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A cylindrical container equipped with a manometer is inverted and pressed into water. The differential height of the manometer and the force needed to hold the container in place are to be determined

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Since pression in $A$ is the same as pression in $B$ i got

But since i cant find the value of $d$ i am unable to solve the rest of the exercice. Any suggestions?

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  • $\begingroup$ What d? You have D given. But why would you need that? H is related to the pressure by the SG. $\endgroup$
    – Solar Mike
    Sep 26, 2021 at 14:20
  • $\begingroup$ because the pression in A is p(atm)+ρ(SG)*gh + ρ(air)*gd and i know that p(atm)=0 but dont know the value of d $\endgroup$
    – pascal
    Sep 26, 2021 at 14:27
  • $\begingroup$ Well since you have just been handed the answer, you won’t learn much since you didn’t work for it. $\endgroup$
    – Solar Mike
    Sep 26, 2021 at 15:02
  • $\begingroup$ but why i dont need the value of d? $\endgroup$
    – pascal
    Sep 26, 2021 at 15:14
  • $\begingroup$ Where is d on your diagram? $\endgroup$
    – Solar Mike
    Sep 26, 2021 at 15:16

2 Answers 2

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Assume the tank is weightless, then

$F = 20\gamma_w A$

$20\gamma_w = \gamma_{mf}h$,

$SG = \dfrac{\gamma_{mf}}{\gamma_w} = 2.1$

Note the tank is pressurized internally by the floating bottom lid "A".

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the force need to hold the cylinder down is

$$F=\pi \frac{25^2}{4} 20/1000cm*1.033kg/atm- \text{weight of the cylinder}$$

the difference between the two sides of the manometer is

$$h=20cm/2.1=9.523cm$$

Edit

responding to OP's comment on how to measure the effective weight of partially submerged cylinder after deducting buoyancy.

$$W_{submerged cylinder}= W_{dry} - \pi *25*20*t$$

  • where t is the thicknesses of the cylinder. and weight is in grams.
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  • $\begingroup$ how you got those expressions ? $\endgroup$
    – pascal
    Sep 26, 2021 at 15:25
  • $\begingroup$ the pressure of the water at depth d = d*\Rho*g $\endgroup$
    – kamran
    Sep 26, 2021 at 16:32
  • $\begingroup$ and how do you calculate the weight of the cylinder since its partially immerse? (i know its not 6*9.81) $\endgroup$
    – pascal
    Sep 26, 2021 at 19:04
  • $\begingroup$ the weight of cylinder is dry weight - 25pi* thickness* 20 grams. $\endgroup$
    – kamran
    Sep 26, 2021 at 19:31
  • $\begingroup$ and how is that equal to 65N (which is the value I should have) ? if i do what you said i got 1570N $\endgroup$
    – pascal
    Sep 27, 2021 at 11:55

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