Fluids (Find the Force Exerted)

[Bcisewski]

Need some assistance on what I am missing
"A tank contains mercury, whose density is 13 600 kg/m3. Find the force exerted by the mercury on a circular plug at the bottom of the tank. The plug has a diameter of 2.54 cm, and is located 37.6 cm below the surface of the mercury". The answer is 25.4 N

1) I have tried p=F/A ending with A*p=F or 2.54*136000=F (Wrong)
2) and F=phgA and again the number to big
3) Finally I tried P2=P1+pgh with P1=(1.01x10^-5) again too large of an answer

Am I missing something from try #1?

Thanks for any assistance

 

[cepheid]

Your first attempt is wrong because you used the density in the formula for pressure! (it's ok...an honest mistake).

pressure (letter 'p'): force per unit area

p = \frac{F}{A}

density (letter 'rho'): mass per unit volume

\rho = \frac{m}{V}

I'm curious...where did you get the answer 25.4 N? Is it given in the book? Because it doesn't specify what the pressure is above the surface of the tank. Assuming it was atmospheric pressure, and using the third formula you tried, I got the wrong answer. But assuming it (p1) was zero, I got the following:

p_2 = 0 + \rho g h

F = pA = (\rho g h)(\pi r^2) = (13 600 kg/m³ )(9.81 N/kg)(0.376m)(\pi)((0.0254m)/2)² = 25.4 N

That's the right answer, but the method seems a little off...does the problem give more info about the pressure above the tank?

 

[Clausius2]

Your first attempt is wrong because you used the density in the formula for pressure! (it's ok...an honest mistake).

pressure (letter 'p'): force per unit area

p = \frac{F}{A}

density (letter 'rho'): mass per unit volume

\rho = \frac{m}{V}

I'm curious...where did you get the answer 25.4 N? Is it given in the book? Because it doesn't specify what the pressure is above the surface of the tank. Assuming it was atmospheric pressure, and using the third formula you tried, I got the wrong answer. But assuming it (p1) was zero, I got the following:

p_2 = 0 + \rho g h

F = pA = (\rho g h)(\pi r^2) = (13 600 kg/m³ )(9.81 N/kg)(0.376m)(\pi)((0.0254m)/2)² = 25.4 N

That's the right answer, but the method seems a little off...does the problem give more info about the pressure above the tank?

You solved it right, cepheid. Below the plug, there is atmospheric pressure too. So that, the force exerted by the atmosphere is canceled at the two sides.