theta=\:\frac{{mg}}{\mathrm{2}}\:. \\ $$$${If}\:\:\mu_{\mathrm{1}} \neq\mathrm{0}\:,\:\mu_{\mathrm{2}} \:=\mathrm{0}\:\: \\ $$$${for}\:{the}\:{about}\:{to}\:{slip}\:{case} \\ $$$${f}_{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:\:\:{N}_{\mathrm{1}} =\mathrm{0}\:\:,\:\theta=\frac{\pi}{\mathrm{2}} \\ $$$${N}_{\mathrm{2}} ={mg}\:. \\ $$$${Hence}\:\:\left(\mathrm{3}\right),\:\left(\mathrm{4}\right)\:{are}\:{correct}\:. \\ $$
Commented by Tinkutara last updated on 19/Oct/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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