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I have been trying for some time now, and seem to be unable to come up with the correct answer. The setup for the problem is as follows: Screenshot of problem

After finding a similar question and rechecking the buoyancy formula and Archimedes' Principle, I can't seem to determine where I've gone wrong.

So far, I have tried calculating out the difference if force using two methods (which, honestly are probably the same method): a. Calculating the difference in pressures and b. using buoyancy. I'll spare the details of the first method, but the second method is as follows:

Since the block is stationary, the net force on it is 0. Therefore, it is an accurate statement that $F_N + F_B - mg = 0$.

$F_N = mg - F_B$

$F_N = hA\rho_bg - hA\rho_wg$

$F_N = hAg (\rho_b - \rho_w)$

$F_N = 0.3*0.5*9.8*(6.7*10^3-2.5*10^3)$

$F_N = 0.6174 * 10^4$

Which is obviously not the correct answer, according to the website. (side note, 13.85 was just a random number I entered to get the answer).

So my question is, what is the proper way to do a buoyancy problem like this? Or is there somewhere I'm going wrong?

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closed as off-topic by John Rennie, Aaron Stevens, Jon Custer, Buzz, Gert Jan 18 at 19:51

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    $\begingroup$ perhaps you need to assume that there is no fluid between the floor and the block, in such a case there is no buoyancy but only preassure at the top. $\endgroup$ – Wolphram jonny Jan 18 at 5:33
  • $\begingroup$ That might be the case, I'll have to try that tomorrow. However, the next problem has the same setup, but asks "How much does the pressure measured under the block exceed atmospheric pressure?" which makes little sense if there is no pressure under the block. $\endgroup$ – Tropingenie Jan 18 at 6:17
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The problem with your solution is, that you are forgetting about the liquid- and air column above the box. Therefore your normal force is too small. The total amount of force exerted on the bottom area is given by the weight of the box (call it $F_b$), the weight of the liquid column above the box (call it $F_l$) and the weight of the air column above the liquid column (call it $F_a$). We than have $$F_N = F_b + F_l +F_a $$$$= hA\rho_bg + (y-h)A\rho_lg + p_aA$$$$ = 66490.3N$$

where $\rho_b$ denotes the density of the block, $\rho_l$ denotes the density of the liquid and $p_a$ denotes the atmospheric pressure.

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  • $\begingroup$ I've marked this answer as correct, since it's essentially what I did to get the right answer. After discussing with a colleage, it turns out that Wolphram jonny's suggestion was indeed correct. Thus, in using Archimedes' principle I was accidentally factoring in a nonexistent "pressure up." Once I got the free-body diagram correct, it was a simple matter of using deriving the same formulas as in this answer, and solving the problem. $\endgroup$ – Tropingenie Jan 18 at 18:58

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