# Buoyancy problem from NEXT STEP

Discussion in 'MCAT Study Question Q&A' started by nai54, Aug 2, 2015.

1. ### nai54 2+ Year Member

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What is the tension on the rope in the diagram assuming the density of water is 1g/cm^3, the density of the object is 2g/cm^3 and its volume is 10cm^3. The picture is a ball completely submerged in water with a string attached to holding it in the water.

I started off by doing Tension=mg- Fb--> T=density(o)V(o)g-density(f)V(o)g and plugged in the numbers. Am I doing this wrong?

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3. ### theonlytycrane 2+ Year Member

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The object is more dense than water so it would sink if it wasn't attached to the string holding it in the water.

The volume of the object is 10 cm^3. 10 cm^3 * 2g / cm^3 = 20g = .02kg. The Force of gravity on the object (down) is .02kg * 10 m/s^2 = .2N.

The buoyant force of water = mass of water displaced * gravity. Water is incompressible so the mass of water displaced = volume of the object * density of water = 10 cm^3 * 1g / cm^3 = 10g = .01kg. The buoyant force (up) is .01kg * 10 m/s^2 = .1N.

We have .2N down and .1N up, so the tension in the string should be .1N (up) to keep the object from sinking.

I'm not sure how the key got 0.19N

#2
Last edited: Aug 2, 2015
4. ### nai54 2+ Year Member

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hmm yes that makes sense. That answer seemed a little off it might be a mistake. Does it matter that the units are g/cm^3? Should it be converted to kg/m^3 since g is m/s^2?

5. ### theonlytycrane 2+ Year Member

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The units for volume of the object cancel with density but you're right- Newton = kg * m / s^2. I need to convert g to kg. Editing my post above. Nice catch

#4
Last edited: Aug 2, 2015
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