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Question-22481




Question Number 22481 by ajfour last updated on 19/Oct/17
Commented by ajfour last updated on 19/Oct/17
Q. 22479 (Solution)
Q.22479(Solution)
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 .
Whentheladderisabouttoslip:Ifμ10,μ20f2=μ2N2=N1.(i)f1+N2=μ1N1+N2=mg(μ1μ2+1)N2=mgN2=mg1+μ1μ2consideringτorqueaboutcentre:N2(l2)cosθ=f2(l2)sinθ+f1(l2)cosθ+N1(l2)sinθN2(1μ2tanθ)=N1(μ1+tanθ)..(iii)Ifμ1=0,μ20thenfrom(ii)wegetN2=mg,andN1=μ2mg.Thenfrom(iii)wegetmg(1μ2tanθ)=N1tanθmgN1tanθ=N1tanθorN1tanθ=mg2.Ifμ10,μ2=0fortheabouttoslipcasef2=0N1=0,θ=π2N2=mg.Hence(3),(4)arecorrect.
Commented by Tinkutara last updated on 19/Oct/17
Thank you very much Sir!
ThankyouverymuchSir!

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