Question Number 139342 by BHOOPENDRA last updated on 26/Apr/21
Commented by BHOOPENDRA last updated on 26/Apr/21
$${mr}.{W}\:{sir}\:{help}\:{me}\:{out}\:{this} \\ $$
Answered by mr W last updated on 26/Apr/21
Commented by mr W last updated on 26/Apr/21
$${if}\:{n}\:{masses}\:{are}\:{in}\:{balance}\:{about}\:{the} \\ $$$${rotation}\:{point}\:\left({origin}\right),\:{then} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{m}_{{k}} {x}_{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} =\mathrm{0} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{m}_{{k}} {y}_{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} =\mathrm{0} \\ $$$${in}\:{our}\:{case}\:{m}_{\mathrm{5}} ,\:\theta_{\mathrm{5}} \:{are}\:{unknown}. \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} +{m}_{\mathrm{5}} {r}_{\mathrm{5}} \mathrm{cos}\:\theta_{\mathrm{5}} =\mathrm{0} \\ $$$$\Rightarrow{m}_{\mathrm{5}} \mathrm{cos}\:\theta_{\mathrm{5}} =−\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} }{{r}_{\mathrm{5}} }\:\:\:\:…\left({i}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} +{m}_{\mathrm{5}} {r}_{\mathrm{5}} \mathrm{sin}\:\theta_{\mathrm{5}} =\mathrm{0} \\ $$$$\Rightarrow{m}_{\mathrm{5}} \mathrm{sin}\:\theta_{\mathrm{5}} =−\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} }{{r}_{\mathrm{5}} }\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)^{\mathrm{2}} +\left({ii}\right)^{\mathrm{2}} : \\ $$$${m}_{\mathrm{5}} =\frac{\sqrt{\left(\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} \right)^{\mathrm{2}} +\left(\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} \right)^{\mathrm{2}} }}{{r}_{\mathrm{5}} ^{\mathrm{2}} } \\ $$$$\left({ii}\right)/\left({i}\right): \\ $$$$\mathrm{tan}\:\theta_{\mathrm{5}} =\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} }{\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} } \\ $$$$\theta_{\mathrm{5}} =\mathrm{tan}^{−\mathrm{1}} \left(\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} }{\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} }\right)\:\:\:\left(\:+\pi\:\right) \\ $$$$\begin{array}{|c|c|c|c|c|c|}{{k}}&\hline{{m}_{{k}} }&\hline{{r}_{{k}} }&\hline{\theta_{{k}} }\\{\mathrm{1}}&\hline{\mathrm{220}}&\hline{\mathrm{0}.\mathrm{22}}&\hline{\mathrm{0}°}\\{\mathrm{2}}&\hline{\mathrm{330}}&\hline{\mathrm{0}.\mathrm{18}}&\hline{\mathrm{40}°}\\{\mathrm{3}}&\hline{\mathrm{220}}&\hline{\mathrm{0}.\mathrm{24}}&\hline{\mathrm{120}°}\\{\mathrm{4}}&\hline{\mathrm{240}}&\hline{\mathrm{0}.\mathrm{32}}&\hline{\mathrm{265}°}\\{\mathrm{5}}&\hline{?}&\hline{\mathrm{0}.\mathrm{20}}&\hline{?}\\\hline\end{array} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} =\mathrm{60}.\mathrm{809479} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} =\mathrm{7}.\mathrm{399973} \\ $$$$\Rightarrow{m}_{\mathrm{5}} =\frac{\sqrt{\left(\mathrm{60}.\mathrm{809479}\right)^{\mathrm{2}} +\left(\mathrm{7}.\mathrm{399973}\right)^{\mathrm{2}} }}{\mathrm{0}.\mathrm{20}^{\mathrm{2}} }=\mathrm{1531}.\mathrm{452}{kg} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{5}} =\frac{\mathrm{7}.\mathrm{399973}}{\mathrm{60}.\mathrm{809473}} \\ $$$$\Rightarrow\theta_{\mathrm{5}} =\mathrm{6}.\mathrm{983}° \\ $$
Commented by BHOOPENDRA last updated on 26/Apr/21
$${thankyou}\: \\ $$
Answered by TheSupreme last updated on 26/Apr/21
$$\underset{{i}} {\sum}{m}_{{i}} \rho_{{i}} {e}^{{i}\theta_{{i}} } =\mathrm{0} \\ $$$${m}_{\mathrm{5}} {e}^{{i}\theta_{\mathrm{5}} } =−\left(\underset{{i}} {\sum}{m}_{{i}} \rho_{\mathrm{1}} {e}^{{i}\theta_{{i}} } \right)/\rho_{\mathrm{5}} \\ $$$${m}_{\mathrm{5}} {cos}\left(\theta_{\mathrm{5}} \right)=−\left(\Sigma{m}_{{i}} \rho_{{i}} {cos}\left(\theta_{{i}} \right)\right)/\rho_{\mathrm{5}} \\ $$$${m}_{\mathrm{5}} {sin}\left(\theta_{\mathrm{5}} \right)=−\left(\Sigma{m}_{{i}} \rho_{{i}} {sin}\left(\theta_{{i}} \right)\right)/\rho_{\mathrm{5}} \\ $$$${m}_{\mathrm{5}} {cos}\left(\theta_{\mathrm{5}} \right)={A} \\ $$$${m}_{\mathrm{5}} {sin}\left(\theta_{\mathrm{5}} \right)={B} \\ $$$${tan}\left(\theta_{\mathrm{5}} \right)=\frac{{B}}{{A}} \\ $$$${m}_{\mathrm{5}} =\frac{{A}}{{cos}\left(\theta_{\mathrm{5}} \right)}=\frac{{A}}{\:\sqrt{\frac{\mathrm{1}}{\frac{{B}^{\mathrm{2}} }{{A}^{\mathrm{2}} }+\mathrm{1}}}}=\frac{{A}^{\mathrm{2}} }{\:\sqrt{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} }} \\ $$
Commented by BHOOPENDRA last updated on 26/Apr/21
$${thankyou} \\ $$