# Fluids: Pressure

1. Nov 25, 2007

### koolkris623

1. The problem statement, all variables and given/known data
A brick weighs 15.0 N and is resting on the ground. Its dimensions are 0.203 m 0.0890 m 0.0570 m. A number of the bricks are then stacked on top of this one. What is the smallest number of whole bricks (including the one on the ground) that could be used, so that their weight creates a pressure of at least one atmosphere on the ground beneath the first brick? (Hint: First decide which face of the brick is in contact with the ground.)

pressure = force / area
density = Mass / volume
1atm = 101300 Pa
Volume = l* w* h

3. The attempt at a solution
well what i did was i found the area of the face to be .203 * .0890..then i divided 15/ Area to get pressure then i did 101300 / pressure to get number of bricks..this doesnt work

2. Nov 25, 2007

### Shooting Star

To get max pressure, the area should be the least. Do you think you have found that correctly?

After that, total weight/area = pressure on the ground.