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
$${is}\:{it}\:{on}\:{the}\:{earth}? \\ $$$${how}\:{can}\:{a}\:{men}\:{weight}\:{only}\:\mathrm{7}\:{kg}\:{and} \\ $$$${how}\:{can}\:{he}\:{climb}\:{a}\:{ladder}? \\ $$
Answered by mrW2 last updated on 28/Mar/18
Commented by mrW2 last updated on 28/Mar/18
![N_2 L sin θ+f_2 L cos θ=Mg (L/2) cos θ+mg s cos θ N_2 L(sin θ+μ_2 cos θ)=(((ML)/2)+ms)g cos θ ⇒N_2 =(((((ML)/2)+ms)g cos θ)/(L(sin θ+μ_2 cos θ))) N_1 L cos θ−f_1 L sin θ=Mg (L/2) cos θ+mg (L−s) cos θ N_1 L(cos θ−μ_1 sin θ)=[ ((ML)/2)+m(L−s) ]g cos θ ⇒N_1 =(([ ((ML)/2)+m(L−s) ]g cos θ)/(L(cos θ−μ_1 sin θ))) N_2 =f_1 =μ_1 N_1 ⇒(((((ML)/2)+ms)g cos θ)/(L(sin θ+μ_2 cos θ)))=((μ_1 [ ((ML)/2)+m(L−s) ]g cos θ)/(L(cos θ−μ_1 sin θ))) ⇒((ML+2ms)/((sin θ+μ_2 cos θ)))=((μ_1 (ML+2mL−2ms))/((cos θ−μ_1 sin θ))) ⇒ML(cos θ−μ_1 sin θ)+2ms(cos θ−μ_1 sin θ)=μ_1 (ML+2mL)(sin θ+μ_2 cos θ)−2mμ_1 s(sin θ+μ_2 cos θ) ⇒2ms(1+μ_1 μ_2 )cos θ=(ML+2mL)(μ_1 sin θ+μ_1 μ_2 cos θ)−ML(cos θ−μ_1 sin θ) ⇒2ms(1+μ_1 μ_2 )cos θ=ML(2μ_1 sin θ+μ_1 μ_2 cos θ−cos θ)+2mμ_1 L(sin θ+μ_2 cos θ) ⇒(s/L)=(((M/m)[2μ_1 sin θ+(μ_1 μ_2 −1) cos θ]+2μ_1 (sin θ+μ_2 cos θ))/(2(1+μ_1 μ_2 )cos θ)) ⇒(s/L)=(((M/m)[2μ_1 tan θ+μ_1 μ_2 −1]+2μ_1 (tan θ+μ_2 ))/(2(1+μ_1 μ_2 ))) ⇒(s/L)=((2[2×(1/2)tan 40°+(1/2)×(1/3)−1]+2×(1/2)(tan 40°+(1/3)))/(2(1+(1/2)×(1/3)))) ⇒(s/L)=((9 tan 40°−4)/7)=0.507 ⇒s=0.507×7=3.55 m](Q32608.png)
$${N}_{\mathrm{2}} \:{L}\:\mathrm{sin}\:\theta+{f}_{\mathrm{2}} \:{L}\:\mathrm{cos}\:\theta={Mg}\:\frac{{L}}{\mathrm{2}}\:\mathrm{cos}\:\theta+{mg}\:{s}\:\mathrm{cos}\:\theta \\ $$$${N}_{\mathrm{2}} {L}\left(\mathrm{sin}\:\theta+\mu_{\mathrm{2}} \:\mathrm{cos}\:\theta\right)=\left(\frac{{ML}}{\mathrm{2}}+{ms}\right){g}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{N}_{\mathrm{2}} =\frac{\left(\frac{{ML}}{\mathrm{2}}+{ms}\right){g}\:\mathrm{cos}\:\theta}{{L}\left(\mathrm{sin}\:\theta+\mu_{\mathrm{2}} \:\mathrm{cos}\:\theta\right)} \\ $$$${N}_{\mathrm{1}} \:{L}\:\mathrm{cos}\:\theta−{f}_{\mathrm{1}} \:{L}\:\mathrm{sin}\:\theta={Mg}\:\frac{{L}}{\mathrm{2}}\:\mathrm{cos}\:\theta+{mg}\:\left({L}−{s}\right)\:\mathrm{cos}\:\theta \\ $$$${N}_{\mathrm{1}} {L}\left(\mathrm{cos}\:\theta−\mu_{\mathrm{1}} \:\mathrm{sin}\:\theta\right)=\left[\:\frac{{ML}}{\mathrm{2}}+{m}\left({L}−{s}\right)\:\right]{g}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{N}_{\mathrm{1}} =\frac{\left[\:\frac{{ML}}{\mathrm{2}}+{m}\left({L}−{s}\right)\:\right]{g}\:\mathrm{cos}\:\theta}{{L}\left(\mathrm{cos}\:\theta−\mu_{\mathrm{1}} \:\mathrm{sin}\:\theta\right)} \\ $$$$ \\ $$$${N}_{\mathrm{2}} ={f}_{\mathrm{1}} =\mu_{\mathrm{1}} {N}_{\mathrm{1}} \\ $$$$\Rightarrow\frac{\left(\frac{{ML}}{\mathrm{2}}+{ms}\right){g}\:\mathrm{cos}\:\theta}{{L}\left(\mathrm{sin}\:\theta+\mu_{\mathrm{2}} \:\mathrm{cos}\:\theta\right)}=\frac{\mu_{\mathrm{1}} \left[\:\frac{{ML}}{\mathrm{2}}+{m}\left({L}−{s}\right)\:\right]{g}\:\mathrm{cos}\:\theta}{{L}\left(\mathrm{cos}\:\theta−\mu_{\mathrm{1}} \:\mathrm{sin}\:\theta\right)} \\ $$$$\Rightarrow\frac{{ML}+\mathrm{2}{ms}}{\left(\mathrm{sin}\:\theta+\mu_{\mathrm{2}} \:\mathrm{cos}\:\theta\right)}=\frac{\mu_{\mathrm{1}} \left({ML}+\mathrm{2}{mL}−\mathrm{2}{ms}\right)}{\left(\mathrm{cos}\:\theta−\mu_{\mathrm{1}} \:\mathrm{sin}\:\theta\right)} \\ $$$$\Rightarrow{ML}\left(\mathrm{cos}\:\theta−\mu_{\mathrm{1}} \:\mathrm{sin}\:\theta\right)+\mathrm{2}{ms}\left(\mathrm{cos}\:\theta−\mu_{\mathrm{1}} \:\mathrm{sin}\:\theta\right)=\mu_{\mathrm{1}} \left({ML}+\mathrm{2}{mL}\right)\left(\mathrm{sin}\:\theta+\mu_{\mathrm{2}} \:\mathrm{cos}\:\theta\right)−\mathrm{2}{m}\mu_{\mathrm{1}} {s}\left(\mathrm{sin}\:\theta+\mu_{\mathrm{2}} \:\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\mathrm{2}{ms}\left(\mathrm{1}+\mu_{\mathrm{1}} \mu_{\mathrm{2}} \right)\mathrm{cos}\:\theta=\left({ML}+\mathrm{2}{mL}\right)\left(\mu_{\mathrm{1}} \mathrm{sin}\:\theta+\mu_{\mathrm{1}} \mu_{\mathrm{2}} \:\mathrm{cos}\:\theta\right)−{ML}\left(\mathrm{cos}\:\theta−\mu_{\mathrm{1}} \:\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\mathrm{2}{ms}\left(\mathrm{1}+\mu_{\mathrm{1}} \mu_{\mathrm{2}} \right)\mathrm{cos}\:\theta={ML}\left(\mathrm{2}\mu_{\mathrm{1}} \mathrm{sin}\:\theta+\mu_{\mathrm{1}} \mu_{\mathrm{2}} \:\mathrm{cos}\:\theta−\mathrm{cos}\:\theta\right)+\mathrm{2}{m}\mu_{\mathrm{1}} {L}\left(\mathrm{sin}\:\theta+\mu_{\mathrm{2}} \:\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\frac{{s}}{{L}}=\frac{\frac{{M}}{{m}}\left[\mathrm{2}\mu_{\mathrm{1}} \mathrm{sin}\:\theta+\left(\mu_{\mathrm{1}} \mu_{\mathrm{2}} −\mathrm{1}\right)\:\mathrm{cos}\:\theta\right]+\mathrm{2}\mu_{\mathrm{1}} \left(\mathrm{sin}\:\theta+\mu_{\mathrm{2}} \:\mathrm{cos}\:\theta\right)}{\mathrm{2}\left(\mathrm{1}+\mu_{\mathrm{1}} \mu_{\mathrm{2}} \right)\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\frac{{s}}{{L}}=\frac{\frac{{M}}{{m}}\left[\mathrm{2}\mu_{\mathrm{1}} \mathrm{tan}\:\theta+\mu_{\mathrm{1}} \mu_{\mathrm{2}} −\mathrm{1}\right]+\mathrm{2}\mu_{\mathrm{1}} \left(\mathrm{tan}\:\theta+\mu_{\mathrm{2}} \right)}{\mathrm{2}\left(\mathrm{1}+\mu_{\mathrm{1}} \mu_{\mathrm{2}} \right)} \\ $$$$ \\ $$$$\Rightarrow\frac{{s}}{{L}}=\frac{\mathrm{2}\left[\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\mathrm{40}°+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}\right]+\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}\:\mathrm{40}°+\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$$\Rightarrow\frac{{s}}{{L}}=\frac{\mathrm{9}\:\mathrm{tan}\:\mathrm{40}°−\mathrm{4}}{\mathrm{7}}=\mathrm{0}.\mathrm{507} \\ $$$$\Rightarrow{s}=\mathrm{0}.\mathrm{507}×\mathrm{7}=\mathrm{3}.\mathrm{55}\:{m} \\ $$
Commented by byaw last updated on 31/Mar/18
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$