Rhett Allain is an Associate Professor of Physics at Southeastern Louisiana University. He
enjoys teaching and talking about physics. Sometimes he takes things apart and can't
put them back together.
Why does it not matter how powerful the vacuum is?
How does a vacuum cleaner work anyway?
How does the vacuum cleaner work?
I know this is not HowStuffWorks.com, but I guess I should show some stuff. The cool thing about a vacuum cleaner (think shop vac so that I don't have to deal with the brushes) is that it is just one thing - a fan. The fan essentially moves air out of the vacuum part of the cleaner so that it "sucks". Here is a diagram for a vacuum cleaner with no hose (no air input).
So, the fan pushes air out of the chamber. It does this by hitting the air with the fan blades. The result is less air inside the chamber than outside. The wooden board on the cover of the hose is pushed there not by the air inside the vacuum, but rather by all the air outside (due to the atmosphere).
In a normal use, air comes in that part of the vacuum - hopefully bringing dirt and stuff with it. Note that I left out important parts of the vacuum cleaner if you want to use it to clean - namely a filter. But..how does the guy use the vacuum cleaner to lift himself? Here is a diagram of a board similar to what the guy created and stuck to a ceiling.
Let me assume that the mass of the board is small enough to be ignored. If this is the case, I can look at three forces acting on the board.
The tension in the string from the big weight hanging on the board (the load)
A force pushing up due to the atmosphere (which is outside the space between the board and the ceiling)
A force pushing down on the board due to pressure of the air inside the space between the board and the ceiling
All of these forces are in the vertical direction - so it is a one-dimensional situation. If I know the pressure of some gas, then the force it exerts on a surface would be:
Here if the pressure is in Pascals and the area is in meters2, then the force will be in Newtons. If the board is in equilibrium, then:
The Pvacuum is the pressure inside the vacuum chamber (just to be clear). From the video, the goal was to calculate the area of the board such that it could support his weight. In the video, Jem claims that even cheap vacuums "pull with a fifth of a vacuum". He also says that is "a suction force of a kilo for every 5 square centimeters". Yes, he made some mistakes there - but the real question: is he talking about the net force? Ok, let me start with 1/5th of a vacuum. Maybe he means that the net pressure on the board is 1/5th of what it would be if the inside of the chamber had no pressure. In this case the pressure inside would be:
If I look at this the other way, (5 kilos per 5 cm2) what would the vacuum pressure be? First, it should be noted that 5 kg, is the mass. The weight (and force) would be 49 Newtons. The area in m2 would be 5 x 10-4 m2.
Problem. 20,000 Pa is not 2,000 Pa. Ok, one more try. I found this in the video:
I have another shot that looks a little better. This is his gauge that he sticks in vacuum cleaners to test them. I am pretty sure it reads about -5.2 in Hg - this is a pressure measurement (inches of mercury). According to my unit conversions, this would be around 18,000 Pa. Ok - so I think the problem is a confusion between pressure and gauge pressure.
Gauge pressure is the difference in pressure between the inside of the thing and the atmosphere. So, a gauge pressure of zero would mean that the pressure inside the thing and outside the thing (whatever that thing may be) is the same.
The gauge reads 18,000 Pa - obviously, the gauge pressure. This agrees with the 1/5th of a vacuum thing. I think the 5 kilo thing he was saying was how regarding atmospheric pressure.
How big of an area would be needed to support a 70 kg person?
This would be a square about 20 cm x 20 cm. Seems about right.
Why can you use a cheapo vacuum cleaner?
In terms of "suction", what is the difference between a cheap and an expensive vacuum cleaner? They might have about the same pressure in the chamber (when the hose is blocked), but the expensive one will have a much higher flow rate. It will probably have a bigger, or faster fan. This is important if you are using the vacuum to actually clean stuff because it is this flow of air that takes dirt with it. If you are using the vacuum to fulfill your childhood Spider-Man dreams, a cheap one will do fine.
How is it possible that cheap old vacuum can lift that guy? Answer: it doesn't. First, the board does not really move - so no work is done on the person (Work = F dot Delta r). All the vacuum has to do is move some of the air out of the space between the board and the wall.
I am seriously considering building something like this - just to hang on the ceiling.
Perhaps an alternate title could be "hey, lets have class outside." I think I understand why students say this, but my standard answer is "no". Oh, but there is XYZ that we need to do. Here is the point I am trying to make - class is for students. Class is not for me. Students pay for classes, so they should get them.
Here is the other point. If a student chooses not to come to class, that is the student's choice. I am ok with that. Maybe it is not a great idea, but these are adults. There can be a problem. What if the class has attendance as a grade? What if the class will give a pop-bonus-quiz if attendance is low (which is essentially the same thing as attendance for a grade)? Well, I don't do these things nor do I encourage them.
There is only one thing in my classes that you have to worry about missing - the final exam. If you don't take that, there isn't much I can do to evaluate you.
Note: The following is a repost from some time ago. Today is the officially day to celebrate Dr. Seuss, so here is my Seuss-related post. Enjoy.
In the second Cat in the Hat book (I think it is the second one), the Cat reveals that he has more smaller cats under his hat. They are labeled A - Z with Z being so small you can't even see. Question: What is the sequence of sizes for successive cats? How big would Cat Z be?
Here is the first picture that Cat reveals Cat A. It is not trivial to measure their relative sizes because they are in different positions. I drew two circles, one around each head and looked at the circle sizes.
So, Cat has a head that is 165 px tall and Cat A has a head 61 px tall. Let me call the total length of Cat = 1 c, then Cat A would be: (assuming the two cats have the same proportions):
If Cat B were the same proportion of Cat A that Cat A is of Cat B, then Cat B would be:
If all the rest of the Cats are of the same proportion, then I can calculate the height of Cat Z.
So, could you see Cat Z? If Cat in the Hat is 1.5 meters tall, then Cat Z would be 8.9x10-12 meters. Could you see this? The most common thing to compare small visible things is hair. Hair can be around 50x10-6 meters. So, Cat Z is WAY smaller than a hair. In fact, you could fit 5 million Cat Zs across one small hair.
But wait. There is more
There are other pictures of the Cats in Cat in a Hat's hat. Here is the next one. It shows Cat, Cat A, and Cat B.
Using the same technique as before, this gives Cat = 136 pixels, Cat A = 63 pixels, Cat B = 47 pixels. So:
Ok, so Cat A is a little different (I will assume that is close before - or close enough). Cat B, however, doesn't fit the pattern I used before. So maybe each successive Cat is not just 0.37 times smaller than the previous. I could explore this further if only I had more data. I do! Here is the next picture from the book.
This gives:
Ok. This is a little odd. It doesn't really disagree with previous pictures, but this seems to indicate that Cat A is not related to Cat in the Hat like B and C are related to A. B and C seem to indicate that each successive cat is around 0.72 times smaller. I wish I had even more data. POOF! I do. There is another picture.
So, I am going to say (except for Cat A), each successive cat is about 0.8 times smaller. If I still use Cat A as 0.37 times smaller than the Cat in the Hat (which is 1.5 meters tall), then Cat Z would be:
This is small, but clearly not too small to see. What if I go with a lower limit of 0.7? In that case:
Indeed smaller - around hair sized. Too small to see? I don't think so. I guess Cat in the Hat lied to the kids. He is such a liar. Maybe there really wasn't a Cat Z. Maybe Cat Z went to the New Zoo by Gerald McGrew, who knows. "hey look kids, here is Cat Z. I know you can't see him, he is just too small to see."
Don't trust the Cat in the Hat.
Follow up questions.
How is it that all those cats fit under Cat in the Hat's hat?
If all the cats stood on top of each other, how tall would they be?
What would the ratio of successive cats have to be in order for Cat Z to be too small to see?
There was a Buzz Out Loud episode in the not-to-distant past where the discussion of youtube came up. I can't remember the exact details, but the main point was that it would be impossible for google (the owners of youtube) to review all of the videos that were submitted.
And here is my calculation. How much would it cost google a year to do this? First, I need to pick some variables (and I will first do this symbolically - then you can put in your own values if it makes you happy).
Submission Rate (s): This is how many minutes of video are submitted each minute (so the units would be minutes/minute). Just as an example, if only one person was using youtube and streamed their whole life - s would be 1 minute/minute.
Pay Rate (p): How much do you pay a reviewer per hour (units: US Dollars/hour)
Reviewer efficiency (e): How much of a reviewer's time is spent switching videos and stuff. This is a unitless quantity that must have a value over 1. For example, if a reviewer spent 10% of the time on stuff not related to reviewing, then e = 1.1. Also, this probably would depend a lot on the system that delivers videos to the reviewers.
Number of Reviewers (n): I think this one is clear.
Movie speed (v): I just thought of this one. What if the reviewers watch a video at a compressed rate. Movie speed of 1.5 means that a 1 minute video would only take 0.67 minutes.
Now for the calculation. No wait, how about a diagram?
The main point of this diagram is that whatever videos come in, have to be reviewed. The Law of Conservation of Videos.
Ok - let me start with an example. Suppose that 20 minutes of video are submitted every minute. If there was only one reviewer (with no movie speed ups) it would take the reviewer 20 minutes to review that video. So, what would happen for the next 20 minutes? You would need 20 more reviewers, where each one would start to review a movie and be unavailable for at least 20 minutes. In terms of my variables, this would be:
Note that the units work out fine since both s and n really don't have any units. Now, what if I want to include the efficiency? The greater the percent of time spent on administration stuff, the more people you would need. Then:
Units are still ok since efficiency also is unitless. What about the movie speed? The greater the movie speed for reviewing movies, the fewer reviewers google would need.
Units are still ok. To get the cost, I need n. Note that n is not how many people they need to hire, but how many people they need working (at any given time - assuming the video submission rate is constant).
So, how much would this cost a year? If the payrate (P) is in dollars per hour, then the yearly cost would be:
Now for the estimates
P = $15/hr. I know this is much greater than minimum wage, but I am assuming there are other benefits costs and stuff.
s = 20 min/min. Seems this is what they said on Buzz Out Loud. Of course, you could probably find a better value for this.
e = 1.05. Total guess here. So 5% of the reviewing time is wasted. But maybe this should be even lower. What are the reviewers going to waste their time on? Watching youtube? I guess they could still play solitaire.
v = 1.5. I picked 1.5 times for the playback speed because this is what the ipod does for podcasts if you want to listen faster. Works ok.
With these values, I get a yearly cost to google of 1.8 million dollars.
Update
Ok, I guess my guess of 20 minutes of video every minute was pretty wild. According to Youtube's fact sheet, they claim that 20 HOURS of video are uploaded each minute. This would change my estimate to 108 million dollars each year.
Thanks to the readers that pointed out my poor guess.
I ride my bike and mostly the wind makes me unhappy. On a very few days the wind is with me on the way to work and then changes so that it is with me again. But most days the wind is fairly constant. So, if the wind is constant then shouldn't everything even out? (Even Stephen).
Assumptions:
Let me start with the assumption that I (a mere mortal) can output at a constant power (but not 57,000 Watts like some people). I will also assume an air resistance force that is proportional to the square of the relative air speed. Here is a diagram.
A couple of quick things to point out. First, the net force on the bike is the zero vector. This is because it is traveling at a constant speed. I am really not going to have to worry about the vertical forces on the bike - they don't really do anything (Yes, I know I should have draw two forces for the upward force from the ground, one on each tire). The friction force is essentially from me (the rider). This includes the internal friction from the gears and stuff. There are two velocities. The vbike is the velocity of the bike relative to the ground. The vair-rel is the velocity of the air relative to the bike. This second velocity is what will be used in the air resistance force. If there is no wind, vair-rel = - vbike. If there is wind (say vwind) then:
Maybe I should have started with vbike-air instead of vair-bike - especially since the sign doesn't really matter. (here is a tutorial on relative velocities) So, in terms of my original stuff and the velocity of the wind (which is the velocity of the air with respect to the ground):
Just a quick check: if vwind = 0 m/s, then vair-rel = - vbike. If I am riding at the same speed as the wind (and in the direction of the wind) then the relative air speed would be zero (vector). Good enough for me.
Back to power
I am going to call the power output (including the internal losses in the bike due to friction and stuff) Prider. But, what I need is a connection between this and the frictional force that pushes the bike. So, suppose the bike moves a distance s. What work would this friction force do on the bike if I consider the bike to be a particle?
Since the friction force and the displacement are in the same direction, the work is positive. If I want the power (and I do) then I can write: (because I am lazy, I am going to write Ff for the friction force instead of Ffriction - also it is because I am really thinking of this as the force the rider exerts on a point-particle system)
If the biker is riding at a constant speed, then the air resistance force is equal in magnitude to the "friction" force. I am using the following model for the magnitude of the air resistance force.
Since the density of air, the cross-sectional area and the drag coefficient are constant, I replaced all those with the constant K. Since the air resistance is equal to the frictional force:
I need the vair-rel in terms of the wind speed. So:
The wind speed and the velocity of the bike are both vectors - obviously, it matters if you are riding in the same direction or opposite direction as the wind. But, this is really a 1-dimensional problem, so I can take the horizontal components of these vectors and make it look like:
So the sign on these velocity components matters. Also, I got rid of the absolute value signs since I am squaring that. The wind can be positive (tail wind) or negative (head wind). It seems like this will work. Now, what I really want is to solve for the velocity of the bike in terms of the wind and the power from the biker.
This is a 3rd order polynomial for vbike - and you know what? Cubic equations kind of suck to deal with. Instead of dealing this symbolically, I will go ahead and determine some values for these constants.
Let me start with the case of no wind. My brother cycles a lot and has a PowerTap. He estimates that I would be at about 200 watts going about 20 mph (9 m/s). So, from this I can get a value for Ffriction which will give me the value of Fair. What I really want is K:
Now for the fun stuff. I need to solve that cubic equation for different values of the wind speed. Here is a method that I am going to use. Now for a graph. This is the bike rider speed as a function of wind speed (I randomly chose to go from wind speed of -15 m/s to +15 m/s where the +15 means the wind is in the same direction as the rider). One more note - 15 m/s is really fast (over 30 mph). You probably shouldn't ride your bike if it is this windy outside.
Remember my initial point (I know it was quite some time ago) - the wind has more of a negative impact than a positive one. Let me plot the magnitude of the change in rider speed due to the wind.
In terms of speed adjustment, you can see I was wrong. What was I thinking? Take a look at, say, 8 m/s wind. If it is going with the rider then it will increase the rider's speed by about 6 m/s. If it is going against the rider, it will decrease the rider's speed by only a little over 4 m/s. I am not sure I have a good explanation for why this is the case - so instead I will make another argument to show that I am correct.
Suppose this is a round trip and the wind is constant in magnitude and direction for the full round trip. Then, I will go faster when going with the wind and slower against the wind. How about I calculate the time for a round trip for 5 km one way with different wind speeds?
See. So even though a rider may 'gain' more speed with the wind, the trip takes longer. Really, this is a classic intro-physics problem (but usually with an air plane where the speed difference with and against the wind is the same). The answer is that it takes longer with the wind because the rider will spend more time on the slow part than on the fast part. This means the average speed is not something close to the speed with zero wind.
One more thing - how fast would the wind have to be to not be able to go at all?
From this plot, even in a 90 mph wind you would still be going forward (though not really fast). I am not going to put too much weight on this calculation because I know some weird things can happen with cubic equations when the sign of the result changes.
I thought this was an interesting idea. One of my colleagues does not collect homework. Instead, the students turn in a sheet that lists which homework problems they worked on and how much time they spent on homework. By doing this, the students get a small homework grade - independent of how much time they claim they spent. I am pretty sure it is not a large percentage of the total grade.
Along with this "report", the students are supposed to keep notes on the homework they claim they did - just for reference. Theoretically, the instructor could verify that the student did the problems reported.
So, is this is good idea? Well, it is certainly not a bad idea. The best thing about this method is that reinforces the idea that homework is for the student. It helps students become more aware of how much time they are spending on the material outside of class.
I am sure there are students that over report the number of hours they are working on homework - but they are clearly just fooling themselves.
What is not so great about this method? I think the biggest downside is that students don't get feedback on their homework. Well, they get feedback if they ask the instructor - which they should.
Can you fall faster than terminal velocity? That is the question.
Air Resistance
Air resistance is a force exerted on an object as it moves through some stuff - air in this case. The magnitude is usually modeled as:
Rho is the density of the stuff the object is moving through
A is the cross sectional area of the object
C is the drag coefficient of the object - this depends on the shape (a cone would be different than a flat disk)
v is the magnitude of the velocity of the object
The direction of this air resistance force is in the opposite direction as the velocity.
Terminal Velocity
Here is a diagram of a sky diver that just jumped out of a stationary balloon.
Here there is the gravitational force (weight) and a small air resistance force. The air resistance is small because the jumper just started falling and is not moving too fast. The net force is in the downward direction. Since this is in the same direction as the velocity, the speed increases.
My last Olympics post may have been a little complicated. I am going to try to make this one a little easier. In this post, I want to look at the landing portion of a ski jump. This could apply to THE ski jump, but there are some things in that even that make it a little more complicated (but I might come back to that in another post). For this case, I will consider the freestyle event - aerials. I didn't search too long, but here is a nice short video.
First, a quick estimation of how high they are "falling" from on the way down. In that video, the jumper takes about 1.5 seconds to get from the highest point to the landing. Using some kinematic equations, (or the work-energy principle) I can find a couple of useful things. How high? How fast?
And how fast?
These two numbers are really just for reference. And here is the big deal. What if you jumped off an 11 meter building? That would be bad, right? (although some people can jump off things like this - here is my dangerous, parkour, jumping calculator) So, this really has to do with acceleration. Acceleration is the change in velocity - let me write it like this:
Note that acceleration is a vector and so is the velocity. In short, a vector has both magnitude and direction (here is the long version of vectors).
Now I am going to draw a diagram for a person jumping something like an aerial, but on flat ground.
The kind folks over at BloggingHeads TV were nice enough to offer me an opportunity to discuss some science stuff with Ed Grabianowski from io9.com. Here it is:
Rhett Allain is an Associate Professor of Physics at Southeastern Louisiana University. He
enjoys teaching and talking about physics. Sometimes he takes things apart and can't
put them back together.
