The Pressure is Getting to Me... - Manometers and Buoyancy

Solutions:

Sphere problem #3 on the handout from today:
a) When the sphere is not submerged, T - mg = 0 so m = T/g = 2 kg.

Once the sphere is submerged, T + Fb - mg = 0 so Fb = mg - T. Since Fb = rho*g*Vsub, V = (mg - T)/(rho*g) = .00153 m^3

b) density is m/V = 2 kg/(.00153 m^3) = 1307 kg/m^3

Book Problems:
22. Apparent mass comes from apparent weight. If you know the apparent weight (Fn) you can divide by g to get the apparent mass.

In this case, Fn = m_apparent * g = 60.6 N

Drawing a FBD of the rock at the bottom of the container, Fn + Fb - mg = 0. Fb = mg - Fn = rho*g*V. This enables us to get that the volume is .002016 m^3.

The density is therefore 8.2 kg/.002016 m^3 = 4067 kg/m^3


24. Answer: 3.0 x 10⁴ kg
Here's the FBD:

Since the balloon is in equilibrium, Fb - m_cargo*g - m_balloon*g - m_Helium*g = 0.

We can find the buoyancy force because we can find the volume of a sphere from the radius, and the density of air is 1.29 kg/m^3 (from p. 276). The same thing is true for knowing the mass of the Helium.

(I think in tutoring, I forgot about the mass of the Helium...)

The only unknown is the mass of the cargo, so you can solve!


26. Using the same method as in #9, we can find that the density of the metal is 8.94 x 10³ kg/m^3. From page 276, we can see that this is copper.

27. Same method, again. Density = 1.03 x 10³ kg/m^3

29. Finding the definition of specific gravity (SG) is a bit tricky in the chapter, which is why coming to tutoring made this a bit easier.

SG is the ratio of a density and the density of water. For example, if the SG = 0.79 for the alcohol, it means that the density is 790 kg/m^3.

Using this info, this is another problem that is along the lines of 9, 26, and 27. You should find that the density of the wood is 850 kg/m^3, which means that its specific gravity is 0.85.

30. This problem is a bit tricky because it's hard to picture what's happening. If we say that the iceberg is a cylinder, with cross section A, we can say that the part below the water is of depth D, and the part above the water is height H. The fraction the problem is asking for is H/(H+D).

Knowing the SG of both the iceberg and the ocean water, we know the densities of both. From a FBD, we can see that Fb - mg = 0.

Fb = rho_ocean*g*V_submerged = rho_ocean*g*A*(D)

mg = rho_ice*g*V = rho_ice*g*A*(H+D)

Setting these equal to each other, we get that D/(H+D) = rho_ocean/rho_ice = .895. The fraction above the water is then 1 - D/(H+D) = .105