1 A. A 2011 penny is thrown into a pool of water for luck. What is the force act

Question

1 A. A 2011 penny is thrown into a pool of water for luck. What is the force acting on the penny if it lands horizontally on the bottom of the 50 cm deep pool? An extra lucky toss has the penny land vertically along the side of the pool at the bottom. What is the force acting on the penny in this case? Assume that the temperature of the water is 10°C and the diameter of the penny is 19.05 mm.

B. A pressurized tank contains oil (SG = 0.90) and has a square, 0.6 m by 0.6 m plate bolted to its side, as shown below. When the pressure gauge on the top of the tank reads 50 kPa, what is the magnitude and location of the resultant force on the attached plate? The outside of the tank is at atmospheric pressure.

Explanation / Answer

Penny Question

Assumption is that the thickness of the penny is negligible.

In the first instance the pressure acting on the penny will be;

P = x g x h ; water at 10C = 1000 kg/m^3 (Wikipedia has it as 999.70 , 1000 is a good approx)

= 1000 x 9.81 x .5

= 4905 Pa

using P = F/A; F = P x A

F = 4905 x ( x (0.01905 x .5)^2) = 1.398 N

In the second instance

P = x g x h' ; where h ' = distance from the centriod to surface, for a circle its centriod is = D/2

= 1000 x 9.81 x ( .5 - 0.01905/2 )

// Distance from surface to bottom = 0.5 so from surface to centriod = .5 - .019505

= 4811.56 Pa

using P = F/A; F = P x A

F = 4811.56 x ( x (0.01905 x .5)^2) = 1.371 N

In both cases atmospheric pressure is not included if you have to include such add answers to the following

F = 100 000 x ( x (0.01905 x .5)^2) = 28.50 N // 100kPa = P atm

Question 2

In the tank the pressure gauge records the vapor pressure in the tank so the pressure at any depth in the tank will be

P = Pgauge + P atm + ( x g x h) // the last pressure is due to the liquid where = SG x water and h is the depth below the surface of the liquid.

Using the pressure at the depth of the plates centriod

P = 50 000 + 100 000 + ( 1000 x 0.9 x 9.81 x (2+0.3)

= 170306.7 Pa// without Patm include P = 70306.7 N

F = PA

= 170306.7 x (L^2)

= 170306.7 x .36

= 61310.41 N // without Patm include F = 25310.412 N

Send me feedback if there is anything wrong.

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