A-uniform-ladder-of-weight-W-and-length-2a-rest-in-limiting-equilibrium-with-one-end-on-a-rough-horizontal-ground-and-the-other-end-on-a-rough-vertical-wall-The-coefficient-of-friction-between-the-l

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Question Number 76579 by Rio Michael last updated on 28/Dec/19

$$\mathrm{A}\mathrm{uniform}\mathrm{ladder}\mathrm{of}\mathrm{weight}W\mathrm{and}\mathrm{length}$$$$2a\mathrm{rest}\mathrm{in}\mathrm{limiting}\mathrm{equilibrium}\mathrm{with}\mathrm{one}$$$$\mathrm{end}\mathrm{on}\mathrm{a}\mathrm{rough}\mathrm{horizontal}\mathrm{ground}\mathrm{and}\mathrm{the}$$$$\mathrm{other}\mathrm{end}\mathrm{on}\mathrm{a}\mathrm{rough}\mathrm{vertical}\mathrm{wall}.$$$$\mathrm{The}\mathrm{coefficient}\mathrm{of}\mathrm{friction}\mathrm{between}\mathrm{the}\mathrm{ladder}$$$$\mathrm{and}\mathrm{the}\mathrm{ground}\mathrm{and}\mathrm{between}\mathrm{the}\mathrm{ladder}\mathrm{and}$$$$\mathrm{the}\mathrm{wall}\mathrm{are}\mathrm{respectively}\mu \mathrm{and}\lambda .\mathrm{If}\mathrm{the}\mathrm{ladder}$$$$\mathrm{makes}\mathrm{an}\mathrm{angle}\theta \mathrm{with}\mathrm{the}\mathrm{ground}\mathrm{where}\tan \theta =\frac{5}{12},$$$$\mathrm{a})\mathrm{show}\mathrm{that}5\mu +6\lambda \mu -6=0.$$$$\mathrm{b})\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\lambda \mathrm{and}\mu \mathrm{given}\mathrm{that}\lambda \mu =\frac{1}{2}.$$

Answered by mr W last updated on 28/Dec/19

Commented by mr W last updated on 28/Dec/19

$$f_1=\mu N_1$$$$f_2=\lambda N_2$$$$N_1+f_2=W\Rightarrow N_1+\lambda N_2=W\dots \left(i\right)$$$$N_2-f_1=0\Rightarrow N_2-\mu N_1=0\dots \left(ii\right)$$$$L=lengthofladder$$$$W\times \frac{L}{2}\cos \theta +f_1\times L\sin \theta -N_1\times L\cos \theta =0$$$$W\cos \theta +2(\mu \sin \theta -\cos \theta )N_1=0\dots \left(iii\right)$$$$\left(i\right)-\left(ii\right)\times \lambda :$$$$N_1(1+\mu \lambda )=W$$$$putthisinto\left(iii\right):$$$$N_1(1+\mu \lambda )\cos \theta +2(\mu \sin \theta -\cos \theta )N_1=0$$$$(1+\mu \lambda )\cos \theta +2(\mu \sin \theta -\cos \theta )=0$$$$2\mu \sin \theta =(1-\mu \lambda )\cos \theta$$$$\Rightarrow \tan \theta =\frac{1-\mu \lambda }{2\mu }=\frac{5}{12}$$$$\Rightarrow 5\mu +6\mu \lambda -6=0$$$$$$$$if\mu \lambda =\frac{1}{2}$$$$5\mu +6\times \frac{1}{2}-6=0$$$$\Rightarrow \mu =\frac{3}{5}$$$$\Rightarrow \lambda =\frac{1}{2\mu }=\frac{5}{6}$$

Commented by Rio Michael last updated on 28/Dec/19

$$\mathrm{thanks}\mathrm{sir}$$

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