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I tried to solve it but I don't know how to bring the Area A in this equation. I used the area of the small vessel, but this one shouldn't be in the equation. Thanks a lot

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  • $\begingroup$ The question text says density, mass and area are given - where? $\endgroup$
    – Solar Mike
    Mar 27 '19 at 16:17
  • $\begingroup$ Isn't the weight of the water displaced equal to the mass (added) of the floating object? In that case, the weight of water = density*H*(A-a), where a is the cross-sectional area of the floating vessel. $\endgroup$
    – JohnHoltz
    Mar 27 '19 at 16:45
  • $\begingroup$ $\Delta H$ is not the increase in depth of the small cylinder (h3-h2 in your image). $\Delta H$ is how high the fluid in the large cylinder rises. Since the total volume of the fluid in the large cylinder remains constant, and the small cylinder is submerged an additional h3-h2, you can calculate how high the water rises in the large cylinder. Does that help? $\endgroup$
    – JohnHoltz
    Mar 29 '19 at 3:13
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This question is asking you to describe how buoyancy works using cylinders because their volumes are easy to calculate.

First, some things we know:

  1. to be buoyant means an object has displaced a mass of fluid equal to it’s own mass.
  2. density is expressed as mass/volume, so if you know a fluid’s mass and density then you can find its volume by dividing mass by density or $\rho = m / V$, therefore $V = m / \rho$
  3. you know the mass of fluid being displaced - it’s exactly the mass being added to the inner cylinder
  4. the volume of a cylinder is its cross sectional area multiplied by its height, or $V = A \cdot H$

If you’re with me so far then putting all this together for your question looks like this:

$$\begin{align} V_{water} &= \dfrac{m}{\rho} \\ A_{total} &= A_{outer} - A_{inner} \end{align}$$

And given #4 above:

$$V_{water} = A_{total} \cdot H_{\delta}$$

Dividing by area to solve for change in height:

$$H_{\delta} = \dfrac{V_{water}}{A_{total}}$$

And substituting known values:

$$H_{\delta} = \dfrac{\frac{m}{\rho}}{A_{outer} - A_{inner}}$$

Did that explanation help you understand where area fits in?

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Since the mass of the fluid and its density are constant, the volume of the fluid is constant. Use that to find a formula that relates the change in height of the fluid on the side of the small cylinder to the change in height of the fluid on the side of the large cylinder.

If the small cylinder lowers by 1 unit, the fluid must rise on the big cylinder by (Aouter-Ainner)/A inner. So the level rises on the inner cylinder by 1 + (Aouter - Ainnier)/Ainner.

Thus the change in the level of the outer cylinder is [(Aouter-Ainner)/Ainner]/[1 + (Aouter-Ainner)/Ainner]. This can be further simplified with algebra.

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