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Question Number 32529 by byaw last updated on 27/Mar/18

Commented by byaw last updated on 27/Mar/18

$$\mathrm{Please}\mathrm{help}$$

Commented by mrW2 last updated on 27/Mar/18

$$isitontheearth?$$$$howcanamenweightonly7kgand$$$$howcanheclimbaladder?$$

Answered by mrW2 last updated on 28/Mar/18

Commented by mrW2 last updated on 28/Mar/18

$$N_{2}L\sin \theta +f_{2}L\cos \theta =Mg\frac{L}{2}\cos \theta +mgs\cos \theta$$$$N_{2}L(\sin \theta +{\mu }_2\cos \theta )=(\frac{ML}{2}+ms)g\cos \theta$$$$\Rightarrow N_2=\frac{(\frac{ML}{2}+ms)g\cos \theta }{L(\sin \theta +{\mu }_2\cos \theta )}$$$$N_{1}L\cos \theta -f_{1}L\sin \theta =Mg\frac{L}{2}\cos \theta +mg(L-s)\cos \theta$$$$N_{1}L(\cos \theta -{\mu }_1\sin \theta )=[\frac{ML}{2}+m(L-s)]g\cos \theta$$$$\Rightarrow N_1=\frac{[\frac{ML}{2}+m(L-s)]g\cos \theta }{L(\cos \theta -{\mu }_1\sin \theta )}$$$$$$$$N_2=f_1={\mu }_1{N}_1$$$$\Rightarrow \frac{(\frac{ML}{2}+ms)g\cos \theta }{L(\sin \theta +{\mu }_2\cos \theta )}=\frac{{\mu }_1[\frac{ML}{2}+m(L-s)]g\cos \theta }{L(\cos \theta -{\mu }_1\sin \theta )}$$$$\Rightarrow \frac{ML+2ms}{(\sin \theta +{\mu }_2\cos \theta )}=\frac{{\mu }_1(ML+2mL-2ms)}{(\cos \theta -{\mu }_1\sin \theta )}$$$$\Rightarrow ML(\cos \theta -{\mu }_1\sin \theta )+2ms(\cos \theta -{\mu }_1\sin \theta )={\mu }_1(ML+2mL)(\sin \theta +{\mu }_2\cos \theta )-2m{\mu }_{1}s(\sin \theta +{\mu }_2\cos \theta )$$$$\Rightarrow 2ms(1+{\mu }_1{\mu }_2)\cos \theta =(ML+2mL)({\mu }_1\sin \theta +{\mu }_1{\mu }_2\cos \theta )-ML(\cos \theta -{\mu }_1\sin \theta )$$$$\Rightarrow 2ms(1+{\mu }_1{\mu }_2)\cos \theta =ML(2{\mu }_1\sin \theta +{\mu }_1{\mu }_2\cos \theta -\cos \theta )+2m{\mu }_{1}L(\sin \theta +{\mu }_2\cos \theta )$$$$\Rightarrow \frac{s}{L}=\frac{\frac{M}{m}[2{\mu }_1\sin \theta +({\mu }_1{\mu }_2-1)\cos \theta ]+2{\mu }_1(\sin \theta +{\mu }_2\cos \theta )}{2(1+{\mu }_1{\mu }_2)\cos \theta }$$$$\Rightarrow \frac{s}{L}=\frac{\frac{M}{m}[2{\mu }_1\tan \theta +{\mu }_1{\mu }_2-1]+2{\mu }_1(\tan \theta +{\mu }_2)}{2(1+{\mu }_1{\mu }_2)}$$$$$$$$\Rightarrow \frac{s}{L}=\frac{2[2\times \frac{1}{2}\tan 40°+\frac{1}{2}\times \frac{1}{3}-1]+2\times \frac{1}{2}(\tan 40°+\frac{1}{3})}{2(1+\frac{1}{2}\times \frac{1}{3})}$$$$\Rightarrow \frac{s}{L}=\frac{9\tan 40°-4}{7}=0.507$$$$\Rightarrow s=0.507\times 7=3.55m$$

Commented by byaw last updated on 31/Mar/18

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}$$

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