# Pressure of stacked bricks question

1. Oct 17, 2006

### Sabres151

I'm having difficulty with the part b.

Question: A stack of lead bricks measures 2.00m by 3.00m by 4.00m
a) Find the mass of the bricks.
b) Find the maximum and minimum possible pressures on the floor under the bricks in Pa and atm.

a)
A = 2m(3m)(4m)
A = 24m^3
mass = V(D)
Weight Density of lead = 11,340 kg/m^3
m = 24m^3(11,340 kg/m^3)
m = 272,160 kg.

b).....I'm thrown off by figuring minimum and maximum. Do I start with
P = F/A ??

2. Oct 18, 2006

### OlderDan

By A I think you mean V (volume). The units for your density suggest that is mass density, not weight density. If so, there has to be a g in your calculations. I don't see an area calculation in your work, bu tyou need area to find pressure. The stack has the same wight no matter how you arrange the bricks. The maximum pressure will be with the smallest area in contact with the floor, and the minuimum pressure will be with the greatest area in contact with the floor.

3. Oct 18, 2006

### Sabres151

Dan, thank you for the response. Yes, you are correct, by A, I should have V. And...should the calculation for the mass of the bricks be using weight density rather than mass density? I used mass density...

4. Oct 18, 2006

### OlderDan

You need the weight of the stack of bricks to find the pressure. The pressure would be different on the moon than it is on earth. It is easy to get the weight from the total mass, so you have all the information you need.

5. Oct 18, 2006

### Sabres151

Ok, now I'm really lost. For Area, I have 24 meters cubed. To find the mass of the bricks I was using the Mass Density of lead (11,340kg/m^3).
Since m = V x D I have a mass of 272,160kg.

6. Oct 18, 2006

### OlderDan

24meters cubed is the volume, and that is what you need to find the mass. How do you find the weight when you know the mass? The stack of bricks is in the shape of a rectange. Each face of the rectangle has an area. Any one of the faces could be on the bottom. The pressure is the weight divided by the area.