Calculating Buoyancy Force: A Question Explained

AI Thread Summary
The discussion centers on calculating the new scale reading when a 20g piece of copper is suspended in water without touching the bottom. The buoyancy force is determined by the volume of water displaced, which is less than the weight of the copper since the piece sinks. The tension in the string holding the copper affects the scale reading, as the buoyant force reduces the effective weight on the scale. Participants clarify that the scale's reading will increase by the weight of the water displaced, not the weight of the copper itself. Ultimately, understanding the relationship between buoyancy and the forces involved is crucial for solving the problem.
cathliccat
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I have a question that reads "A container of water is placed on a scale and the scale reads 120g. Now a 20 g piece of copper (specific gravity = 8.9) is suspended from a thread and lowered into the water but does not touch the bottom of the container. What will the scale now read round off to the nearest whole number?"

This has to do with buoyancy force, but I'm not sure what to do here. The buoyancy force is equal to the amount displaced, so is it equal to 20g?

Can someone help me get started on this one?
 
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Originally posted by cathliccat
This has to do with buoyancy force, but I'm not sure what to do here. The buoyancy force is equal to the amount displaced, so is it equal to 20g?
No.
Hint: how much water was displaced by the copper?
 
The weight should be less than the weight of the copper coin because the coin sinks, and there is still tension on the string.

If you think of it in terms of the force of the string, then you initially have the string holding the entire weight of the coin(*), but when the coin is in the water, there is a bouyant force that acts to reduce the amount of force on the string. That buyant force is being exerted by the water, so the change in the force on the scale is equal to the change in the force on the string.

(*)Technically, there is a buyant force due to air, but it is so small that it is negligible in most situations.
 
I still don't know what to do. The water displaced: would that be 120-20 = 100?
 
Can you calculate the volume of the coin?
 
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