Calculating weight of underwater object given normal force

[mm2424]

Homework Statement

A glass ball of radius 2 cm sits at the bottom of a container of milk that has a density of 1.03 g/cm^3. THe normal force on the ball from the container's lower surface has a magnitude of 9.48 * 10^-2 N. What is the mass of the ball?

Homework Equations

Normal force + buoyant force = Weight

The Attempt at a Solution

My prof supplied the answer to this problem, so I know we're supposed to calculate the buoyant force (which I understand how to do) and add it to the normal force. What I don't understand is that when I first saw this question, I assumed the following: the milk and the atmosphere above the ball are all pushing down on the ball. The normal force therefore must contend with not only the weight of the ball but also the weight of the milk and air above. So I figured to answer the problem we'd have to take the normal force, add the buoyant force, then subtract the force from the milk and air above. However, the problem doesn't give you enough information to do that, since we don't know the height of the milk above the ball...so I assume my reasoning is wrong. Can someone explain why this problem is simpler than I initially thought?

Thanks!

 

[netgypsy]

OK so you realize that you have to find the sum of the forces and since it isn't moving, set the expression equal to zero or you can set the up forces = the down forces.
A carton of milk is not an 8 foot deep pool so don't worry about any downward pressure of the milk. So you have the gravitational weight of the ball downward, the normal force upward and the buoyant force exerted by the amount of milk displaced by the ball. So first you have to find the buoyant force. Can you do that?

 

[mm2424]

Hi, thanks. yes, I understand how to calculate the upward force due to the displaced milk. Just so I understand, though, if this problem entailed a ball on the bottom of an 8 foot deep pool of milk, would I then have to do something with the pressure of the milk and the atmosphere above? And in this problem, is the pressure of the atmosphere above negligible?

 

[netgypsy]

Good question. On a ball the pressure from the water is uniform if I remember correctly. And it doesn't compress so the water isn't any heavier at that depth since liquids don't compress. This isn't my area of expertise but my gut feeling is you would do it the same way. It's gases way down that cause serious problems, not solids. The problem with people diving is that we breathe gases and have gases in our ears and lungs and so on and they compress and expand and their solubility changes and so on.

 

[asteriatic]

I would like to clarify that the weight of an object is equal to its mass multiplied by the acceleration due to gravity, not just its mass. Therefore, the weight of the glass ball can be calculated by multiplying its mass by the acceleration due to gravity (9.8 m/s^2).

In this problem, we are given the normal force acting on the ball, which is equal to the weight of the ball. We also know the density of the milk and the radius of the ball. Using the buoyant force equation (buoyant force = density of fluid x volume of displaced fluid x acceleration due to gravity), we can calculate the buoyant force acting on the ball.

Now, to find the mass of the ball, we can use the weight equation (weight = mass x acceleration due to gravity) and substitute the known values for weight (normal force + buoyant force) and acceleration due to gravity. Solving for mass, we can find that the mass of the glass ball is 0.000307 kg.

To address your initial confusion, the normal force is only concerned with the weight of the object itself, not the weight of any other substances above it. The buoyant force takes into account the weight of the displaced fluid, which in this case is the milk. Therefore, we do not need to consider the weight of the milk and air above the ball in this problem.

I hope this explanation helps to clarify any confusion. Remember to always use the appropriate equations and units when solving physics problems.