Question-22481 Tinku Tara June 4, 2023 Others 0 Comments FacebookTweetPin Question Number 22481 by ajfour last updated on 19/Oct/17 Commented by ajfour last updated on 19/Oct/17 Q.22479(Solution) Answered by ajfour last updated on 19/Oct/17 Whentheladderisabouttoslip:Ifμ1≠0,μ2≠0f2=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)sinθ⇒N2(1−μ2tanθ)=N1(μ1+tanθ)…..(iii)⇒Ifμ1=0,μ2≠0thenfrom(ii)wegetN2=mg,andN1=μ2mg.Thenfrom(iii)wegetmg(1−μ2tanθ)=N1tanθ⇒mg−N1tanθ=N1tanθorN1tanθ=mg2.Ifμ1≠0,μ2=0fortheabouttoslipcasef2=0⇒N1=0,θ=π2N2=mg.Hence(3),(4)arecorrect. Commented by Tinkutara last updated on 19/Oct/17 ThankyouverymuchSir! Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: A-ladder-of-mass-m-is-leaning-against-a-wall-It-is-in-static-equilibrium-making-an-angle-with-the-horizontal-floor-The-coefficient-of-friction-between-the-wall-and-the-ladder-is-1-and-that-betwNext Next post: 0-pi-2-cos-2x-ln-sin-x-dx-pi-4-solution-1-0-pi-2-2cos-2-x-1-ln-sin-x-dx-2-0-pi-2-cos-2-x-ln-sin-x-dx-0-pi-2- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.