A solid sphere of aluminum (density 2.7 g/cm^3) is gently dropped into a deep ocean. (The density of ocean water is approximately 1.03 g/cm^3.) Calculate the sphere's acceleration at the point where it is completely submerged into the ocean. As the sphere drops deeper, does the acceleration increase or decrease compared to the acceleration beneath the surface?

Question

anonymous · Answer

The buoyant force on a submerged object Is equal to the force of gravity on the fluid displaced by that object.  In equation form that is given by:$$F _{b}=\rho _{f}Vg$$where Fb is the buoyant force, ρf is the density of the displaced fluid; V is the volume of the displaced fluid; and g is the acceleration of gravity.

anonymous · Answer

The net force on the object will be the difference between the force of gravity on the object  and the buoyant force acting on the object:$$F _{net}=F _{g}-F _{b}=mg-\rho _{f}Vg$$

anonymous · Answer

We can rewrite that and get:$$ma _{net}=mg-\rho _{f}Vg$$where m is again the mass of the object; and anet is the net acceleration the object experiences.

anonymous · Answer

Do you see now how to solve the equation?

anonymous · Answer

As for the rest of the problem, as the object sinks, the pressure exerted by downward on the object by the fluid above it increases with depth:$$p-\rho _{f}gh$$That means as the sphere sinks the net force becomes:$$F _{net}=F _{pressure}+F _{g}-F _{buoyant}$$If we expand this we get:$$ma _{net}=\frac{ \rho _{f}gh }{ A _{object} }+mg-\rho _{g}Vg$$where h is the depth under water; and Aobject is the surface area over which the pressure is pushing downward.  Note that the water over the top half of the sphere is pushing downward while the buoyant force is being exerted on the bottom half of the sphere.  $$Looking at the equation yousho$$

anonymous · Answer

Looking at the equation you should see that if the sphere sinks far enough there should come a point where Fnet=0 or in terms of the equation:$$\frac{ \rho _{f}gh }{ A _{object} }+mg=\rho _{f}Vg$$

anonymous · Answer

I get 6.06m/s ,however the answer says 6.2m/s. I plugged in, rho object V object for m in mass , ma=mg−ρfVg .

anonymous · Answer

Let's rewrite the equation I gave in a response earlier.  That equation was:$$ma _{net}=mg-\rho _{f}Vg$$Now we're not give the mass of the sphere, m, but we know that the mass can be given by:$$m=\rho _{Al}V _{sphere}$$So now our earlier equation becomes:

anonymous · Answer

$$\rho _{Al}V _{sphere}a _{net}=\rho _{Al}V _{sphere}g-\rho _{f}V _{sphere}g$$which can be simplified to yield:$$\rho _{Al}a _{net}=\rho _{Al}g-\rho _{f}g$$Rearrange that and solve for ante and we get:$$a _{net}=\frac{ \rho _{Al}g-\rho _{f}g }{ \rho _{Al} }$$

anonymous · Answer

That answer should be 6.06 m/s^2 if you use g=9.8m/s^2.  If you use g=10m/s^2, a common approximation, the answer is 6.2 m/s^2.

anonymous · Answer

Thank you.

anonymous · Answer

You're welcome.