Equilibrium 7990

At the end of one arm of the equilibrium scales, which are in equilibrium, a lead body with a volume V1 is suspended in the air. At the end of the other arm is an aluminum body with a volume V2. The balance arms have sizes l1 and l2, lead density h1 = 11,340 kg / m³, aluminum density h2 = 2,700 kg / m³. If we change the position of the bodies and immerse them in water, equilibrium occurs again (h3 = 1000 kg / m³). Do not consider the buoyancy of the air.

a) What is the ratio of arm sizes l1: l2?
b) What is the volume ratio of V2: V1 bodies?

Correct answer:

a =  1.2034
b =  5.0543

Step-by-step explanation:

$$\begin{gathered} h_1=11340{\text{kg/m}}^3 \\ {h}_2=2700{\text{kg/m}}^3 \\ {h}_3=1000{\text{kg/m}}^3 \\ {M}_1=M_2 \\ {l}_1\cdot m_1=l_2\cdot m_2 \\ {l}_1\cdot h_1\cdot V_1=l_2\cdot h_2\cdot V_2 \\ {l}_2\cdot (h_1-h_3)\cdot V_1=l_1\cdot (h_2-h_3)\cdot V_2 \\ a=l_1:l_2 \\ b=V_2:V_1 \\ a\cdot h_1=b\cdot h_2 \\ (h_1-h_3)=a\cdot (h_2-h_3)\cdot b \\ a\cdot h_1=b\cdot h_2 \\ (h_1-h_3)=a\cdot (h_2-h_3)\cdot b \\ (11340-1000)=(b\ast 2700/11340)\ast (2700-1000)\ast b \\ (11340-1000)=(b\cdot 2700/11340)\cdot (2700-1000)\cdot b \\ -404.761905b^2+10340=0 \\ 404.761905b^2-10340=0 \\ p=404.761905;q=0;r=-10340 \\ D=q^2-4pr=0^2-4\cdot 404.761905\cdot (-10340)=16740952.3908 \\ D>0 \\ {b}_{1,2}=\frac{-q\pm \sqrt{D}}{2p}=\frac{\pm \sqrt{16740952.39}}{809.52381} \\ {b}_{1,2}=\pm 5.0542934558565 \\ {b}_1=5.0542934558565 \\ {b}_2=-5.0542934558565 \\ \text{ Factored form of the equation: } \\ 404.761905(b-5.0542934558565)(b+5.0542934558565)=0 \\ b=b_1=5.0543\doteq 5.0543 \\ a=b\cdot h_2/h_1=5.0543\cdot 2700/11340=1.2034 \end{gathered}$$

Our quadratic equation calculator calculates it.

$b=b_1=5.0543$

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