A Metal Block submereged in water from a spring scale

In summary, the conversation was about a physics problem involving a 14.0kg metal block suspended in water. The forces acting on the top and bottom of the block were discussed, as well as the reading on the spring scale and the difference between the forces on the bottom and top of the block. The conversation also touched on finding the pressure of the water and the area of the top and bottom of the block. The final part of the conversation focused on setting up an equation to solve for the cable tension and the importance of considering the direction of forces when subtracting them.
  • #1
Fittleroni
23
0

Homework Statement



* 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.


Homework Equations



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

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.
 
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  • #2
Fittleroni said:
*Po = 1.0130 x 10^5
That's atmospheric pressure. Good, you'll need it.
p = m/v,
v= .10 x .10 x .12 = 0.0012
m = 14
Therefore p = 11666
That's the density of the metal block. You don't need it.
g= 9.81
h = 0.1
Therefore P = 112744
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
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.
 
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  • #4
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  • #5
So...

The density of water is 1g/cm^3, so to find the weight I would multiply that by 9.81m/s/s?
 
  • #6
The weight of what? Find the pressure due to the water.
 
  • #7
I don't know what I am doing, this is the last question on my assignment, and I don't have a clue how to solve this. How do I find the fluid density?
 
  • #8
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
The bottom of the block is 17.05cm below the surface of the water, is this correct?
 
  • #10
1000*0.0505*9.81 = 495.4 = P
Am I supposed to add P0 = 1.0130 105 N/m2
 
  • #11
Fittleroni said:
How do I find the fluid density?
You already have the density of water (you gave it in post #5): it's 1 gm/cm^3 or 1000 kg/m^3.

Fittleroni said:
The bottom of the block is 17.05cm below the surface of the water, is this correct?
Right.

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

[tex]P = P_{atm} + \rho_{water} g h[/tex]
 
  • #12
Allright then, so that's 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
Fittleroni said:
How would I find the area of the top and bottom of the block.
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.)

Is this the area that I will be using the F=PA equation?
Yep. Be sure to use proper units: Force in Newtons, area in m^2, pressure in N/m^2.
 
  • #14
Great, so I got 1029N for the F(bottom), which was correct. How could I find the reading on the spring scale, wouldn't it be 1029N?
 
  • #15
Fittleroni said:
How could I find the reading on the spring scale, wouldn't it be 1029N?
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
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
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
Fittleroni said:
The forces must add up to zero.
Right.
Are these forces F(top)=1017, F(bottom)=1029,
In what direction do they act?
& F(g)=9.81?
F_g = mg
Tension = sqrt (m/u)?
No. Just call it T. That's the scale force which is what you're going to solve for.
How would I set up an equation?
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
Would it be -1017 + 1029 -137 +F4 = 0
 
  • #20
Fittleroni said:
Would it be -1017 + 1029 -137 +F4 = 0
Looks good. (I get slightly different values for the water pressure force--1018 & 1030--but that's no big deal.) Solve for F4.
 
  • #21
Thanks. I got F4 = 125. Which also was correct. One last question, for finding the difference between the forces on the bottom and the top of the block, I tried subtracting F1 from F2, and this did not work. I thought the difference was just a mere subtraction.
 
  • #22
Sure it's just a subtraction, but signs count! What did you get?
 
  • #23
I got 12
 
  • #24
Or is it 2046?
 
  • #25
Perhaps what they want the difference between the total force acting on the top and the bottom. Realize that while the water pushes down on the top, there's also the tension force pulling up on the top. What's the total force on the top of the block?
 
  • #26
I don't know. So lte see, the F(top) is equal to 1017N, and there is a 125 N force pulling up. So would it be 1017N-125N?
 
  • #27
Or sorry, 1017N + 125N?
 
  • #28
Fittleroni said:
I don't know. So lte see, the F(top) is equal to 1017N, and there is a 125 N force pulling up. So would it be 1017N-125N?
That's right. The net force acting on the top of the block is 1017N -125N downward.

You know the force on the bottom. So find the difference.
 
  • #29
Allright, thank you for basically tutoring me on this question. You're a great teacher.
 

1. What is the purpose of submerging a metal block in water using a spring scale?

The purpose of submerging a metal block in water using a spring scale is to measure the buoyant force acting on the block. This can help determine the density of the metal block and the specific gravity of the fluid.

2. How does the spring scale work in this experiment?

The spring scale works by measuring the amount of force needed to stretch or compress a spring. In this experiment, the spring scale is attached to the metal block and the force of the buoyant force is measured.

3. What factors affect the accuracy of the measurements in this experiment?

The accuracy of the measurements in this experiment can be affected by several factors, such as the accuracy of the spring scale, the accuracy of the measurements of the metal block, and any external forces that may affect the buoyant force, such as air currents or vibrations.

4. How can this experiment be used to determine the density of the metal block?

By measuring the buoyant force acting on the metal block and knowing the specific gravity of the fluid, the density of the metal block can be calculated using the formula density = specific gravity x density of the fluid.

5. What other applications can this experiment have in the field of science?

This experiment can have various applications in the field of science, such as determining the specific gravity of different fluids, studying the buoyancy of objects, and understanding the principles of Archimedes' principle. It can also be used to demonstrate the concept of density and its importance in various scientific fields.

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