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Find the center of mass of a block of wood in oil

  • #1
A small square plank of oak floats in a beaker half full of water. The piece of oak is 7 cm on a side and 3 cm thick and floats on its side as shown in Figure P10.80.
Density of oak 750kg/m^3
Density of water 1000kg/m^3

There were two parts to the problem, the first is:
Find the location of the center of mass of the plank with respect to the water’s surface. Someone else gave me the equation, so I was able to get this part, but I can't find the equation myself anywhere.

s=side length in meters
w=width in meters
?(oak)=750
?(water)=1000

center mass= w-[(s^2)*w*750/(1000*(s^2))]

Second part (I need help here) is: A very light vegetable oil is poured slowly into the beaker, so that the oil floats on the water without mixing and until the oak plank is a few centimeters below the surface of the oil. If the density of the oil is 640 kg/m3, calculate the new position of the center of mass of the plank with respect to the water’s surface.

For the first part, I got .0075m. As I stated, someone just gave me the equation so I got the answer, but it didn't help me with understanding how to solve the problem. So I am completely lost and if anyone could just point me in the right direction, it would be appreciated.
 

Answers and Replies

  • #2
gneill
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Hi Angie_Agnes, welcome to Physics Forums. In future, please use and follow the posting template for the homework forums.

To derive the equation for the location of the center of mass with respect to the water surface you need to know a couple of facts:

1. The block will sink into the water until it displaces its own weight in water.
2. The center of mass of the block of oak is located half way between the horizontal faces of the block.

So you need to determine the amount of water displaced by the block versus how much of the block is submerged. Solve for the submersion depth that balances the weight of water displaced with the weight of the block. Adjust for the location of the C.O.M. with respect to the bottom of the block.

Another way to look at floatation or buoyancy is to consider that it is the force due to the liquid pressure on the bottom surface of the block that supports it in the water. In effect the block "sits on" a surface that is pressing upwards with a certain pressure. The block will sink until the pressure x surface area of the block's bottom is equal to the weight of the block. This way of looking at buoyancy might help you for the second part of the problem. Do you know a formula for the pressure versus depth in a liquid?
 
  • #3
The equations I know of:
P0=P(atm) or P at a reference point
d=density
V=Volume

P=P0+dgh

Force of buoyancy(Fb): Fb=dVg

I found the volume of submersion and got .000172 m by using d(oil)/d(oak)=V(oak)/(Vsubmerged) I don't know how to adjust for COM though.
 
  • #4
gneill
Mentor
20,792
2,770
Since the density of oil is less than that of oak the block will want to sink (completely) in oil. That is, the buoyancy provided by the oil alone is not enough to support the block. Thus it will still sink somewhat into the water, too.

Consider the case where the oil is carefully poured until the level of the oil just reaches the height of the block as it floats in the water. The block's top surface will be level with the oil surface at some distance h1 from the water surface (so h1 is the depth of the oil) and the blocks bottom surface will a distance h2 below the water level. Note that the thickness of the block, w, is the sum of h1 and h2.

Using your expression P = P0 + d*g*h, can you write an expression for the pressure at bottom of the block? You can ignore the P0 due to the atmospheric pressure, since it presses down equally on everything and will cancel out in the balanced equations. Initially you can write the expression in terms of h1, h2, and the densities.

attachment.php?attachmentid=40413&stc=1&d=1319814036.gif
 

Attachments

  • #5
P=ρgh -> P=1000*9.8*h2
 
  • #6
gneill
Mentor
20,792
2,770
P=ρgh -> P=1000*9.8*h2
You're forgetting the weight of the oil on top of the water...

Start with the pressure at the bottom of the oil layer (at the oil-water interface). That'll be your Po for the descent into the water layer.
 
  • #7
P=640*9.8*h1+1000*9.8*h2
h1=.03-h2
P=640*9.8*(.03-h2)+1000*9.8*h2
P=188.16-16072h2=P
 
  • #8
gneill
Mentor
20,792
2,770
P=640*9.8*h1+1000*9.8*h2
h1=.03-h2
P=640*9.8*(.03-h2)+1000*9.8*h2
P=188.16-16072h2=P
I think you might have miscalculated the h2 coefficient when you wrote your last line. Check your expansion.

Once that's fixed, solve for h2 in terms of P. What must P be in order to support the block?
 
  • #9
Thanks for helping me with this, you have been very patient.
I now have:

h2=(p-188.16)/3528

Now for what P needs to be to support the block, is it buoyant force? so it would have to equal mass x gravity of the block. Which would be (.11025*9.8) = 1.08045.

h2=(1.08-188.16)/3528 -> -.053m
 
  • #10
gneill
Mentor
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Very close! But P is a pressure, not a weight. What pressure is required to support the block?
HINT: The block's weight is distributed over its bottom surface area.
 
  • #11
ahhh yes, makes sense- P=F/A so... P=mg/a
P=1.08045/(.07^2) = 220.5
so applying that to the equation: h2=(220.5-188.16)/3528=.009167m
 
  • #12
gneill
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That looks good! So now you've got h2, the depth of the block's bottom below the water surface. Can you now find the location of the block's center of mass in relation to the water surface?
 
  • #13
is it just h2*.5? .0045835m
 
  • #14
gneill
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Look at the geometry. The center of mass of the block is a fixed distance from its top and bottom.
 
  • #15
.03-h2=h1 = .020833m
-> .03/2= .015m
.020833-.015=.005833m ??
 
  • #16
gneill
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Yup. Now all you need to do is express the result in cm and use a suitable number of significant digits. :smile:
 
  • #17
Thanks so much, you are awesome. Too late for me to turn in the homework, but I at least understand the problem now.The picture definitely helped.
 
  • #18
gneill
Mentor
20,792
2,770
Thanks so much, you are awesome. Too late for me to turn in the homework, but I at least understand the problem now.The picture definitely helped.
Glad I could be of help!
 

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