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Atmospheric pressure must also be accounted for in dealing with the pressure inside a fluid. So the pressure p, at a depth h in a fluid is p = p0 + ρgh. This means p - p0 = Δp = ρgh. P is absolute pressure. The difference ∆p = p - p0 is called gauge pressure. It is the pressure of the fluid, independent of the atmospheric pressure, p0. Note that pressure is measured in N/m^2 or pascal (Pa) with 1 Pa = 1 N/m^2. Part 5: Pascal's Principle Choose the third configuration/container. Set gravity = 9.8 m/s^2, atmosphere ON, fluid density = 1000 kg/m^3, and choose grid. Place a pressure gauge along the 2 meter mark on each side and record the pressure reading. Then click and drag a 250 kg weight in the well on the left. As you drag it, a dashed receiving box will appear above the fluid on the left side. Drop it in. Record the new pressure on both sides. Record the pressures for 250, 500, 750, and 1000 kg. Remove the pressure gauges and return the masses to the top. Consider the next measurements approximations. Place the vertical ruler against the fluid on the left so that the 3 centimeter mark is set at the top of the fluid (narrow tube). Each centimeter is equivalent to one meter of depth, with 5 small markings at 0.2 centimeters. Again, place the masses sequentially in the dashed boxes and record the change in depth on the left side. Return the masses to the top. Place the vertical ruler on the right side (wide tube) with the 3 centimeter mark at the top of the fluid. Again, place the masses in the dashed boxes sequentially and record the increase in the height of the fluid. Create a Data Table to Show Your Results: Mass [kg] | Pressure In (left tank) Pin | Pressure Out (right tank) Pout | Depth Decrease (m) | Height Increase (m) | Depth/Height 0 | 0 | 0 | 0 250 | | | 500 | | | 750 | | | 1000 | | | Compare the pressure changes on the left and right sides as the masses are added to the fluid. How does the addition of masses affect the changes in pressure on both sides? How does this relate to Pascal's Law? Calculate the ratio of depth change on the left to height change on the right. After each mass increase, record it. What pattern do you see? Assume that each of these sides is a tube and the cross-sectional line you see is the diameter. The approximate measurement of the diameters on the left and right are 1.75 cm and 8 cm respectively. By approximately how much of a factor does the volume increase on the right compare to the volume decrease on the left side? Now calculate the cross-sectional area of the left and right tubes in centimeters squared using the approximate diameter values provided. Right (Larger)____ Left (smaller)_ Recall that ΔP = F x A. Based on this, determine how much greater the force upward on the right tube must be compared to the force downward on the left side. Why is this information so important to mechanics at auto repair shops?

          Atmospheric pressure must also be accounted for in dealing with the pressure inside a fluid. So the pressure p, at a depth h in a fluid is p = p0 + ρgh. This means p - p0 = Δp = ρgh. P is absolute pressure. The difference ∆p = p - p0 is called gauge pressure. It is the pressure of the fluid, independent of the atmospheric pressure, p0. Note that pressure is measured in N/m^2 or pascal (Pa) with 1 Pa = 1 N/m^2.

Part 5: Pascal's Principle
Choose the third configuration/container. Set gravity = 9.8 m/s^2, atmosphere ON, fluid density = 1000 kg/m^3, and choose grid. Place a pressure gauge along the 2 meter mark on each side and record the pressure reading. Then click and drag a 250 kg weight in the well on the left. As you drag it, a dashed receiving box will appear above the fluid on the left side. Drop it in. Record the new pressure on both sides. Record the pressures for 250, 500, 750, and 1000 kg. Remove the pressure gauges and return the masses to the top. Consider the next measurements approximations.

Place the vertical ruler against the fluid on the left so that the 3 centimeter mark is set at the top of the fluid (narrow tube). Each centimeter is equivalent to one meter of depth, with 5 small markings at 0.2 centimeters. Again, place the masses sequentially in the dashed boxes and record the change in depth on the left side. Return the masses to the top. Place the vertical ruler on the right side (wide tube) with the 3 centimeter mark at the top of the fluid. Again, place the masses in the dashed boxes sequentially and record the increase in the height of the fluid.

Create a Data Table to Show Your Results:
Mass [kg] | Pressure In (left tank) Pin | Pressure Out (right tank) Pout | Depth Decrease (m) | Height Increase (m) | Depth/Height
0 | 0 | 0 | 0
250 | | |
500 | | |
750 | | |
1000 | | |

Compare the pressure changes on the left and right sides as the masses are added to the fluid. How does the addition of masses affect the changes in pressure on both sides? How does this relate to Pascal's Law? Calculate the ratio of depth change on the left to height change on the right. After each mass increase, record it. What pattern do you see?

Assume that each of these sides is a tube and the cross-sectional line you see is the diameter. The approximate measurement of the diameters on the left and right are 1.75 cm and 8 cm respectively. By approximately how much of a factor does the volume increase on the right compare to the volume decrease on the left side? Now calculate the cross-sectional area of the left and right tubes in centimeters squared using the approximate diameter values provided.

Right (Larger)______________________ Left (smaller)___________________

Recall that ΔP = F x A. Based on this, determine how much greater the force upward on the right tube must be compared to the force downward on the left side. Why is this information so important to mechanics at auto repair shops?

Added by Jose C.

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