All AP Physics 2 Resources
Example Questions
Example Question #12 : Pressure
Consider the diagram of a hydraulic lift shown below.
Based on this diagram, which of the following statements is true?
None of these statements are true because the pressure and force on both the left and right side are equal
The pressure on the left side is greater than the pressure on the right side
The force on the right side is greater than the force on the left side
The force on the left side is greater than the force on the right side
The pressure on the right side is greater than the pressure on the left side
The force on the right side is greater than the force on the left side
In this question, we're shown a hydraulic lift. A lift such as this functions to transmit a smaller force into a larger force via an incompressible liquid. From the diagram shown, we're asked to determine the relative values of the pressure and force on the right and left sides.
A hydraulic lift is able to transmit a small force into a larger force due to the liquid it contains being unable to compress. What this means is that when a force is applied to a given area of the liquid, this pressure is transmitted to all parts of the liquid. Therefore, at any given point, the pressure at all points within the lift will be equal. Thus, we can rule out the answer choices that list pressure.
But what about the forces? If the pressure is the same everywhere in an incompressible liquid, does this also mean the force is the same everywhere? The answer is no. Since pressure depends on the ratio of force to area, equal pressure does not mean equal force, because the area on which the force is acting must be taken into account. We can show this with the following expressions.
As can be seen from the above expression, the output force is greater than the input force by a factor equal to the ratio of the areas on which each force acts. In other words, to maintain constant pressure, a greater area will mean a greater force. Thus, from the diagram, we can see that the force on the right () will be greater than the force on the left ().
Example Question #11 : Pressure
A girl is standing on two feet. She then stands on one foot. How does the pressure on her foot compare to when she is standing on two feet?
It is doubled
Impossible to determine
It is quadrupled
None of these
It is cut in half
It is doubled
Pressure is defined as force divided by area
When the girl is standing on one foot, the force of gravity pulling her down stays the same, but area is cut in half. Thus, pressure is doubled.
Example Question #13 : Pressure
Paul weighs . What must be the surface area of his shoe if he uses it to try to kill an ant? Assume the shoe applies uniform force, he can only apply as much force as his weight, and his other foot is not in contact with the ground. The ant cannot withstand any pressure greater than .
Pressure is force divided by area. The force Paul applies is . The rest is just algebra.
Example Question #21 : Pressure
A glass is filled with water. The gauge pressure at the top of the glass is zero and the gauge pressure at the bottom is . A second glass with half the height and twice the diameter of the first glass is also filled with water. What is the gauge pressure at the bottom of the second glass?
Pressure so the diameter changing does not matter but half the height changes the pressure of the second glass to .
Example Question #51 : Fluid Statics
Suppose that a barometer is filled with water, as shown in the diagram below. If this barometer's column has a height of , what would be the height of a barometer that used mercury instead of water?
Note: Mercury has a specific gravity of .
For this question, we're asked to see how the height of a barometer's column will change once the liquid changes from water to mercury.
Barometers are a measure of atmospheric pressure. As the ambient pressure pushes down on the surface of the liquid, this pressure is transmitted throughout the liquid, such that it causes the liquid level in the water to rise. As the ambient pressure increases, the column in the barometer climbs.
We'll need to use the equation for pressure as it relates to density and height.
Since the ambient pressure is the same in both instances, we can set these terms equal to each other. That is to say, it doesn't matter which liquid is being used in the column, because both of them will feel the same pressure from the atmosphere.
Next, we can rearrange the above expression to isolate the term for the height in the mercury column.
Finally, we can plug in the terms given to us in the question stem to arrive at our answer.
Example Question #21 : Pressure
Determine the pressure below the surface of freshwater.
None of these
Pressure is equal to force over area:
The force will be due to the gravity of the water above. The mass of the water will be the density times the volume.
The volume will be equal to the base area times the length
Combining equations
Plugging in values:
Example Question #51 : Fluid Statics
Consider a bucket 1m tall. The bucket is completely filled, with 0.5m of oil on the top and 0.5m of water below it. Write an expression for the pressure at a point 10cm from the bottom of the bucket, in terms of and atmospheric pressure
Pressure in a static fluid varies along the axis in which gravity acts. In other words, the deeper one goes, the higher the pressure is. This can be expressed as:
, where h represents the depth. Since we are examining a point 0.1m from the bottom, this is a depth of 0.9m; 0.5 of which is through oil and 0.4 of which is through water.
Example Question #51 : Fluids
What is the mass of air inside a small freezer with dimensions by by ?
Write the density formula.
The density of air is:
Find the volume of the freezer in meters. Converting the dimensions from to gives.
Solve for the mass of the air.
Example Question #51 : Fluid Statics
Suppose that a cube composed of a mysterious substance is hanging from a string while also submerged in water. If the cube is not moving and has a side length of 8cm and the tension in the string is equal to 15N, what is the specific gravity of this mysterious substance?
To answer this question, it's best to start by drawing a free body diagram so that we can identify the forces acting on the cube. Since the cube is submerged in water but not accelerating, we know that we are dealing with a situation in which there is no net force acting on the cube. In the upward direction we have the force caused by the tension in the string as well as the buoyant force of the displaced water. In the downward direction we have the weight of the cube itself.
In order to calculate the buoyant force, we'll need to calculate the volume of the cube in order to determine the amount of water displaced. Since the cube is fully submerged, the volume of the cube will be exactly equal to the volume of displaced water.
Next, plug this value into the above equation for the mass of the cube.
Now that we've found the mass of the cube, we can divide this value by its volume in order to obtain its density.
And lastly, since we have the density of the cube, we can calculate its specific gravity by dividing its density by the density of water.
Example Question #53 : Fluid Statics
A raft is observed floating in a fresh-water lake with of its volume submerged. If the raft were floated in the Great Salt Lake, how much of its volume would be submerged? The density of fresh water is . Due to its high salt content, the density of the Great Salt Lake is .
Use the information given to find the density of the raft. Since the raft is at rest, the forces on it add to zero. Therefore the magnitude of the gravity force is equal to the magnitude of the buoyant force:
The mass of the raft is equal to its density times its volume. We start using subscripts to keep track of which densities and volumes we mean:
The 's cancel, and the is of the raft's total volume:
Setting the magnitudes equal again at the Great Salt Lake:
. Cancel the 's and solve for the ratio of submerged volume to total volume of the raft:
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