# A Metal Block submereged in water from a spring scale

1. Nov 22, 2007

### Fittleroni

1. The problem statement, all variables and given/known data

* A 14.0kg block of metal measuring 12cm x 10cm x 10cm is suspended from a scale and immersed in water. The 12.0 cm dimension is vertical and the top of the block is 5.05 cm below the surface of the water.

(a) What are the forces acting on the top and on the bottom of the block? (Use P0 = 1.0130 105 N/m2.)
Ftop =
Fbottom =

(b) What is the reading of the spring scale?

(c) Find the difference between the forces on the bottom and the top of the block.

2. Relevant equations

*P = Po + pgh ( for part A)
And for part (c) I know the difference is equal to the buoyant force.

3. The attempt at a solution

*Po = 1.0130 x 10^5
p = m/v,
v= .10 x .10 x .12 = 0.0012
m = 14
Therefore p = 11666
g= 9.81
h = 0.1
Therefore P = 112744
F=PA
A(top)=0.1
Ftop= 11274 x 0.1 = 11274
Which is wrong, so after using up all my tries for my web assign, I don't know where to go. I cannot answer for Fbottom using that formula because it is not correct. I think the problem may lie in the numbers I am using for h, and A.
:surprised:surprised:surprised:surprised:surprised:surprised

2. Nov 22, 2007

### Staff: Mentor

That's atmospheric pressure. Good, you'll need it.
That's the density of the metal block. You don't need it.
Not sure what you did here.

What's the pressure 5.05 cm below the surface of the water? (What's the density of water?) Hint: The total pressure at any point is atmospheric pressure plus the pressure due to the water depth.

3. Nov 22, 2007

### Fittleroni

How do I find the pressure?

below the surface of the water, and why don't I need the density?
Also (h) is the 10cm converted to meters.
And the P is from the equation *P = Po + pgh ( for part A). I figured I could solve it because I have all the variables.

Last edited: Nov 22, 2007
4. Nov 22, 2007

### Staff: Mentor

The static pressure due to a given depth of fluid is given by $\rho g h$. $\rho$ is the density of the fluid, not of the submerged object.

Read more here: Static Fluid Pressure

5. Nov 22, 2007

### Fittleroni

So....

The density of water is 1g/cm^3, so to find the weight I would multiply that by 9.81m/s/s?

6. Nov 22, 2007

### Staff: Mentor

The weight of what? Find the pressure due to the water.

7. Nov 22, 2007

### Fittleroni

I don't know what I am doing, this is the last question on my assignment, and I dont have a clue how to solve this. How do I find the fluid density?

8. Nov 22, 2007

### Fittleroni

The pressure exerted by a static fluid depends only upon the depth of the fluid, the density of the fluid, and the acceleration of gravity. So, I know the depth 5.05cm, (for the top of the block), g = 9.81 m/s/s, and the density is.............

9. Nov 22, 2007

### Fittleroni

The bottom of the block is 17.05cm below the surface of the water, is this correct?

10. Nov 22, 2007

### Fittleroni

1000*0.0505*9.81 = 495.4 = P
Am I supposed to add P0 = 1.0130 105 N/m2

11. Nov 22, 2007

### Staff: Mentor

You already have the density of water (you gave it in post #5): it's 1 gm/cm^3 or 1000 kg/m^3.

Right.

The total pressure at any point a distance "h" below the water surface is given by:

$$P = P_{atm} + \rho_{water} g h$$

12. Nov 22, 2007

### Fittleroni

Allright then, so thats settled. Thanks.
How would I find the area of the top and bottom of the block. Is this the area that I will be using the F=PA equation?

13. Nov 22, 2007

### Staff: Mentor

You have the dimensions of the block. Area = length X width. (They tell you which one is the vertical, so don't use that one.)

Yep. Be sure to use proper units: Force in Newtons, area in m^2, pressure in N/m^2.

14. Nov 22, 2007

### Fittleroni

Great, so I got 1029N for the F(bottom), which was correct. How could I find the reading on the spring scale, wouldnt it be 1029N?

15. Nov 22, 2007

### Staff: Mentor

No, you have to figure it out step by step. The spring scale reads the tension in the cable attached to the block. Identify all the forces acting on the block (I count four--the cable tension is one of them). What must those forces add up to since the block is in equilibrium? Set up an equation and solve for the cable tension.

16. Nov 22, 2007

### Fittleroni

The forces must add up to zero. Are these forces F(top)=1017, F(bottom)=1029, & F(g)=9.81? How would I set up an equation? Tension = sqrt (m/u)?

17. Nov 22, 2007

### Fittleroni

So the three forces are 1017(Ftop) 1029(Fbottom), & 9.81(Fg). Set them to equal zero. And is this how I solve for the tension in the cable? T= sqrtm/u or sqrt m/g?

18. Nov 22, 2007

### Staff: Mentor

Right.
In what direction do they act?
F_g = mg
No. Just call it T. That's the scale force which is what you're going to solve for.
F_1 + F_2 + F_3 + F_4 = 0

But make sure you use correct signs for each force: If it points up, make it positive; if it points down, make it negative.

19. Nov 22, 2007

### Fittleroni

Would it be -1017 + 1029 -137 +F4 = 0

20. Nov 22, 2007

### Staff: Mentor

Looks good. (I get slightly different values for the water pressure force--1018 & 1030--but that's no big deal.) Solve for F4.