A Metal Block submereged in water from a spring scale

[Fittleroni]

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.

 

[Doc Al]

*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.

 

[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.

 

[Doc Al]

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: http://hyperphysics.phy-astr.gsu.edu/hbase/pflu.html#fp"

 

[Fittleroni]

So...

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

 

[Doc Al]

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

 

[Fittleroni]

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?

 

[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....

 

[Fittleroni]

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

 

[Fittleroni]

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

 

[Doc Al]

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.

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:

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

 

[Fittleroni]

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?

 

[Doc Al]

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.

 

[Fittleroni]

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?

 

[Doc Al]

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.

 

[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)?

 

[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?

 

[Doc Al]

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.

 

[Fittleroni]

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

 

[Doc Al]

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.

 

[Fittleroni]

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.

 

[Doc Al]

Sure it's just a subtraction, but signs count! What did you get?

 

[Fittleroni]

I got 12

 

[Fittleroni]

Or is it 2046?

 

[Doc Al]

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?

 

[Fittleroni]

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?

 

[Fittleroni]

Or sorry, 1017N + 125N?

 

[Doc Al]

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.

 

[Fittleroni]

Allright, thank you for basically tutoring me on this question. You're a great teacher.