Calculate the buoyant force exerted by the water on the sphere

In summary, a hollow, plastic sphere with a volume of 0.700 M^3 is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake with a tension of 760 N. The buoyant force exerted on the sphere by the water is 6860 N and the mass of the sphere is 622 kg. When the cord breaks and the sphere rises to the surface, 88.9% of its volume will still be submerged.
  • #1
MAPgirl23
65
0
A hollow, plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.700 M^3 and the tension in the cord is 760 N.

Calculate the buoyant force exerted by the water on the sphere. Take the density of water to be 1000 kg/m^3 and the free fall acceleration to be 9.80 m/s^2.
**for this I set up Bouyant = density *Volume*gravity = 6860 N

What is the mass of the sphere? Take the density of water to be 1000kg/m^3 and the free fall acceleration to be 9.80m/s^2 .
**here I used Buoyant = mg+T and solved for m, so answer for m= 622 kg

The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged? Express your answer as a percentage.
**For the sphere to be at rest, mg = B. There is no tension anymore, since the cord was broken. I express B as a function of the volume of the sphere that is still submerged:

mg = B --> rho*V*g = 0.7 how do I express it as a percentage? I tried 70%
 
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  • #2
MAPgirl23 said:
A hollow, plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.700 M^3 and the tension in the cord is 760 N.

Calculate the buoyant force exerted by the water on the sphere. Take the density of water to be 1000 kg/m^3 and the free fall acceleration to be 9.80 m/s^2.
**for this I set up Bouyant = density *Volume*gravity = 6860 N

What is the mass of the sphere? Take the density of water to be 1000kg/m^3 and the free fall acceleration to be 9.80m/s^2 .
**here I used Buoyant = mg+T and solved for m, so answer for m= 622 kg

The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged? Express your answer as a percentage.
**For the sphere to be at rest, mg = B. There is no tension anymore, since the cord was broken. I express B as a function of the volume of the sphere that is still submerged:

mg = B --> rho*V*g = 0.7 how do I express it as a percentage? I tried 70%

The first two look good. The mg on the left side of your last equation involves the mass of the sphere you already calculated. The buoyant force will be rho*V*g, but rho is the density of water and V is the volume of water displaced. This does not equal .700. Use the mg = B part to find B and use B = rho*V*g to find V. The ratio of that V (volume of displaced water) to the volume of the sphere is the fraction submerged.
 
  • #3
622 kg * 9.8 = 6095.6 N

6095.6N / (1000 * 9.8) = 0.622

0.622 / 0.700 = 88.9% is this correct?
 
  • #4
MAPgirl23 said:
622 kg * 9.8 = 6095.6 N

6095.6N / (1000 * 9.8) = 0.622

0.622 / 0.700 = 88.9% is this correct?

Looks good to me.
 

1. How do you calculate the buoyant force exerted by water on a sphere?

The buoyant force exerted by water on a sphere can be calculated using the formula Fb = ρVg, where Fb is the buoyant force, ρ is the density of water, V is the volume of the displaced water, and g is the acceleration due to gravity.

2. What factors affect the buoyant force on a sphere in water?

The buoyant force on a sphere in water is affected by the density of water, the volume of the displaced water, and the acceleration due to gravity. It is also affected by the density and volume of the sphere itself.

3. How does the buoyant force change if the density of the sphere is increased?

If the density of the sphere is increased, the buoyant force will increase as well. This is because the denser the object, the more water it will displace, leading to a greater buoyant force.

4. Can the buoyant force ever be greater than the weight of the sphere?

Yes, the buoyant force can be greater than the weight of the sphere if the density of the sphere is less than the density of water. In this case, the sphere will float and the buoyant force will be greater than its weight, allowing it to stay afloat.

5. How does the shape of the sphere affect the buoyant force?

The shape of the sphere does not affect the buoyant force, as long as the volume and density of the sphere remain the same. The buoyant force is only affected by the volume of the displaced water and the density of the object.

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