Physlet Examples
Always go through these before starting the problems
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With fluids, instead of discussing forces, we usually talk about pressure, the force per unit area or P = F/A. This is because the direction of the force a liquid exerts on its container depends on the shape of the container (force is normal to the surface of the container) and the size of the container. Pressure is not a vector (no direction) and does not depend on the size of the container Position is given in meters and pressure is given in pascals).
Why does pressure increases as a function of depth. Assume the
blue liquid is water (density 1000 kg/m3). Pick a
point to measure the pressure somewhere in the upper tube. If
the dimension of the container into the screen (the dimension you
cannot see) is 1 m, what is the volume of water above the point you
picked? What is the mass and thus the weight of the water at
that point? For example, consider a depth of 3 m. The
pressure is 29,400 N/m2. The volume of water above
this point is a cylinder of volume 9.4 m3. The mass
of the water is the volume times the water's density, or 9,400 kg,
and therefore the weight of the water is 92,120 N.
What is the force downward at that point? We find this by
dividing the weight by the cross-sectional area of the column
of water at that point, which is 3.14 m2. This
pressure should be equal to the pressure reading. The units of
pressure are N/m2 = pascals (abbreviated Pa).
Actually, this is the gauge pressure, not the absolute pressure,
because we assumed P = 0 at the top of the water column when the
pressure (due to the atmosphere) is actually around 1 x 105
Pa. The absolute pressure then would be the pressure at the
top due to the atmosphere added to the pressure due to the weight of
the water. All of this comes together in the equation:
P = P0 + ρgy,
where P0 is the pressure at the top, ρ is the density of
the liquid, g is acceleration due to gravity and y is the depth of
the liquid. |
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Example of a hydraulic lift. Notice that because the pressure is the same at the same height in the fluid, the force required for the hand to support the mass is much less than the weight of the mass (10 times less). This is because the areas of the gray circular "lids" on the top of the oil are different by a factor of 10. Now calculate the pressure exerted by the green mass and the pressure exerted by the force arrow. They should be the same. The ratio of the areas is the ratio of the forces required for equilibrium. Enter a new value for the mass (between 100 and 300 kg) and try it. |
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The buoyant force on an object (whether or not it floats) is due to the pressure difference between the bottom of the object and the top of the object. If the object is going to be buoyed up, the bottom of the object is subjected to a greater pressure than the top of the object. Remember that pressure is force/area and so, if the pressure at the bottom of an object (submerged under a liquid) is greater than the pressure on top (still in air), there is a net upward force. This animation shows a block lowered into a liquid and then floating. The density of the block can be changed by click-dragging in the block on the upper left of the animation (click-drag at the white-gray interface of the box). The graph in the lower right of the animation shows the pressure difference (P) as a function of depth (Y). When the net upward force, the buoyant force, is equal to the weight of an object (the white and gray striped block), it floats in the gray liquid. In addition, the amount of fluid displaced when the block is in the fluid is shown in the block to the left. FB = ρliquidgVliquid displaced = ρliquidgVsubmerged part of object . |
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