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.)

Question

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

bobpursley · Answer

pressure= total weight/area I will be happy to critique your thinking.

AL · Answer

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

bobpursley · Answer

smallest number of bricks will mean the greatest pressure, or smallest area. The side of the brick giving the smallest area will be .0890 x .0570, wouldn't it? The weight on that face gives the greatest pressure.

AL · Answer

oo ok cool thanks