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 of 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)=(b 2700/11340) (27001000) b  (113401000)=(b 2700/11340) (27001000) b 404.7619047619b2+10340=0 404.7619047619b210340=0 b1,2=±10340/404.7619047619=±5.054293457 b1=5.054293457 b2=5.054293457  b=b1=5.05435.0543  a=b h2/h1=5.0543 2700/11340=1.2034

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