Under Pressure - Fluid Properties

Remember, knowing the definitions and equations is CRUCIAL in fluid mechanics since the tough parts come from applying them

Question 3. The pressure is the same in each case, as well as the area, and this makes the force exerted on the bottom face of the container the same for all three. There is some downward pressure exerted on the diagonal sides in the second and third container. This makes up for the additional weight of water in the second and third container.

Problems:

9. This is straight plug-and-chug - the force is just P*A = 3.08 x 10⁵ N. The force on the bottom of the table is the same, as the atmospheric pressure is the same and the area is as well.


11. Make sure you convert the area into square meters! 200 cm^2 *(1 m/100 cm)^2 = .02 m^2

The total force is 4*P*A = 19,200 N = mg. Thus m = 1959 kg.


16. The key to this problem is in the hint - the pressure is the same at both a and b.

rho_oil*g*H1 + P₀ = rho_water*g*H2 + P₀

rho_oil = 654 kg/m^3

17. The total height of the water over the bottom of the hill is 5.0 m + 100 m sin 60 = 91.6 meters.

The total pressure at the bottom is given by the equation we derived today:

P = Patm + rho*g*h which is the absolute pressure at the bottom.

If we now subtract Patm from both sides, we get that the gauge pressure is rho*g*h = 8.98 x 10⁵ Pa.

18. This is similar to 17. P_absolute = Patm + rho*g*h so the gauge pressure is P_absolute - Patm = rho*g*h = 4.02 x 10⁵ Pa

19.
a) The volume of the water in the tube is pi*R^2*H = 3.39 x 10^-4 m^3. (Don't forget to convert to meters!) Since rho = M/V, M = rho*V = 0.339 kg.

b) The pressure on the inside surface of the barrel lid is given by Patm + rho*g*h. The pressure on the outside surface of the lid is Patm. Thus the net pressure will be the gauge pressure, rho*g*h. The force is therefore equal to the pressure times the area: rho*g*h*A = 1000 kg/m^3 * 9.8 m/s^2 * 12 m * pi*(.2 m)^2 = 1.48 x 10⁴ N