a.) Estimate how much pressure a cat puts exerts on you if it stands on you with all four paws. Give your answer in English units.
b.) How much more pressure do you feel at the bottom of a 6.0 m deep diving pool compared to the pressure you feel on land?
Give your answer in MKS units.
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In part a) of this problem, you are asked to estimate the pressure exerted by a cat. The definition of pressure involves force and area; both are numbers that you can estimate for a cat. Therefore, you don’t need to do anything more complicated than work with the definition of pressure.
In part b) of this problem, you are asked for the pressure exerted by a fluid at a given depth. Pressure and depth of a fluid are directly related through the definition of pressure of a fluid. Therefore, part b) is also a definition problem using the expression for pressure exerted by a fluid.
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There is no need for a picture in most definition problems, and this is one of them.
In part a), you can estimate force (weight) and area information and are asked for the closely-related pressure. A picture of a cat will not provide any additional insight or organization beyond what is already present in the problem.
In part b), you are given information about depth of a fluid and are asked for the closely-related pressure due to that fluid. You may certainly draw a picture of a swimming pool if it helps to visualize the problem, but a drawing will not provide any organization beyond what is already present in the problem. -
In equation form, pressure is defined as
P = F/A
This is the only relation you need for part a) of this problem.
The pressure exerted by a fluid is given by
Pfluid = ρfluidgh
This is the only relation you need for part b) of this problem.
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a) P = F/A
The force exerted by a cat is equal to the weight of a cat. I personally know house cats that weigh as little as 5 lb. and as much as 18 lb. so I will estimate the weight of a cat to be 12 lb. Any reasonable value that you select is fine.
The area over which a cat exerts pressure is the area of its paws. Each paw has an area of around 1 square inch, so
P ≈ 12 lb/ 4 sq. in.
P ≈ 3 lb/sq. in.
b) Pfluid = ρfluidgh
Pwater = ρwatergh
Pwater = (1000 kg/m3)(9.8 m/s2)(6.0 m)
Pwater = 59,000 Pa
The symbol h in this equation is counter-intuitive, as it refers to the amount of water above you (your depth) and not height. h is depth.
The pressure that you feel when you are in a fluid depends on the amount and density of all fluid above you. Air is a fluid, and so even if we aren’t submersed in liquid, we feel a certain amount of pressure due the air: P0. Under water, the total pressure that we feel is the pressure of the air (P0) plus the pressure due to the water, ρfluidgh. This problem asks only for the pressure due to the water so we only need the second term. It is perfectly fine to use Pfluid = P0 + ρfluidgh and then to subtract out P0 at the end of the problem to find the amount of pressure the water exerts beyond that exerted by the air.
(1 sq. in/paw)(4 paws) = 4 sq. in.
The problem asks for the pressure at the bottom of a diving pool; the reasonable assumption is that the pool is filled with water. You can find the density of water in your textbook or on the internet.
The symbol “≈” means “approximately equal to.” In this problem, you need to estimate both force and area, so your answer is a reasonable approximation.
Why do you call this a definition problem rather than a fluids problem?
Why do you call part b) a definition problem rather than a fluids problem?
Problems involving buoyant force or Bernoulli’s equation, do, indeed, involve the pressure of a fluid. However, in those cases you may not deal directly with pressure at all (buoyant force problems) or you may relate how pressure changes to other factors such as the motion of the fluid (Bernoulli’s Law.) In part b) of this problem, you are asked only to relate the pressure in a fluid to the depth of that fluid—a direct link given by the definition of the density of a fluid. You are not looking at changes in other variables from one location to another
My book uses “Pfluid = P0 + ρfluidgh” for the pressure of a fluid. Why didn’t you include P0? (same box as for “Pfluid” below)
The pressure that you feel when you are in a fluid depends on the amount and density of all fluid above you. Air is a fluid, and so even if we aren’t submersed in liquid, we feel a certain amount of pressure due the air: P0. Under water, the total pressure that we feel is the pressure of the air (P0) plus the pressure due to the water, ρfluidgh. This problem asks only for the pressure due to the water so we only need the second term. It is perfectly fine to use Pfluid = P0 + ρfluidgh and then to subtract out P0 at the end of the problem to find the amount of pressure the water exerts beyond that exerted by the air.
How does h relate to depth? (same box as for “h” below)
The symbol h in this equation is counter-intuitive, as it refers to the amount of water above you (your depth) and not height. h is depth.
What are the English units for pressure? (same box as for “English units” below)
The English unit for force is the pound (lb) and for area is square foot or square inch. Therefore, English units of pressure are lb/sq.ft. or lb/sq.in.
How do you know the MKS units for pressure are Pascals? (bring up same box as “Pa” below)
The MKS unit of pressure is the Pascal.
1 Pa = 1 N/m2 = 1 kg/m s2. The Pascal is a small unit—the pressure exerted by the atmosphere at sea level is 101,000 Pa. In this problem, Pwater = (1000 kg/m3)(9.8 m/s2)(6.0 m) = 59,000 kg (m2/m3) (/s2) = 59,000 kg/m s2 = 59,000 Pa.What does the "≈" symbol mean? (bring up same box as "≈" below)
The symbol “≈” means “approximately equal to.” In this problem, you need to estimate both force and area, so your answer is a reasonable approximation.
How do you know to use the density of water and not some other fluid? (same box as "Pwater" below)
The problem asks for the pressure at the bottom of a diving pool; the reasonable assumption is that the pool is filled with water. You can find the density of water in your textbook or on the internet.
My book uses "Pfluid = P0 + ρfluidgh" for the pressure of a fluid. Why didn't you include P0? (same box as for "Pfluid" on the Select the Relation page)
The pressure that you feel when you are in a fluid depends on the amount and density of all fluid above you. Air is a fluid, and so even if we aren’t submersed in liquid, we feel a certain amount of pressure due the air: P0. Under water, the total pressure that we feel is the pressure of the air (P0) plus the pressure due to the water, ρfluidgh. This problem asks only for the pressure due to the water so we only need the second term. It is perfectly fine to use Pfluid = P0 + ρfluidgh and then to subtract out P0 at the end of the problem to find the amount of pressure the water exerts beyond that exerted by the air.
Why do you call part b) a definition problem rather than a fluids problem?
Problems involving buoyant force or Bernoulli’s equation, do, indeed, involve the pressure of a fluid. However, in those cases you may not deal directly with pressure at all (buoyant force problems) or you may relate how pressure changes to other factors such as the motion of the fluid (Bernoulli’s Law.) In part b) of this problem, you are asked only to relate the pressure in a fluid to the depth of that fluid—a direct link given by the definition of the density of a fluid. You are not looking at changes in other variables from one location to another
The English unit for force is the pound (lb) and for area is square foot or square inch. Therefore, English units of pressure are lb/sq.ft. or lb/sq.in.
The MKS unit of pressure is the Pascal.
1 Pa = 1 N/m2 = 1 kg/m s2. The Pascal is a small unit—the pressure exerted by the atmosphere at sea level is 101,000 Pa. In this problem, Pwater = (1000 kg/m3)(9.8 m/s2)(6.0 m) = 59,000 kg (m2/m3) (/s2) = 59,000 kg/m s2 = 59,000 Pa.V = volume = l x w x h = Area x h
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Each part of this problem is merely a definition problem. Pressure is defined as force/area, regardless of the cause of the force. In the particular case of a fluid, force is the weight of the fluid above you:
Pfluid = (mg)fluid above you/A = (ρVg)fluid above you/A
= ρfluidg(V/A) = ρfluidg(A h/A) = ρfluidgh
and so in the case of a fluid the definition of pressure can be rewritten as ρfluidgh.
Both parts of the problem are straightforward, single step, calculations once the correct version of the definition is identified. Because the unit Pascal is probably not a familiar one, you can see if you answer to part b) makes sense by comparing your answer to a known pressure. In this case, the pressure due to 6.0 m of water is a little over half atmospheric pressure (101,000 Pa) so it is reasonable.