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Question Number 73210 by TawaTawa last updated on 08/Nov/19

Commented by mr W last updated on 09/Nov/19

$(2) i s n o t d e f i n e d . n o u n i q u e s o l u t i o n .$
Commented by TawaTawa last updated on 09/Nov/19

$God bless you sir$
	Answered by mind is power last updated on 08/Nov/19
	$let A=center of m_1 let B=center of m_2 AB=0.8m CM of(m_1 ,m_2 )=D such that m_1 .AD=m_2 DB⇒DB=((15)/(25))AD AD+DB=0.8⇒AD=0.5m 2) let G=center of Rod wich is midpoint since the road is uniform R.road =h=0CM of (m_1 ,R,m_2 )=CM(CM(m_1 ,m_2 ),R(15)) CM(m1,m2)=D(M=40) DR=0.5−0.4=0.1 ⇒Dh.40=15.hR⇒hR=((40)/(15))=(8/3)Dh Dh+hR=DR=0.1⇒Dh=((0.3)/(11)) ⇒Ah=((0.3)/(11))+0.5=((5.8)/(11))=0.527..m 2) let A point (0,h) let B(0,t)⇒∣t−h∣=l C(h,0) h^2 +t^2 =l^2 his representation in coordinat plan of our problem C,2m,Aand B mass=m find G suche mGA^→ +mGB^→ +2mGC^→ =0 ′vector′ ⇒(4m)GA^→ =−mAB^→ −2mAC^→ ⇒GA^→ =((−AB^→ )/4)−2((AC^→ )/4) AB^→ =(0.t−h)=(0,l) AC^→ =(h,−h) GA^→ =(−(h/2),−l+(h/2)) ⇒AG^→ =((h/2),l−(h/2)) G(x,y)⇒ { ((x=(h/2))),((y−h=l−(h/2)⇒y=l+(h/2))) :} we can go forther in case CB=l ⇒(h^2 +t^2 )=l^2 =∣h−t∣^2 =h^2 +t^2 −2ht⇒ht=0 CA=l⇒2h^2 =l^2 ⇒h=+_− (l/(√2)) we have 4 case if we want$
	$let A = center of m_{1}$ $let B = center of m_{2}$ $AB = 0 . 8 m$ $CM of (m_{1}, m_{2}) = D$ $such that m_{1} . AD = m_{2} DB \Rightarrow DB = \frac{15}{25} AD$ $AD + DB = 0.8 \Rightarrow AD = 0 . 5 m$ $2) let G = center of Rod wich is midpoint since the road is uniform$ $R . road$ $= h = 0 CM of (m_{1}, R, m_{2}) = CM (CM (m_{1}, m_{2}), R (15))$ $CM (m 1, m 2) = D (M = 40)$ $DR = 0.5 - 0.4 = 0.1$ $\Rightarrow Dh . 40 = 15 . hR \Rightarrow hR = \frac{40}{15} = \frac{8}{3} Dh$ $Dh + hR = DR = 0.1 \Rightarrow Dh = \frac{0.3}{11}$ $\Rightarrow Ah = \frac{0.3}{11} + 0.5 = \frac{5.8}{11} = 0.527 . . m$ $2)$ $let A point (0, h)$ $let B (0, t) \Rightarrow∣ t - h ∣= l$ $C (h, 0) h^{2} + t^{2} = l^{2}$ $his representation in coordinat plan of our problem$ $C, 2 m, Aand B mass = m$ $find G suche$ $mG \vec{A} + mG \vec{B} + 2 mG \vec{C} = 0^{'} {vector}^{'}$ $\Rightarrow (4 m) G \vec{A} = - mA \vec{B} - 2 mA \vec{C}$ $\Rightarrow G \vec{A} = \frac{- A \vec{B}}{4} - 2 \frac{A \vec{C}}{4}$ $A \vec{B} = (0 . t - h) = (0, l)$ $A \vec{C} = (h, - h)$ $G \vec{A} = (- \frac{h}{2}, - l + \frac{h}{2})$ $\Rightarrow A \vec{G} = (\frac{h}{2}, l - \frac{h}{2})$ $G (x, y) \Rightarrow {\begin{cases} x = \frac{h}{2} \\ y - h = l - \frac{h}{2} \Rightarrow y = l + \frac{h}{2} \end{cases}$ $we can go forther$ $in case CB = l$ $\Rightarrow (h^{2} + t^{2}) = l^{2} =∣ h - t ∣^{2} = h^{2} + t^{2} - 2 ht \Rightarrow ht = 0$ $CA = l \Rightarrow {2 h}^{2} = l^{2} \Rightarrow h = \underline{+} \frac{l}{\sqrt{2}} we have 4 case if we want$
	Commented by TawaTawa last updated on 08/Nov/19

	$God bless you sir, Thanks for your time$
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