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Buoyant Force & Acceleration

  • Thread starter AG1189
  • Start date
5
0
1. Homework Statement
-A thin spherical shell of mass 0.300 kg and diameter 0.250 m is filled with alcohol ( = 806 kg/m3). It is then released from rest on the bottom of a pool of water. Find the acceleration of the alcohol filled shell as it rises toward the surface of the water.

2. Homework Equations
FB-(M1+M2)g=(M1+M2)a


3. The Attempt at a Solution
M1=0.3kg, r=0.125 m, V= (4*pi*r^3) /3,
M2 = pV = (806*V), FB = pVg = (1000)(V)(9.8)
. After finding FB, I plugged it back into the equation and got 2.211 m/s^2. This answer was wrong, could someone please tell me if I am using the wrong equation, or if I just screwed something up.
 
175
0
1. Homework Statement
-A thin spherical shell of mass 0.300 kg and diameter 0.250 m is filled with alcohol ( = 806 kg/m3). It is then released from rest on the bottom of a pool of water. Find the acceleration of the alcohol filled shell as it rises toward the surface of the water.

2. Homework Equations
FB-(M1+M2)g=(M1+M2)a


3. The Attempt at a Solution
M1=0.3kg, r=0.125 m, V= (4*pi*r^3) /3,
M2 = pV = (806*V), FB = pVg = (1000)(V)(9.8)
. After finding FB, I plugged it back into the equation and got 2.211 m/s^2. This answer was wrong, could someone please tell me if I am using the wrong equation, or if I just screwed something up.
It looks like you might have just done something wrong in your calculations...I set it up like this:

[tex]
\sum{F} = \rho_{water} V g - \rho_{alcohol} V g - mg = (m + \rho_{alcohol} V) a
[/tex]

[tex]
a = \frac{g(\frac{4}{3} \pi r^3 (\rho_{water} - \rho_{alcohol}) - m)}{m + \rho_{alcohol} \frac{4}{3} \pi r^3}
[/tex]

and I got a number a little lower than yours. Maybe it was rounding? Generally I like to get everything in one equation and then plug it all in at once to reduce the chance of that sort of mistake.
 
5
0
:)

it worked out, thank you very much. I got 1.829711 m/s^2, I think thats what you got. :smile: :smile:
 

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