# What mass of water can be added to the box before it sinks?

Tuesday, September 28th, 2010 at
1:40 pm

A hollow cubical box, 0.29 m on a side, with walls of negligible thickness floats with 30 % of its volume submerged.

What mass of water can be added to the box before it sinks?

**Tagged with:** cubical • Mass Of Water • sinks

**Filed under:**
Kitchen

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To answer this question you should first understand the answer to another:

Why does the box float?

The box floats because it displaces 30% of its volume of water. The water surrounding the box pushes upwards against the box with a force equal to the weight of the volume of water displaced.

We can assume that the box is neither sinking nor rising, so the weight of the box must equal the upthrust of the surrounding water. So we can calculate the mass of the box:

mass of box*g – mass of displaced water*g = 0

mass of box – (0.29^3)*(30%)*(density of water) = 0

mass of box = 7.3167 kg

So the box will sink when the weight of water added to the box, plus the weight of the box, is greater than the weight of the water displaced by the whole box.

So for the box to stay afloat:

mass of added water + mass of box < (0.29^3)*(density of water)

mass of added water < 17.0723 kg

So about 17 kg can be added before it sinks. (with density of water taken as 1000 kg m^-3)

A more direct way to answer is to say that 0.3 of the the empty box is submerged, leaving 70% of the box above the water, so the mass of water that can be supported is the amount of water that would fill that 70% of the box.

Mass of added water = 0.7*(0.29^3)*(density of water).