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Tuesday, 17 June 2014 13:22

F2-25: BALANCE PARADOX - BUOYANCY WITH CROSSOVER

f2-25a

Additional Info

  • ID Code: F2-25
  • Purpose: Present buoyancy in a paradoxical way.
  • Description: The balance is initially at equilibrium with a mass hanging from an arm on the left pan in balance with the water beaker on the right pan, as seen in the photograph at the left above. Q: If the mass is allowed to hang into the beaker of water, how does this effect the balance? In particular, what, if anything must be done to restore equilibrium? (Note that there are a 100 gram weight and two 50 gram weights available at the lower left of th picture, and these weights can be added to either side of the balance to restore equilibrium.) A: Because the volume of the block is 50 cm^3, the weight on the left side is reduced by 50 grams when the block is submerged in the water. Conversely, the weight on the right side is increased by 50 grams, the reaction force on that pan. To restore equilibrium, 100 grams must be added to the left pan, as seen in the photograph at the right above.
  • Availability: Available
Read 2281 times Last modified on Monday, 31 August 2020 16:05