How Much Lead Can a Tin Cup Carry Without Sinking?

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To determine how much lead a tin cup can carry without sinking, the buoyant force must equal the total weight of the cup and the lead. The tin cup has a volume of 0.012 m^3 and a mass of 0.13 kg, resulting in a downward force of 860 N. The buoyant force can be calculated using the weight of the water displaced by the cup, which is equal to the volume of the cup multiplied by the density of water. By setting the equation Ftin + Flead = Fwater, the maximum mass of lead can be derived. This analysis utilizes Archimedes' Principle to ensure the cup remains afloat.
r3dxP
A tin cup has a total volume of .012m^3 and mass of .13kg. How many grams of lead shot of density 11.4g/cm^3 could it carry without sinking in water?

the question i have here is what is equal to what and how to solve this?
so far.. i have.. Ftin = Vpg = .012m^3 * 7310kg/m^3 * 9.8m/s^2 = 860N
the tin alone pushes down 860N onto the water.
how can i find the force pushed by the water in the sink? cause once i find this force, i can set the equation.. Ftin + Flead = Fwater
once i get the Flead, i can solve for mass by Flead / g..
 
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And Archimedes' Principle rears its attractive head for the second time in twenty-four hours.

The buoyant force will equal the weight of the water displaced by the cup.
 
how can i solve for the buoyant force?
 
You know the total volume of the cup - you therefore know the maximum amount of water it can displace, yes? Just find the weight of that much water and Bob's your uncle.

Clear?
 
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