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I am having trouble understanding the reasoning for a solution for a buoyant forces question.

The density of a human body can be calculated from its weight in air $W_{\text{air}}$, and its weight while submersed in water $W_w$. The density of a human body is proportional to:

Solution: $$\frac{W_{\text{air}}}{W_{\text{air}}-W_{\text{w}}}$$

Why does $F_{\text{buoy}}=W_{\text{air}}-W_{\text{w}}$? Why do we have to divide $W_{\text{air}}$ by $F_{\text{buoy}}$?

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  • $\begingroup$ This might help - Buoyant force on partially submerged bodies $\endgroup$
    – mmesser314
    Jul 18 at 23:08
  • $\begingroup$ @mmesser314 Thanks but this doesn't quite help me as it doesn't explain why $F_b=W_{\text{air}}-W_{\text{w}}$ $\endgroup$ Jul 18 at 23:12
  • $\begingroup$ No, but it does gives some hints as to how to think about the problem. Another hint is that the buoyant force is the total force on the submerged object. It is the vector sum of the forces of gravity and the upward force from the water. $\endgroup$
    – mmesser314
    Jul 18 at 23:16
  • $\begingroup$ @mmesser314 This is what I am confused about. If water is an upward force, then why is it $W_{\text{air}}-W_{\text{w}}$? Shouldnt it be $W_{\text{w}}-W_{\text{air}}=F_b$? $\endgroup$ Jul 18 at 23:20
  • $\begingroup$ With the weights defined as the reading on a scale, your formula gives the ratio of the density of an object which sinks, to that of water. It would not work for a human who can float (and has no weight in water). (It assumes that the density of air is negligible.) (The buoyant force is defined as the decrease in weight provided by the fluid.) $\endgroup$
    – R.W. Bird
    Jul 22 at 18:17
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The Bouyant Force is (volume of the body submerged)(density of the fluid)(gravitational acceleration) or in short $Vdg$.

Now the weight in fluid (say $x$) is nothing but the net downward force on the body inside the fluid which will be $Mg-Vdg$. So we have $x=Mg-Vdg$ and therefore $Vdg=Mg-x$ that is Bouyant force is (Weight in air)-(Weight in water).

Hope this helps.

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  • $\begingroup$ Thanks for your answer! Why are we subtracting $Mg$ by $Vdg$? is $Mg$ an upward force? $\endgroup$ Jul 19 at 0:28
  • $\begingroup$ @SamKirkiles No, Mg is the downward force and Vdg is the upward force. So the net downward force will be Mg-Vdg $\endgroup$ Jul 19 at 1:36

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