# Aquarium Pressure Problem

## Homework Statement

An aquarium tank is 100 cm long, 35 cm wide, and 40 cm deep. It is filled to the top

a)What is the force of the water on the bottom (100 cm x 35 cm) of the tank?

b) What is the force of the water on the front window (100 cm x 40 cm) of the tank?

## The Attempt at a Solution

For part a, can I just use I find the mass of the water and use mg? Actually I think I might have to use integration, but could someone explain it to me?

Last edited:

Related Introductory Physics Homework Help News on Phys.org
Dick
Homework Helper
Yes, for part a you can just use mg. Why? Think of a force diagram. For part b you have to integrate pressure over area.

For part b, I'm pretty sure I have to use integration.
P=Po + ro*g*y where y is the depth
P= (101300Pa) + (1000kg/m^3)(9.80)d

Say I cut the front surface into little rectangles--each with width .35 and height dy
So would I multiply the above by .35dy and take the integral evaluated from 0 to 40 ? Please help me set up this integral. Thanks

Dick
Homework Helper
You don't cut it into little rectangles, it's one big rectangle. Just integrate. P= (101300Pa) + (1000kg/m^3)(9.80)y. Where y is depth and x is the distance coordinate across the front. You'll want to integrate this dx*dy. What's the range of x and y? BTW it only asks for the pressure of the water, so you might want to drop the 101300Pa. That's from the air. The question is a little ambiguous about that. Try it both ways to make sure.

Last edited:
I don't understand what you mean by integrate this dx*dy.

Would I take the integral from 0 to .40 of (1000kg/m^3)(9.80)y * (1.0m) dy?

Dick
Homework Helper
I don't understand what you mean by integrate this dx*dy.

Would I take the integral from 0 to .40 of (1000kg/m^3)(9.80)y * (1.0m) dy?
That works also. If you haven't done double integration don't worry about it.

Ok so is this right: Integral of 1000*9.80*y dy

= 4900 y^2 evaluated from 0 to .4

4900* (.4)^2 = 784 N

Also why is it that I drop the pressure from the air--is it because the force from the air is down, but the force of the water on the front of the glass is horizontal?

Dick