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I just want to check my understanding of this:

Suppose I have a balance scale, with a small dense weight in the left pan and a large less-dense weight in the right pan. It happens to balance.

Now I move my scale and my weights into a vacuum. I expect it to tilt to the right.

My question is whether the following constitutes a full and accurate explanation of why it tips to the right:

1) The upward air pressure on the bottom of the left pan is equal to the upward air pressure on the bottom of the right pan. This equality still holds when I move to the vacuum (where both pressures are zero).

2) The gravitational force on both sides is unchanged when I move to the vacuum.

3) The downward air pressure on the top of the left pan is greater than the downward air pressure on the top of the right pan, because the large weight on the right displaces a bunch of air. When I move to the vacuum, this pressure disappears on both sides, which is a bigger reduction on the left than it is on the right.

4) This accounts for all the forces, and shows that the net reduction in downward force on the left side is larger than the net reduction in downward force on the right side. Therefore the scale tilts to the right.

Is that story both complete and correct?

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Since in air, the two bodies balance on the scale, equal downward forces are acting on both of them. Let subscripts 1,2, refer to body of lower and higher density respectively (both are assumed denser than air). Then balancing requires: \begin{align} (\rho_1-\rho_{air})V_1g & =(\rho_2-\rho_{air})V_2g \\ \end{align} Since $\rho_1>\rho_2$ we must have $V_1<V_2$. Now mass of the two bodies are $m_1=\rho_1V_1$ and $m_2=\rho_2V_2$. Substitution into above equation and rearranging gives: \begin{align} m_1-m_2 & =\rho_{air}(V_1-V_2)<0 \\ \therefore\quad m_1<m_2 \end{align} Therefore in vacuum the the scale will tip towards the side of lower density body, because there is more weight on that side ($m_1g<m_2g$).

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  • $\begingroup$ I believe this is the same as the argument I gave above. You've ignored my point 1), which is fine because that contributes a net difference of zero on the two sides of the balance anyway. My point 2), in your notation, was that $m_1$ and $m_2$ remain the same in and out of the vacuum, which you've made use of. My point 3) was that $\rho_{air}V_1g<\rho_{air}V_2g$. My point 4) was that the only relevant forces on side $i$ come from $\rho_iV_ig-\rho_{air}V_ig$. So I think you and I are saying exactly the same thing in slightly different words/notation. Or am I missing something? Thanks! $\endgroup$ – WillO Sep 1 '17 at 3:59
  • $\begingroup$ @WillO Discussion in terms of pressure exerted on upper and lower surfaces of the body is confusing to me. Would it alter your argument if the two bodies were of the same height? $\endgroup$ – Deep Sep 1 '17 at 4:03
  • $\begingroup$ No, my argument is based entirely on the amount of air displaced by each body. I am ignoring pressure differences due to height. $\endgroup$ – WillO Sep 1 '17 at 4:07
  • $\begingroup$ @WillO Your argument says that the pan tilts towards right, which is to the side of the body of higher density. But this is not correct as my answer shows. Is that simply a typo? $\endgroup$ – Deep Sep 1 '17 at 4:13
  • $\begingroup$ In other words, my argument was that balancing requires $\rho_{1}V_1g-P_{1up}-\rho_{air}V_1g=\rho_{2}V_2g-P_{2up}-\rho_{air}V_2g$ where $P_{iup}$ is the pressure exerted on the bottom of the pan by air pressure. I then cancelled the $P_{iup}$'s because they are equal, giving your equation. $\endgroup$ – WillO Sep 1 '17 at 4:14
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Lets neglect variations of pressure with height. A uniform pressure of any magnitude exerts no net force or moment on any closed volume. (If it did, the body would move or rotate in an environment that has no privileged direction) Take the closed volume to fit closely around your scale and weights. The force/moment due to air pressure is zero in both cases. Nothing else changes.

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  • $\begingroup$ I apologize if I'm misunderstanding you, but....are you concluding, then, that the scale remains balanced in the vacuum? $\endgroup$ – WillO Sep 1 '17 at 2:48
  • $\begingroup$ Yes, that is my conclusion! $\endgroup$ – Philip Roe Sep 1 '17 at 14:26

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