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cathliccat
Dec2-03, 02:54 PM
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?

Doc Al
Dec2-03, 03:04 PM
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?

NateTG
Dec2-03, 03:05 PM
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.

cathliccat
Dec5-03, 02:34 PM
I still don't know what to do. The water displaced: would that be 120-20 = 100?

NateTG
Dec5-03, 03:28 PM
Can you calculate the volume of the coin?