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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
$${mr}.{W}\:{sir}\:{help}\:{me}\:{out}\:{this} \\ $$
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 (origin), then Σ_(k=1) ^n m_k x_k =Σ_(k=1) ^n m_k r_k cos θ_k =0 Σ_(k=1) ^n m_k y_k =Σ_(k=1) ^n m_k r_k sin θ_k =0 in our case m_5 , θ_5 are unknown. Σ_(k=1) ^4 m_k r_k cos θ_k +m_5 r_5 cos θ_5 =0 ⇒m_5 cos θ_5 =−((Σ_(k=1) ^4 m_k r_k cos θ_k )/r_5 ) ...(i) Σ_(k=1) ^4 m_k r_k sin θ_k +m_5 r_5 sin θ_5 =0 ⇒m_5 sin θ_5 =−((Σ_(k=1) ^4 m_k r_k sin θ_k )/r_5 ) ...(ii) (i)^2 +(ii)^2 : m_5 =((√((Σ_(k=1) ^4 m_k r_k cos θ_k )^2 +(Σ_(k=1) ^4 m_k r_k sin θ_k )^2 ))/r_5 ^2 ) (ii)/(i): tan θ_5 =((Σ_(k=1) ^4 m_k r_k sin θ_k )/(Σ_(k=1) ^4 m_k r_k cos θ_k )) θ_5 =tan^(−1) (((Σ_(k=1) ^4 m_k r_k sin θ_k )/(Σ_(k=1) ^4 m_k r_k cos θ_k ))) ( +π ) determinant ((k,m_k ,r_k ,θ_k ),(1,(220),(0.22),(0°)),(2,(330),(0.18),(40°)),(3,(220),(0.24),(120°)),(4,(240),(0.32),(265°)),(5,?,(0.20),?)) Σ_(k=1) ^4 m_k r_k cos θ_k =60.809479 Σ_(k=1) ^4 m_k r_k sin θ_k =7.399973 ⇒m_5 =((√((60.809479)^2 +(7.399973)^2 ))/(0.20^2 ))=1531.452kg tan θ_5 =((7.399973)/(60.809473)) ⇒θ_5 =6.983°
$${if}\:{n}\:{masses}\:{are}\:{in}\:{balance}\:{about}\:{the} \\ $$$${rotation}\:{point}\:\left({origin}\right),\:{then} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{m}_{{k}} {x}_{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} =\mathrm{0} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{m}_{{k}} {y}_{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} =\mathrm{0} \\ $$$${in}\:{our}\:{case}\:{m}_{\mathrm{5}} ,\:\theta_{\mathrm{5}} \:{are}\:{unknown}. \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} +{m}_{\mathrm{5}} {r}_{\mathrm{5}} \mathrm{cos}\:\theta_{\mathrm{5}} =\mathrm{0} \\ $$$$\Rightarrow{m}_{\mathrm{5}} \mathrm{cos}\:\theta_{\mathrm{5}} =−\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} }{{r}_{\mathrm{5}} }\:\:\:\:…\left({i}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} +{m}_{\mathrm{5}} {r}_{\mathrm{5}} \mathrm{sin}\:\theta_{\mathrm{5}} =\mathrm{0} \\ $$$$\Rightarrow{m}_{\mathrm{5}} \mathrm{sin}\:\theta_{\mathrm{5}} =−\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} }{{r}_{\mathrm{5}} }\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)^{\mathrm{2}} +\left({ii}\right)^{\mathrm{2}} : \\ $$$${m}_{\mathrm{5}} =\frac{\sqrt{\left(\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} \right)^{\mathrm{2}} +\left(\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} \right)^{\mathrm{2}} }}{{r}_{\mathrm{5}} ^{\mathrm{2}} } \\ $$$$\left({ii}\right)/\left({i}\right): \\ $$$$\mathrm{tan}\:\theta_{\mathrm{5}} =\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} }{\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} } \\ $$$$\theta_{\mathrm{5}} =\mathrm{tan}^{−\mathrm{1}} \left(\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} }{\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} }\right)\:\:\:\left(\:+\pi\:\right) \\ $$$$\begin{array}{|c|c|c|c|c|c|}{{k}}&\hline{{m}_{{k}} }&\hline{{r}_{{k}} }&\hline{\theta_{{k}} }\\{\mathrm{1}}&\hline{\mathrm{220}}&\hline{\mathrm{0}.\mathrm{22}}&\hline{\mathrm{0}°}\\{\mathrm{2}}&\hline{\mathrm{330}}&\hline{\mathrm{0}.\mathrm{18}}&\hline{\mathrm{40}°}\\{\mathrm{3}}&\hline{\mathrm{220}}&\hline{\mathrm{0}.\mathrm{24}}&\hline{\mathrm{120}°}\\{\mathrm{4}}&\hline{\mathrm{240}}&\hline{\mathrm{0}.\mathrm{32}}&\hline{\mathrm{265}°}\\{\mathrm{5}}&\hline{?}&\hline{\mathrm{0}.\mathrm{20}}&\hline{?}\\\hline\end{array} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{cos}\:\theta_{{k}} =\mathrm{60}.\mathrm{809479} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{m}_{{k}} {r}_{{k}} \mathrm{sin}\:\theta_{{k}} =\mathrm{7}.\mathrm{399973} \\ $$$$\Rightarrow{m}_{\mathrm{5}} =\frac{\sqrt{\left(\mathrm{60}.\mathrm{809479}\right)^{\mathrm{2}} +\left(\mathrm{7}.\mathrm{399973}\right)^{\mathrm{2}} }}{\mathrm{0}.\mathrm{20}^{\mathrm{2}} }=\mathrm{1531}.\mathrm{452}{kg} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{5}} =\frac{\mathrm{7}.\mathrm{399973}}{\mathrm{60}.\mathrm{809473}} \\ $$$$\Rightarrow\theta_{\mathrm{5}} =\mathrm{6}.\mathrm{983}° \\ $$
Commented by BHOOPENDRA last updated on 26/Apr/21
thankyou
$${thankyou}\: \\ $$
Answered by TheSupreme last updated on 26/Apr/21
Σ_i m_i ρ_i e^(iθ_i ) =0 m_5 e^(iθ_5 ) =−(Σ_i m_i ρ_1 e^(iθ_i ) )/ρ_5 m_5 cos(θ_5 )=−(Σm_i ρ_i cos(θ_i ))/ρ_5 m_5 sin(θ_5 )=−(Σm_i ρ_i sin(θ_i ))/ρ_5 m_5 cos(θ_5 )=A m_5 sin(θ_5 )=B tan(θ_5 )=(B/A) m_5 =(A/(cos(θ_5 )))=(A/( (√(1/((B^2 /A^2 )+1)))))=(A^2 /( (√(A^2 +B^2 ))))
$$\underset{{i}} {\sum}{m}_{{i}} \rho_{{i}} {e}^{{i}\theta_{{i}} } =\mathrm{0} \\ $$$${m}_{\mathrm{5}} {e}^{{i}\theta_{\mathrm{5}} } =−\left(\underset{{i}} {\sum}{m}_{{i}} \rho_{\mathrm{1}} {e}^{{i}\theta_{{i}} } \right)/\rho_{\mathrm{5}} \\ $$$${m}_{\mathrm{5}} {cos}\left(\theta_{\mathrm{5}} \right)=−\left(\Sigma{m}_{{i}} \rho_{{i}} {cos}\left(\theta_{{i}} \right)\right)/\rho_{\mathrm{5}} \\ $$$${m}_{\mathrm{5}} {sin}\left(\theta_{\mathrm{5}} \right)=−\left(\Sigma{m}_{{i}} \rho_{{i}} {sin}\left(\theta_{{i}} \right)\right)/\rho_{\mathrm{5}} \\ $$$${m}_{\mathrm{5}} {cos}\left(\theta_{\mathrm{5}} \right)={A} \\ $$$${m}_{\mathrm{5}} {sin}\left(\theta_{\mathrm{5}} \right)={B} \\ $$$${tan}\left(\theta_{\mathrm{5}} \right)=\frac{{B}}{{A}} \\ $$$${m}_{\mathrm{5}} =\frac{{A}}{{cos}\left(\theta_{\mathrm{5}} \right)}=\frac{{A}}{\:\sqrt{\frac{\mathrm{1}}{\frac{{B}^{\mathrm{2}} }{{A}^{\mathrm{2}} }+\mathrm{1}}}}=\frac{{A}^{\mathrm{2}} }{\:\sqrt{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} }} \\ $$
Commented by BHOOPENDRA last updated on 26/Apr/21
thankyou
$${thankyou} \\ $$

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