Physics with Albegra for the nonlinear thinkers
October 18, 2011 9:15 PM Subscribe
How do the variables on both sides of physics equations not become redundant (specifically involving adding/subtracting forces and centripetal force and acceleration)?
I'm not naturally a math/logic person, but I've found physics pretty easy to figure out due to all the pictures it involves. I can usually picture what's going on my head/draw it out and solve from there, fudging my way through the algebra to the correct answer most of the time. The only place I get hung up, waste time, and can't reconcile an intuitive understanding of the situations is when I'm setting equations equal to one another to solve for a certain variable. I can't seem to wrap my nonlogical brain around how variables on both sides of the equations you set equal to eachother work. Somehow I usually get the right answer, but it's not without a lot of hair pulling, trying to figure out how the textbook gets from point A to point B algebraically.
For example, we know Force = m(a). On the left side of the equations, we add the forces, sometimes with negative signs and sometimes not (it's okay to wait till the end to figure out the sign sometimes, apparently). So we'll add and subtract the external forces (ex: Normal force (N) - gravity force (mg) + static friction (fs) on the left side. According to F = ma, that left side also equal the mass time acceleration in terms of the number yielded, right? Sometimes we set the right side equal to zero and move forces to the opposite side of the equals sign to solve for them, and sometimes we leave "ma" on the right side. What's the method behind that madness? Why do we leave ma on the right sometimes, and other times get rid of the ma to solve for individual forces? That's my first point of confusion.
Here's second example of a problem I had trouble with related to the above issue. Passengers on a roller coaster feel 33% heavier at the bottom of a turn with a radius r. Ffeeling" heavier means that the passenger's apparent weight is 1.33 times the true weight. My textbook said a few chapters back that apparent weight (which is the Normal Force) = mg (which is the true weight) + ma. So N = mg + ma, which I don't entirely understand since normal force is usually -mg. Why do you add mass again ("ma") when you normally have acceleration on the right side of the equation (F = ma)? I do realize you have to take into account whether the object is falling through the surface (-a), flying off the surface (+a) or not accelerating at all vertically. (How that relates to the fact that objects in uniform circular motion are always accelerating since the direction of the velocity vector is always changing, I'm not sure.) For the roller coaster problem, I ended up just going with the fact that normal force and gravity force are both acting on the car, representing them both using "mg" (1.33mg - 1mg). I set those forces equal to centripetal force (mv2/radius of the roller coaster turn) and it worked. I got the speed that's required to get the passengers to feel heavier. It was just such a headache figuring out that dang left side of the equation since N = mg + ma and I couldn't reconcile that with my way of finding the solution.
Can one of you lovely Metafi-ites explain the concept of variable manipulation and setting equations equal to eachother in simple terms? My professor doesn't speak English as a first language and can't always give a straight answer for these types of questions. I hope one of y'all can see where I'm going wrong. Thanks!
I'm not naturally a math/logic person, but I've found physics pretty easy to figure out due to all the pictures it involves. I can usually picture what's going on my head/draw it out and solve from there, fudging my way through the algebra to the correct answer most of the time. The only place I get hung up, waste time, and can't reconcile an intuitive understanding of the situations is when I'm setting equations equal to one another to solve for a certain variable. I can't seem to wrap my nonlogical brain around how variables on both sides of the equations you set equal to eachother work. Somehow I usually get the right answer, but it's not without a lot of hair pulling, trying to figure out how the textbook gets from point A to point B algebraically.
For example, we know Force = m(a). On the left side of the equations, we add the forces, sometimes with negative signs and sometimes not (it's okay to wait till the end to figure out the sign sometimes, apparently). So we'll add and subtract the external forces (ex: Normal force (N) - gravity force (mg) + static friction (fs) on the left side. According to F = ma, that left side also equal the mass time acceleration in terms of the number yielded, right? Sometimes we set the right side equal to zero and move forces to the opposite side of the equals sign to solve for them, and sometimes we leave "ma" on the right side. What's the method behind that madness? Why do we leave ma on the right sometimes, and other times get rid of the ma to solve for individual forces? That's my first point of confusion.
Here's second example of a problem I had trouble with related to the above issue. Passengers on a roller coaster feel 33% heavier at the bottom of a turn with a radius r. Ffeeling" heavier means that the passenger's apparent weight is 1.33 times the true weight. My textbook said a few chapters back that apparent weight (which is the Normal Force) = mg (which is the true weight) + ma. So N = mg + ma, which I don't entirely understand since normal force is usually -mg. Why do you add mass again ("ma") when you normally have acceleration on the right side of the equation (F = ma)? I do realize you have to take into account whether the object is falling through the surface (-a), flying off the surface (+a) or not accelerating at all vertically. (How that relates to the fact that objects in uniform circular motion are always accelerating since the direction of the velocity vector is always changing, I'm not sure.) For the roller coaster problem, I ended up just going with the fact that normal force and gravity force are both acting on the car, representing them both using "mg" (1.33mg - 1mg). I set those forces equal to centripetal force (mv2/radius of the roller coaster turn) and it worked. I got the speed that's required to get the passengers to feel heavier. It was just such a headache figuring out that dang left side of the equation since N = mg + ma and I couldn't reconcile that with my way of finding the solution.
Can one of you lovely Metafi-ites explain the concept of variable manipulation and setting equations equal to eachother in simple terms? My professor doesn't speak English as a first language and can't always give a straight answer for these types of questions. I hope one of y'all can see where I'm going wrong. Thanks!
Think about the meaning of the equals sign. The left and right side may look different, but they are mathematically equivalent. That means we can add or subtract the same quantity from each side, and they're still equivalent. It works for multiplication and division, too.
Starting with: f = ma
Subtract ma from both sides:
f - ma = ma - ma
Lucky for us, a quantity minus itself is zero (additive inverse property), so ma - ma = 0 and
f - ma = 0
You can think of this as "moving" ma from right to left, but really we've subtracted it from both sides. This "moving" is done to simplify the equation and make it easier to solve.
These are basic rules of algebra.
posted by qxntpqbbbqxl at 9:52 PM on October 18, 2011 [1 favorite]
Starting with: f = ma
Subtract ma from both sides:
f - ma = ma - ma
Lucky for us, a quantity minus itself is zero (additive inverse property), so ma - ma = 0 and
f - ma = 0
You can think of this as "moving" ma from right to left, but really we've subtracted it from both sides. This "moving" is done to simplify the equation and make it easier to solve.
These are basic rules of algebra.
posted by qxntpqbbbqxl at 9:52 PM on October 18, 2011 [1 favorite]
Equations, in a fundamental and literal sense, relate two quantities (regardless of the signs, variable names, whatever is going on mathematically) that are equal. The choice about what side to put terms like 'F', "m*a", etc. is arbitrary, as long as both sides are equal. Now, it can sometimes be very difficult in a given situation to figure out what is equal to what, let alone solve for a single variable, but equality is what you must have in order to relate two quantities.
To take "F=ma", that is the equation you use when you know the value of 'm' and 'a', and wish to calculate 'F'. That particular equation (again, it may be helpful to think, at least momentarily, that an equation is just saying that two things are equal) is old and has extensive experimental backing. No one has ever found a situation where, when sufficiently investigated, "F ≠ ma".
As I noted above, sometimes it can be hard to figure out what is, in fact, equal to what. You have noticed that in your roller coaster problem. Oftentimes, even in relatively advanced physics, one makes use of clever frames of reference, simplifying assumptions, approximations, and similar "tricks" to reduce the complexity of the initial statement of the problem in an equation (or collection of equations).
For a person at rest, gravity is attempting to accelerate you in one direction, but the normal force is resisting. Since you're not moving, you know that 'a' must be 0. Therefore, plugging in 0 for 'a', F must be 0. Since we know the force of gravity, therefore the normal force must be equal in magnitude (strength) but opposite. That is stated as "Fgravity = -Fnormal". You could, just as equivalently, state that as "Fgravity + Fnormal = 0". Or substitute in "Fgravity=m*g" and make it "-m*g = Fnormal". It's just symbol manipulation. The "best" way to structure the equation is really up to you, the only rules are algebra and the desire to isolate a given quantity by itself. When you say "Normal force is usually -mg", that's just a convention based on the Y component of gravity being in the negative direction.
posted by wnissen at 9:52 PM on October 18, 2011 [1 favorite]
To take "F=ma", that is the equation you use when you know the value of 'm' and 'a', and wish to calculate 'F'. That particular equation (again, it may be helpful to think, at least momentarily, that an equation is just saying that two things are equal) is old and has extensive experimental backing. No one has ever found a situation where, when sufficiently investigated, "F ≠ ma".
As I noted above, sometimes it can be hard to figure out what is, in fact, equal to what. You have noticed that in your roller coaster problem. Oftentimes, even in relatively advanced physics, one makes use of clever frames of reference, simplifying assumptions, approximations, and similar "tricks" to reduce the complexity of the initial statement of the problem in an equation (or collection of equations).
For a person at rest, gravity is attempting to accelerate you in one direction, but the normal force is resisting. Since you're not moving, you know that 'a' must be 0. Therefore, plugging in 0 for 'a', F must be 0. Since we know the force of gravity, therefore the normal force must be equal in magnitude (strength) but opposite. That is stated as "Fgravity = -Fnormal". You could, just as equivalently, state that as "Fgravity + Fnormal = 0". Or substitute in "Fgravity=m*g" and make it "-m*g = Fnormal". It's just symbol manipulation. The "best" way to structure the equation is really up to you, the only rules are algebra and the desire to isolate a given quantity by itself. When you say "Normal force is usually -mg", that's just a convention based on the Y component of gravity being in the negative direction.
posted by wnissen at 9:52 PM on October 18, 2011 [1 favorite]
I should add that you can write a computer program, feed it a bunch of equations, and ask it to perform algebra for you. The algebraic manipulations are just moving symbols around to given set of rules. That's the beauty of it. There's a whole field of study, called "logic," that deals with the conclusions you can draw from a set of assumptions.
Even the things that look like actual numbers, like "g=9.81... N/kg" are, from the standpoint of algebra, arbitrary. You could set up an equation with "g=2" and come to all kinds of conclusions that would be valid on a world where 'g' was 2. As long as you follow the rules of algebra, the only thing wrong with "g=2" is that those conclusions don't match the conclusions you reach doing experiments on earth.
posted by wnissen at 10:06 PM on October 18, 2011
Even the things that look like actual numbers, like "g=9.81... N/kg" are, from the standpoint of algebra, arbitrary. You could set up an equation with "g=2" and come to all kinds of conclusions that would be valid on a world where 'g' was 2. As long as you follow the rules of algebra, the only thing wrong with "g=2" is that those conclusions don't match the conclusions you reach doing experiments on earth.
posted by wnissen at 10:06 PM on October 18, 2011
Aaaand the roller coaster problem, explained in a manner which should hopefully demystify some of Newton's work. It's a bit of a bastard example for your professor to spring on you, because it centrifugal force can seem awfully spooky until you really have a firm grasp on the math.
So: your weight is the force you exert against the ground. When looking at centrifugal force as a newbie, it helps to look at it backwards: your weight is the force that you feel the ground pushing up against you. Again, Newton's second law:
F (your weight) = m (your mass) * a (your acceleration)
If you're in your roller coaster cart, heading in a straight line across the ground at a constant speed, the acceleration "a" that you feel (along the vertical axis, normal to the earth) is simply "g."
But remember Newton's first law: in order to change your vertical motion (send you up into the loop), something must provide a force in the upward direction. That would be the cart pushing up on you. (And the cart is being pushed up by the track, and so on. But we don't much care about that.)
But if the cart is pushing up into you (applying a force), that makes you weigh more. After all, just up above we defined weight as the force you feel the ground push up on you with. So not only do you have you plain old weight (mg) to contend with, but you also have the cart trying to accelerate you (at rate "a") up into the sky. To accomplish that acceleration, it has to be pressing up against you according to Newton's second law: F = ma.
So you have two components of force pushing up: the normal reaction to gravity (mg) plus the extra from being accelerating up (ma).
Ftotal = m*g + m*a
The reason that I say that it's a bit of a bastard question is because inertial centrifugal force is called a "fictitious force" for a reason. It's not fake or anything, but it is rather hard to describe without getting into the calculus of physics.
Also, note that I defined the positive axis of acceleration straight up: the ground pushing on you. This is purely for convenience. We could make positive be pointing down, do everything in negative numbers, and still get the same result.
And finally, I know that this isn't much in the way of "here are the N rules you follow to determine sign, etc.", but those approaches only get you so far before you hit a problem for which you don't have a hard rule. Basic physics is all about 1. understanding the phenomena (Newton's laws, etc.), and 2. being able to translate those phenomena into real language and space.
posted by introp at 10:16 PM on October 18, 2011 [1 favorite]
So: your weight is the force you exert against the ground. When looking at centrifugal force as a newbie, it helps to look at it backwards: your weight is the force that you feel the ground pushing up against you. Again, Newton's second law:
F (your weight) = m (your mass) * a (your acceleration)
If you're in your roller coaster cart, heading in a straight line across the ground at a constant speed, the acceleration "a" that you feel (along the vertical axis, normal to the earth) is simply "g."
But remember Newton's first law: in order to change your vertical motion (send you up into the loop), something must provide a force in the upward direction. That would be the cart pushing up on you. (And the cart is being pushed up by the track, and so on. But we don't much care about that.)
But if the cart is pushing up into you (applying a force), that makes you weigh more. After all, just up above we defined weight as the force you feel the ground push up on you with. So not only do you have you plain old weight (mg) to contend with, but you also have the cart trying to accelerate you (at rate "a") up into the sky. To accomplish that acceleration, it has to be pressing up against you according to Newton's second law: F = ma.
So you have two components of force pushing up: the normal reaction to gravity (mg) plus the extra from being accelerating up (ma).
Ftotal = m*g + m*a
The reason that I say that it's a bit of a bastard question is because inertial centrifugal force is called a "fictitious force" for a reason. It's not fake or anything, but it is rather hard to describe without getting into the calculus of physics.
Also, note that I defined the positive axis of acceleration straight up: the ground pushing on you. This is purely for convenience. We could make positive be pointing down, do everything in negative numbers, and still get the same result.
And finally, I know that this isn't much in the way of "here are the N rules you follow to determine sign, etc.", but those approaches only get you so far before you hit a problem for which you don't have a hard rule. Basic physics is all about 1. understanding the phenomena (Newton's laws, etc.), and 2. being able to translate those phenomena into real language and space.
posted by introp at 10:16 PM on October 18, 2011 [1 favorite]
You should really check out The Mechanical Universe series of videos (which can be found on google video). It's an old series of lectures on physics produced by Caltech filled with wonderful visual representations of these types of calculations.
posted by zengargoyle at 10:16 PM on October 18, 2011 [1 favorite]
posted by zengargoyle at 10:16 PM on October 18, 2011 [1 favorite]
Sometimes we set the right side equal to zero and move forces to the opposite side of the equals sign to solve for them, and sometimes we leave "ma" on the right side. What's the method behind that madness?
If you know the thing isn't accelerating, then a=0, so ma=0, so your equation looks like (some forces added up)=0. That's convenient because now that 'ma' is gone, it's one less term you have to keep track of. Then you just rearrange terms until the equation looks like (what you want to know)=(some stuff you do know), and you're done. The intermediate forms of the equation don't always have an obvious physical meaning, although if you try you can sometimes figure out a physical meaning of various parts.
Alternately— but completely equivalently— you can say, "I know the thing isn't accelerating, so I know the forces must all balance out," which is another way of saying they add up to 0, but people often skip forward to (some forces)=(some other forces), putting in the minus sign as they do. It's the same procedure, they're just glossing over some steps that they've done a zillion times.
As for the minus signs— you just need to choose an arbitrary direction to be "positive"; it doesn't matter which as long as you're consistent, but as wnissen says some choices make the problem easier to solve. Draw some arrows in the corner of the diagram to remind yourself (I usually do). Then when you're summing forces, make sure you're considering the force acting on the object, rather than the (equal and opposite) force that it is exerting back. Sometimes the opposite force is the one you intuitively think of, which makes it easy to get the sign wrong; just remember that forces come in pairs ("for every action, there is an equal and opposite reaction") and add up the correct half of each pair.
How that relates to the fact that objects in uniform circular motion are always accelerating since the direction of the velocity vector is always changing, I'm not sure
In this problem you're only thinking about the instant you're at the bottom of the turn. Centripetal acceleration is always towards the center of the arc, so at this instant, the center of the arc is straight up, and so is the acceleration. (You can do the math at other points, too, but the forces wouldn't all line up in a line, so you'd need to do vector math— not difficult but don't worry about it yet.)
So, going back to the person sitting still. What do you know about them? You know that gravity is pulling on them with mg. You can arbitrarily decide that up is positive and down is negative. Gravity is pulling down, so that's -mg. You know that they're not moving— so a=0. And you can guess that there's another force acting on them, the normal force, so the whole equation is F=ma --> (-mg)+(normalforce)=ma --> (-mg)+(normalforce)=0 --> normalforce=mg. Okay, we decided that upwards was positive, so the normal force acting on the person is "mg" upwards. (The force acting on the bench they're on, of course, is opposite: "-mg" upwards, or "mg" downwards.)
Now the person on the roller coaster. What do you know? They're accelerating upwards, because they're on a curved track, so a=v2/r. Gravity's pulling on them the same as always (mg downwards = -mg), and the seat is pushing on them (normal force). Again, F=ma --> (-mg)+(normalforce)=ma, but this time a isn't zero so we can't get rid of that term. But we can reshuffle the equation— normalforce=mg+ma=mg+mv2/r. That step is just algebra.
In any physics equation, the stuff on either side is not just numerically equal, it should also be the same kind of thing— the same units. On the left hand, there's a force. On the right hand, there's the sum of a force (mg) and … something else. "mv2/r" isn't a force, but it looks like one mathematically. And intuitively, that's what it is: it's the "centrifugal force", which we know isn't a real physical force, but if you pretend you're stationary in your (actually moving) rollercoaster, then the "centrifugal force" is the mysterious extra force you'd have to have acting on you to explain why you're being pressed into your seat harder.
posted by hattifattener at 10:19 PM on October 18, 2011 [1 favorite]
If you know the thing isn't accelerating, then a=0, so ma=0, so your equation looks like (some forces added up)=0. That's convenient because now that 'ma' is gone, it's one less term you have to keep track of. Then you just rearrange terms until the equation looks like (what you want to know)=(some stuff you do know), and you're done. The intermediate forms of the equation don't always have an obvious physical meaning, although if you try you can sometimes figure out a physical meaning of various parts.
Alternately— but completely equivalently— you can say, "I know the thing isn't accelerating, so I know the forces must all balance out," which is another way of saying they add up to 0, but people often skip forward to (some forces)=(some other forces), putting in the minus sign as they do. It's the same procedure, they're just glossing over some steps that they've done a zillion times.
As for the minus signs— you just need to choose an arbitrary direction to be "positive"; it doesn't matter which as long as you're consistent, but as wnissen says some choices make the problem easier to solve. Draw some arrows in the corner of the diagram to remind yourself (I usually do). Then when you're summing forces, make sure you're considering the force acting on the object, rather than the (equal and opposite) force that it is exerting back. Sometimes the opposite force is the one you intuitively think of, which makes it easy to get the sign wrong; just remember that forces come in pairs ("for every action, there is an equal and opposite reaction") and add up the correct half of each pair.
How that relates to the fact that objects in uniform circular motion are always accelerating since the direction of the velocity vector is always changing, I'm not sure
In this problem you're only thinking about the instant you're at the bottom of the turn. Centripetal acceleration is always towards the center of the arc, so at this instant, the center of the arc is straight up, and so is the acceleration. (You can do the math at other points, too, but the forces wouldn't all line up in a line, so you'd need to do vector math— not difficult but don't worry about it yet.)
So, going back to the person sitting still. What do you know about them? You know that gravity is pulling on them with mg. You can arbitrarily decide that up is positive and down is negative. Gravity is pulling down, so that's -mg. You know that they're not moving— so a=0. And you can guess that there's another force acting on them, the normal force, so the whole equation is F=ma --> (-mg)+(normalforce)=ma --> (-mg)+(normalforce)=0 --> normalforce=mg. Okay, we decided that upwards was positive, so the normal force acting on the person is "mg" upwards. (The force acting on the bench they're on, of course, is opposite: "-mg" upwards, or "mg" downwards.)
Now the person on the roller coaster. What do you know? They're accelerating upwards, because they're on a curved track, so a=v2/r. Gravity's pulling on them the same as always (mg downwards = -mg), and the seat is pushing on them (normal force). Again, F=ma --> (-mg)+(normalforce)=ma, but this time a isn't zero so we can't get rid of that term. But we can reshuffle the equation— normalforce=mg+ma=mg+mv2/r. That step is just algebra.
In any physics equation, the stuff on either side is not just numerically equal, it should also be the same kind of thing— the same units. On the left hand, there's a force. On the right hand, there's the sum of a force (mg) and … something else. "mv2/r" isn't a force, but it looks like one mathematically. And intuitively, that's what it is: it's the "centrifugal force", which we know isn't a real physical force, but if you pretend you're stationary in your (actually moving) rollercoaster, then the "centrifugal force" is the mysterious extra force you'd have to have acting on you to explain why you're being pressed into your seat harder.
posted by hattifattener at 10:19 PM on October 18, 2011 [1 favorite]
Response by poster: 2 quick notes (absorbing your explanations!)--
1. My professor only talks about centripetal force (not centrifugal force?). Centripetal force points toward the center and causes the centripetal acceleration that points toward the center and constantly changes the velocity's direction, right? Why don't our problems reference centrifugal force?
2. The apparent weight when talking about uniform circular motion... is the same as apparent weight when talking about a vertically moving elevator, or no? The apparent weight (N = ma + mg) formula was paired with the drawing of an elevator flying up and down. The roller coaster has acceleration (centripetal acceleration keeping up on the circular path of radian r, right?) always pointing towards the center, so would that make a positive or negative? It can't be zero if it's uniform circular motion, right? Feel free to ignore this second question if it's already been answered above... still reading!
posted by sunnychef88 at 10:27 PM on October 18, 2011
1. My professor only talks about centripetal force (not centrifugal force?). Centripetal force points toward the center and causes the centripetal acceleration that points toward the center and constantly changes the velocity's direction, right? Why don't our problems reference centrifugal force?
2. The apparent weight when talking about uniform circular motion... is the same as apparent weight when talking about a vertically moving elevator, or no? The apparent weight (N = ma + mg) formula was paired with the drawing of an elevator flying up and down. The roller coaster has acceleration (centripetal acceleration keeping up on the circular path of radian r, right?) always pointing towards the center, so would that make a positive or negative? It can't be zero if it's uniform circular motion, right? Feel free to ignore this second question if it's already been answered above... still reading!
posted by sunnychef88 at 10:27 PM on October 18, 2011
Response by poster: **circular path of radius r
posted by sunnychef88 at 10:28 PM on October 18, 2011
posted by sunnychef88 at 10:28 PM on October 18, 2011
Why don't our problems reference centrifugal force?
Because it's not really a force. There is no force pushing you outwards; it's just a figment of your inertia.
The apparent weight when talking about uniform circular motion... is the same as apparent weight when talking about a vertically moving elevator, or no
Yes, it's precisely the same. It's just that in an elevator you're not also moving sideways, and the direction of acceleration isn't constantly changing to keep you going in a circle. But the essential physics there is the same.
posted by hattifattener at 10:32 PM on October 18, 2011 [2 favorites]
Because it's not really a force. There is no force pushing you outwards; it's just a figment of your inertia.
The apparent weight when talking about uniform circular motion... is the same as apparent weight when talking about a vertically moving elevator, or no
Yes, it's precisely the same. It's just that in an elevator you're not also moving sideways, and the direction of acceleration isn't constantly changing to keep you going in a circle. But the essential physics there is the same.
posted by hattifattener at 10:32 PM on October 18, 2011 [2 favorites]
Re your updated question 2: it's virtually impossible to explain in numbers unless you understand trigonometry and some amount vector math. The over-simple summary: the centrifugal/centripetal acceleration is ONLY along the same line as gravity in two places: at the top and at the bottom. At all other times that line from center-to-cart is tilted with respect to vertical, so the concept of "sign" goes right out the window. As in your first example, the force pressing sideways on something has zero impact on its acceleration up or down.
posted by introp at 10:36 PM on October 18, 2011 [1 favorite]
posted by introp at 10:36 PM on October 18, 2011 [1 favorite]
Someone above mentioned drawing a picture and I agree - forces diagrams are awesome. You draw your object (or system of objects), little arrows representing all the forces, and you draw your coordinate system (+x is to the right, +y is up, for example) and then you can put the forces in terms of your coordinates (a force that's pointing to the right is going in the X direction, and one that's pointing to the left is negative in the X direction, etc). Then what you do is add all the forces in the X direction, then add all the forces in the Y direction, to get net force in each direction.
As far as centripetal acceleration - picture the bottom of the arc. At that moment, you're only moving forward; you have no velocity up or down. If you have a ball on a string and you spin it, and you cut or let go of the string, it continues moving in a straight line - an object in motion will stay in motion, object at rest will stay at rest, unless acted upon by an outside force (that's F=ma in words, saying that if force = 0 then acceleration=0 and so the object's velocity does not change). Anyway, in order to have a curve, the ball or cart that is at the bottom of its arc must start moving upward - velocity(up) increases, which means acceleration(up) > 0, which means force(up) > 0. And if you have a full circle, then on the far right side of that circle the ball is only moving vertically. It was moving to the right a moment ago, in a moment it will be moving to the left - in order to make that happen it has to accelerate to the left. Again, toward the center of the circle (perpendicular to the direction of motion).
posted by Lady Li at 11:05 PM on October 18, 2011
As far as centripetal acceleration - picture the bottom of the arc. At that moment, you're only moving forward; you have no velocity up or down. If you have a ball on a string and you spin it, and you cut or let go of the string, it continues moving in a straight line - an object in motion will stay in motion, object at rest will stay at rest, unless acted upon by an outside force (that's F=ma in words, saying that if force = 0 then acceleration=0 and so the object's velocity does not change). Anyway, in order to have a curve, the ball or cart that is at the bottom of its arc must start moving upward - velocity(up) increases, which means acceleration(up) > 0, which means force(up) > 0. And if you have a full circle, then on the far right side of that circle the ball is only moving vertically. It was moving to the right a moment ago, in a moment it will be moving to the left - in order to make that happen it has to accelerate to the left. Again, toward the center of the circle (perpendicular to the direction of motion).
posted by Lady Li at 11:05 PM on October 18, 2011
One thing to keep in mind is that F, m, a, and g are just variables. g is usually 9.8m/s2 (but not always, g varies depending on where on the earth you are, how high up you are, and so on. Not to mention other planets)
m, a and F are whatever defined by the problem you're working on.
But before you think about the problem, go zen. Realize that there are no 'rules' and all those letters just represent different numbers. But, numbers with different units. What you really have to look out for is kilograms, meters, and seconds. So g is measured in meters/(seconds*seconds) because it's acceleration. When you do math on quantities in physics you do math on the units just like. Breaking down the units is really helpful.
For example Lets say for some reason you have to divide velocity by acceleration . Acceleration is m/s2 and velocity is m/s. So (a)/(v) = (m/s)/(m/s2) = s. So you get an amount of time. What does that mean? Well, it here it means that if you have a target velocity of x and you need to accelerate to y then it will take x/y seconds. As the acceleration goes up the amount of time goes up.
That's all done just by looking at the units in the equation, not the variables.
the other thing that's true but you probably don't need to understand for your class (at least at this point) is that all of these quantities are three dimensional vectors in space, not scalar quantities. So for example:
changing our notation to v1 = (0,0,1)m/s and v2 = (1,0,0)m/s you do the regular equations where a = Δv/t. Δv = v2 - v1 which is equal to (-1,0,1)m/s. Now the magnitude of a vector is found by the equation √(x2 + y2 + z2) which in this case is √(2)m/s ~= 1.4142m/s
Now all we do is divide by 5s (because it's still change in velocity over change in time) so we end up with (√(2)m/s)/s = (√(2)m/s2), and as we know anything in m/s2 is acceleration, just like we want.
(To think about this visually, just imagine a rocket that was falling through space but pointed up at a 45° angle. As the rocket fires it engines for 5 seconds, it slows it's fall, but also starts moving to the left. After 5 seconds, it's not falling at all, but is moving in an entirely leftward direction)
So you don't really need to know trig, just linear algebra. However, you need to understand calculus to really understand the concept of instantaneous change in velocity.
Why don't our problems reference centrifugal force?
they're using "apparent weight" instead of "centrifugal force" to make it simpler to understand, I think. Apparent weight is just m * c where c would be the 'centrifugal force'
posted by delmoi at 11:10 PM on October 18, 2011 [1 favorite]
m, a and F are whatever defined by the problem you're working on.
But before you think about the problem, go zen. Realize that there are no 'rules' and all those letters just represent different numbers. But, numbers with different units. What you really have to look out for is kilograms, meters, and seconds. So g is measured in meters/(seconds*seconds) because it's acceleration. When you do math on quantities in physics you do math on the units just like. Breaking down the units is really helpful.
For example Lets say for some reason you have to divide velocity by acceleration . Acceleration is m/s2 and velocity is m/s. So (a)/(v) = (m/s)/(m/s2) = s. So you get an amount of time. What does that mean? Well, it here it means that if you have a target velocity of x and you need to accelerate to y then it will take x/y seconds. As the acceleration goes up the amount of time goes up.
That's all done just by looking at the units in the equation, not the variables.
the other thing that's true but you probably don't need to understand for your class (at least at this point) is that all of these quantities are three dimensional vectors in space, not scalar quantities. So for example:
2. The apparent weight when talking about uniform circular motion... is the same as apparent weight when talking about a vertically moving elevator, or no? The apparent weight (N = ma + mg) formula was paired with the drawing of an elevator flying up and down. The roller coaster has acceleration (centripetal acceleration keeping up on the circular path of radian r, right?) always pointing towards the center, so would that make a positive or negative? It can't be zero if it's uniform circular motion, right? Feel free to ignore this second question if it's already been answered above... still reading!So velocity is a vector, let's say you're going up and you end up going left over 5 seconds (with a constant acceleration during those 5 seconds). So you start with a velocity of x = 0, y = 0, z = 1 and end up with x = 1, y = 0, z = 0. Those two vectors have the same magnitude different directions.
changing our notation to v1 = (0,0,1)m/s and v2 = (1,0,0)m/s you do the regular equations where a = Δv/t. Δv = v2 - v1 which is equal to (-1,0,1)m/s. Now the magnitude of a vector is found by the equation √(x2 + y2 + z2) which in this case is √(2)m/s ~= 1.4142m/s
Now all we do is divide by 5s (because it's still change in velocity over change in time) so we end up with (√(2)m/s)/s = (√(2)m/s2), and as we know anything in m/s2 is acceleration, just like we want.
(To think about this visually, just imagine a rocket that was falling through space but pointed up at a 45° angle. As the rocket fires it engines for 5 seconds, it slows it's fall, but also starts moving to the left. After 5 seconds, it's not falling at all, but is moving in an entirely leftward direction)
So you don't really need to know trig, just linear algebra. However, you need to understand calculus to really understand the concept of instantaneous change in velocity.
Why don't our problems reference centrifugal force?
they're using "apparent weight" instead of "centrifugal force" to make it simpler to understand, I think. Apparent weight is just m * c where c would be the 'centrifugal force'
posted by delmoi at 11:10 PM on October 18, 2011 [1 favorite]
Response by poster: The vector explanation and reminder to use dimensional analysis/watch units was helpful. We do use lots of trig and vector/scalar quantities in addition to algebra. That's the easier part of the math than algebra for the visual-spatial learner, fortunately. Thanks again, all!
posted by sunnychef88 at 11:16 PM on October 18, 2011
posted by sunnychef88 at 11:16 PM on October 18, 2011
so we end up with (√(2)m/s)/s = (√(2)m/s2)
Er, that should be *(√(2)m/s)/5s = (0.2)*(√(2)m/s2) since the acceleration happened over 5 seconds in my example
posted by delmoi at 11:27 PM on October 18, 2011
Er, that should be *(√(2)m/s)/5s = (0.2)*(√(2)m/s2) since the acceleration happened over 5 seconds in my example
posted by delmoi at 11:27 PM on October 18, 2011
Oh yeah - just a footnote since you mention 'always accelerating' as a source of confusion - you are 'accelerating' when your speed decreases or velocity changes direction, same as when speed increases. That's not how the word is colloquially used and it helps to keep the distinction in mind. Similarly, velocity is the vector form of speed and always has a direction attached.
posted by Lady Li at 1:32 AM on October 19, 2011
posted by Lady Li at 1:32 AM on October 19, 2011
Why do we leave ma on the right sometimes, and other times get rid of the ma to solve for individual forces?
Let me start by some generalities before getting to answer your specific question.
An acceleration can be the result of the application of forces onto an object. The resulting acceleration will be proportional to the (vector) sum of the forces. Mathematically, this is embodied by the equation F = ma, where F represents the net force (i.e. vector sum) and m is the constant of proportionality which we call mass and is a measure of inertia (resistance to change in the state of motion).
When the net force is zero, there is no change in the state of motion i.e., the acceleration is zero. In that case, if the initial velocity is zero, it will remain zero thus staying constant; if it is non-zero, it will also stay constant. Velocity equal zero (object at rest) is just a special case for us because we somehow see object at rest to be in a different state than objects that are moving with respect to us. (For physicists, they are one and the same... it's just a matter of picking a frame of reference.)
Now, for your question... Sometimes, problems are set up (by a teacher) such that the acceleration will be zero. Usually, but not always, these problems will also be set up so that the velocity will be zero. These are known as statics problems. In this case, we can't ask you to find the acceleration (since it is zero!); instead, we ask you to use the equations to solve for one of the variables (one of the forces in the problem) that is not given.
When the acceleration is not zero, we talk about "dynamics" problems. Here, we usually do not give you the acceleration, but all the information required to know everything about the forces in the problem. In that case, you have to keep the "ma" in the equation.
Since forces and acceleration are vectors, we usually have a system of equations, one for each coordinate. Often, we can choose our system of coordinates such that the acceleration will have a zero component in all but one coordinate. This often simplifies the algebra.
Now, onto the roller coaster problem...
When two surfaces are in contact with each other, there will (usually) be a force acting on each of them (equal and opposite, blah, blah, blah...). It is convenient to choose a system of axis so that one axis is perpendicular (or normal) to the surfaces and the other parallel to them. The force of contact is then usually written as the sum of two components: one that is in the direction of the perpendicular or normal axis (which is often referred to as "the" normal force) and the other which is in the direction parallel to the surfaces (and often referred to as the friction force).
Small detour...
When someone is sitting in a chair, why don't they fall through it? Gravity is acting on them after all? The reason is that, the chair, in contact with the body of the person (thus, we have two surfaces in contact) exerts a "normal force" in the direction opposite to that of the force of gravity so that the system is perfectly balanced...
Back to the roller coaster problem...
The sum of the forces acting on the person are:
1. gravity --- always in the "down" direction
2. the "normal" force --- pushing away from the surface at the point of contact...
That's it.
What do these forces produce? Ans: an acceleration. If it turns out that this acceleration (parallel to the net forces applied) is perpendicular to the velocity (like it is for a roller coaster at the bottom of a turn), then the acceleration will not change the magnitude of the velocity (aka speed), but only its direction, thus resulting in a circular motion. One can work out the math and show that the magnitude of the acceleration will be a = v^2 / r.
So, the equation describing our problem is
(sum) F = m a
Let's take "up" as the positive direction. The normal force N is in the positive direction; the gravity (mg) is in the negative direction and the acceleration will also be in the positive direction.
Using + and - sign to indicate direction (easier than to draw vectors), I will then have
N - mg = ma = mv^2/r.
That's it for the equations...
Now how do we feel "heavier" or "lighter"? Well, if we are sitting on a chair and both the chair and us are in free fall ... we won't feel the chair pushing us up very much ... The normal force will be zero. On the other hand, if we are sitting on a chair and someone pushes the chair up (accelerates it upwards), we are going to feel that push as increased pressure on our bum, etc. This increased pressure on our bum (transmitted also to our internal organs) makes us feel heavier.
If we are feeling 33% heavier than usual, it is because the normal force is 33% higher than usual. What is it usually? ... Well, usually, when we are sitting, we are at rest and the acceleration is zero, thus:
N_usual - mg = 0
which gives N = mg. Here, we must have N = 1.33 mg. Put that back in the equation above:
1.33 mg - mg = mv^2/r
and you can finally solve the problem! ;-)
posted by aroberge at 6:31 AM on October 19, 2011 [1 favorite]
Let me start by some generalities before getting to answer your specific question.
An acceleration can be the result of the application of forces onto an object. The resulting acceleration will be proportional to the (vector) sum of the forces. Mathematically, this is embodied by the equation F = ma, where F represents the net force (i.e. vector sum) and m is the constant of proportionality which we call mass and is a measure of inertia (resistance to change in the state of motion).
When the net force is zero, there is no change in the state of motion i.e., the acceleration is zero. In that case, if the initial velocity is zero, it will remain zero thus staying constant; if it is non-zero, it will also stay constant. Velocity equal zero (object at rest) is just a special case for us because we somehow see object at rest to be in a different state than objects that are moving with respect to us. (For physicists, they are one and the same... it's just a matter of picking a frame of reference.)
Now, for your question... Sometimes, problems are set up (by a teacher) such that the acceleration will be zero. Usually, but not always, these problems will also be set up so that the velocity will be zero. These are known as statics problems. In this case, we can't ask you to find the acceleration (since it is zero!); instead, we ask you to use the equations to solve for one of the variables (one of the forces in the problem) that is not given.
When the acceleration is not zero, we talk about "dynamics" problems. Here, we usually do not give you the acceleration, but all the information required to know everything about the forces in the problem. In that case, you have to keep the "ma" in the equation.
Since forces and acceleration are vectors, we usually have a system of equations, one for each coordinate. Often, we can choose our system of coordinates such that the acceleration will have a zero component in all but one coordinate. This often simplifies the algebra.
Now, onto the roller coaster problem...
When two surfaces are in contact with each other, there will (usually) be a force acting on each of them (equal and opposite, blah, blah, blah...). It is convenient to choose a system of axis so that one axis is perpendicular (or normal) to the surfaces and the other parallel to them. The force of contact is then usually written as the sum of two components: one that is in the direction of the perpendicular or normal axis (which is often referred to as "the" normal force) and the other which is in the direction parallel to the surfaces (and often referred to as the friction force).
Small detour...
When someone is sitting in a chair, why don't they fall through it? Gravity is acting on them after all? The reason is that, the chair, in contact with the body of the person (thus, we have two surfaces in contact) exerts a "normal force" in the direction opposite to that of the force of gravity so that the system is perfectly balanced...
Back to the roller coaster problem...
The sum of the forces acting on the person are:
1. gravity --- always in the "down" direction
2. the "normal" force --- pushing away from the surface at the point of contact...
That's it.
What do these forces produce? Ans: an acceleration. If it turns out that this acceleration (parallel to the net forces applied) is perpendicular to the velocity (like it is for a roller coaster at the bottom of a turn), then the acceleration will not change the magnitude of the velocity (aka speed), but only its direction, thus resulting in a circular motion. One can work out the math and show that the magnitude of the acceleration will be a = v^2 / r.
So, the equation describing our problem is
(sum) F = m a
Let's take "up" as the positive direction. The normal force N is in the positive direction; the gravity (mg) is in the negative direction and the acceleration will also be in the positive direction.
Using + and - sign to indicate direction (easier than to draw vectors), I will then have
N - mg = ma = mv^2/r.
That's it for the equations...
Now how do we feel "heavier" or "lighter"? Well, if we are sitting on a chair and both the chair and us are in free fall ... we won't feel the chair pushing us up very much ... The normal force will be zero. On the other hand, if we are sitting on a chair and someone pushes the chair up (accelerates it upwards), we are going to feel that push as increased pressure on our bum, etc. This increased pressure on our bum (transmitted also to our internal organs) makes us feel heavier.
If we are feeling 33% heavier than usual, it is because the normal force is 33% higher than usual. What is it usually? ... Well, usually, when we are sitting, we are at rest and the acceleration is zero, thus:
N_usual - mg = 0
which gives N = mg. Here, we must have N = 1.33 mg. Put that back in the equation above:
1.33 mg - mg = mv^2/r
and you can finally solve the problem! ;-)
posted by aroberge at 6:31 AM on October 19, 2011 [1 favorite]
Best answer: This might give you a little study break, and is also relevant to your question about centripetal/centrifugal force. :-)
posted by doctord at 6:55 AM on October 19, 2011
posted by doctord at 6:55 AM on October 19, 2011
This thread is closed to new comments.
Newton's first law (F = ma) allows you to compute, given a force acting on an object, the acceleration of that object. So in this case we could figure the acceleration of the object on the table by:
F = N - mg
F = ma
.. substitute F here..
ma = N - mg
We know the object isn't accelerating at all. It's just sitting there. So a == 0.
m*0 = N - m*g
0 = N - m*g
.. or written another way ...
m*g = N
The force that the table is providing upwards against your object is equal to its mass times the acceleration due to gravity.
Does explaining it that way help understand why we'd move things about, etc.?
posted by introp at 9:47 PM on October 18, 2011