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    can anyone tell me what a standard value for upthrust is and what units my equation should be in. i'm caculation it using U=4/3(pi)r(fluid density)g
    i got the fluid density values from the internet but i'm not sure what units they should be in for SI? other than that i know obviously r is in meters etc. any help would be appreciated. oh the upthrust is for vegetable oil. so olive oil will probrably do as its around the same consistency i'm just looking for a general answer . thanks.
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    (Original post by candiedmango)
    U=4/3(pi)r(fluid density)g
    Where did you get that from? What is U?
    What exactly are you working on? Are you dropping a small ball in a liquid? Then that force would be called buoyancy, and the radius r in your formula would be cubed.
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    yeh ive realised its cubed. U is upthrust. i am dropping a ball bearing of radius 0.95cm into oil.
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    (Original post by candiedmango)
    yeh ive realised its cubed. U is upthrust. i am dropping a ball bearing of radius 0.95cm into oil.
    What are you going to do when you drop it? Do you realise that there is the very important viscous force?

    denisty is in kilograms per metres cubed: \mathrm{kg/m^3}.
    g is in metres per seconds squared: \mathrm{m/s^2}.
    r is, as you say, in metres.

    Therefore the unit of force is \mathrm{m^3\cdot m/s^2 \cdot kg/m^3=kg\cdot m/s^2} which is called a newton.
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    okay then so the current value i have for my fluid density is 0.92 g/cm^3 would that mean to convert it to kg/m^3 it would be 0.92x10^-5. if so that would make my answer for U rediculously small. as it stands i have.

    4/3(pi)x(0.95x10^-2)^3x0.92x9.8 = 3.24x10^-5 so if i was to change it from g/cm^3 this would make the answer around 3.24x10^-10. with the weight of the ball being 0.02798kg, this means that the fluid density is almost exactly the same as the weight.
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    (Original post by candiedmango)
    okay then so the current value i have for my fluid density is 0.92 g/cm^3 would that mean to convert it to kg/m^3 it would be 0.92x10^-5. if so that would make my answer for U rediculously small. as it stands i have.

    4/3(pi)x(0.95x10^-2)^3x0.92x9.8 = 3.24x10^-5 so if i was to change it from g/cm^3 this would make the answer around 3.24x10^-10. with the weight of the ball being 0.02798kg, this means that the fluid density is almost exactly the same as the weight.
    The part in bold is incorrect. Can you recalculate it? If not, show your working so that I can advise you what's incorrect
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    my working is this (how its entered in calc) (((4/3)x(pi))x(0.95x10^-2)^3)x0.92x10^-5x9.8 = 3.237973784x10^-10. not sure how that has gone wrong unless the data i have for fluid density is wrong. the value was obtained from the internet. however i've also found one that is 900 kg/l. but if you put that into the equation then the upthrust is greater than the weight which would mean the value of F would be negative.
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    (Original post by candiedmango)
    my working is this (how its entered in calc) (((4/3)x(pi))x(0.95x10^-2)^3)x0.92x10^-5x9.8 = 3.237973784x10^-10. not sure how that has gone wrong unless the data i have for fluid density is wrong. the value was obtained from the internet. however i've also found one that is 900 kg/l. but if you put that into the equation then the upthrust is greater than the weight which would mean the value of F would be negative.
    I meant that I'd like you to show me how you make the transition from \mathrm{g/cm^3} to \mathrm{kg/m^3}.

    0.92\ \mathrm{g/cm^3} \neq 0.92 \cdot 10^{-5}\ \mathrm{kg/m^3}.
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    i just figured that g to kg is 10^-3 and cm to m is 10^-2. i was mainly asking you to clarify if that was correct before.
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    (Original post by candiedmango)
    i just figured that g to kg is 10^-3 and cm to m is 10^-2. i was mainly asking you to clarify if that was correct before.
    This is correct. But it doesn't lead to the conclusion that g/cm^3 = kg/m^3 x 10^-5.

    Spoiler:
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    \displaystyle \mathrm{\frac{g}{cm^3}=\frac{k g\cdot 10^{-3}}{\left( m \cdot 10^{-2}\right)^{3}}=\frac{kg}{m^3} \cdot \frac{10^{-3}}{10^{-6}}=\frac{kg}{m^3}\cdot 10^3}
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    oh, yes but with 0.92x10^3 as fluid density this leads to U being bigger than the weight. which would lead to a negative value for F or fluid resistance
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    (Original post by candiedmango)
    oh, yes but with 0.92x10^3 as fluid density this leads to U being bigger than the weight. which would lead to a negative value for F or fluid resistance
    No. It would give a negative value of the net force! Fluid resistance has nothing to do with what you're working on in the very moment!

    If upthrust is greater than weight, it simply means that the body will flow on the surface (or move upwards, if it's below surface initially). This is the case when density of the object is smaller than density of the fluid.

    I leave the calculations up to you.
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    but for my calculation i'm using U+F=W so W-U=F. And it moved downwards through the liquid.
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    (Original post by candiedmango)
    but for my calculation i'm using U+F=W so W-U=F. And it moved downwards through the liquid.
    Here F is the net force.

    So if F=W-U, it means that if F is positive, the object will fall; if F is 0, object will float; if F is negative, the object will stay on the surface (or move upwards).

    If the object sinks in the liquid, it means that upthrust is smaller than force. I take it that from your calculations it follows that the object should stay on the surface? Then I suggest that you check your data about the object first, and then recalculate upthrust and weight once again.
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    no, this is based on an experiment i did where all objects sank. This is why i was wondering about my value of upthrust. i know that W is correct so my only fault will be in calculating U.
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    (Original post by candiedmango)
    no, this is based on an experiment i did where all objects sank. This is why i was wondering about my value of upthrust. i know that W is correct so my only fault will be in calculating U.
    Agreed, I'm not gonna deny the experiment.

    That's exactly what I said, recalculate U then.

    If theory still disagrees with the experiment it clearly means that you have erroneous information about either the mass or the radius of the ball, or the density of the liquid.
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    Any recalculated value comes out the same, i believe the error must lie within the density of the liquid. As i found this on the internet.
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    What is the ball made of?

    From the values you have provided,

    m=0.02798 kg,
    r=0.95 cm,

    I've calculated the ball's density to be around 7.8\cdot 10^3\ \mathrm{kg/m^3}.

    It's around ten times smaller than density of iron, and I don't know any metals that would have such density. It's around wood's density, which floats on water, and so could do in oil.

    Edit: Of course the density is around iron's density, not ten times smaller.
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    it was a standard ball bearing so probrably a compound of metals.
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    (Original post by candiedmango)
    it was a standard ball bearing so probrably a compound of metals.
    Edit: Oops, sorry, my mistake. The density of the ball is just around iron's density. Which means that it should sink in the liquid. What is the value of upthrust you've obtained?
 
 
 
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