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:

h1=11340 kg/m3 h2=2700 kg/m3  h3=1000 kg/m3  M1=M2 l1 m1=l2 m2  l1 h1 V1=l2 h2 V2 l2 (h1h3) V1=l1 (h2h3) V2  a=l1:l2 b=V2:V1  a h1=b h2 (h1h3)=a (h2h3) b  a h1=b h2 (h1h3)=a (h2h3) b  (113401000)=(b2700/11340)(27001000)b  (113401000)=(b 2700/11340) (27001000) b 404.761905b2+10340=0 404.761905b210340=0  p=404.761905;q=0;r=10340 D=q24pr=024404.761905(10340)=16740952.3908 D>0  b1,2=2pq±D=809.52381±16740952.39 b1,2=±5.0542934558565 b1=5.0542934558565 b2=5.0542934558565   Factored form of the equation:  404.761905(b5.0542934558565)(b+5.0542934558565)=0  b=b1=5.05435.0543  a=b h2/h1=5.0543 2700/11340=1.2034

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