Calculating weight of underwater object given normal force

AI Thread Summary
To calculate the weight of a glass ball submerged in milk, the normal force and buoyant force must be considered. The normal force acting on the ball is 9.48 * 10^-2 N, while the buoyant force arises from the volume of milk displaced by the ball. The discussion clarifies that the pressure from the milk above is negligible in this scenario, as the container is not deep enough to significantly affect the ball's weight. If the situation involved a deeper body of liquid, atmospheric pressure would need to be accounted for, but in this case, it simplifies to balancing the forces. Understanding these principles allows for the correct calculation of the ball's mass without needing additional information about the liquid's height.
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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!
 
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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?
 
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?
 
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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