Equal and Opposite Reaction?
May 18, 2007 8:09 AM Subscribe
Physics: Difference between two people pulling on each end of a rope, and one person pulling on a rope tied to a tree.
I'm wondering, what exactly the forces involved here are. If in the middle of the rope there was a stress meter or some sort of gauge, would two people of equal strength pulling on each side, with good traction against the ground, be able to apply more force then one person pulling on the rope with the other side tied to a tree?
I intuitively thought that two people pulling would exert more force, but I now am not so sure - and admitted defeat during the gigantic argument I ended up getting in to with my boyfriend over this a couple months ago. I've been curious ever since about the actual mechanics that are involved. I wish I'd saved the bar napkins we were drawing diagrams and counter-diagrams on, they'd make for interesting research material on the effects of alcohol on rational thought...
The crux of this debate we had was over which would put more strain on to the rope, and if in the two different cases you'd be losing more or less force to pulling on the earth itself. Forgive me any lapses in terminology, AP physics was a decade ago.
I'm wondering, what exactly the forces involved here are. If in the middle of the rope there was a stress meter or some sort of gauge, would two people of equal strength pulling on each side, with good traction against the ground, be able to apply more force then one person pulling on the rope with the other side tied to a tree?
I intuitively thought that two people pulling would exert more force, but I now am not so sure - and admitted defeat during the gigantic argument I ended up getting in to with my boyfriend over this a couple months ago. I've been curious ever since about the actual mechanics that are involved. I wish I'd saved the bar napkins we were drawing diagrams and counter-diagrams on, they'd make for interesting research material on the effects of alcohol on rational thought...
The crux of this debate we had was over which would put more strain on to the rope, and if in the two different cases you'd be losing more or less force to pulling on the earth itself. Forgive me any lapses in terminology, AP physics was a decade ago.
In the two person situation you can't exert so much force the other person falls, slips or otherwise loses the grip on the rope. This doesn't apply to the tree, so more force could in principle be applied in that situation.
Otherwise there's no difference in the forces one person exerts in that situation.
posted by edd at 8:16 AM on May 18, 2007
Otherwise there's no difference in the forces one person exerts in that situation.
posted by edd at 8:16 AM on May 18, 2007
Think of it this way. The acceleration on the rope is 0 (cause it ain't going anywhere). Therefor the net force must be zero. Let the force that a person can put on the rope be F. Since person 1 puts force F on the rope the tree must put force -F on the rope. Similarly, since person 1 puts force F on the rope person 2 must put force -F on the rope. So the person and the tree excert the same force.
--Chris
posted by Maastrictian at 8:17 AM on May 18, 2007
--Chris
posted by Maastrictian at 8:17 AM on May 18, 2007
Response by poster: Edd, we were picturing perfectly matched strength.
posted by voidcontext at 8:18 AM on May 18, 2007
posted by voidcontext at 8:18 AM on May 18, 2007
Response by poster: I guess this was just a lot less complicated then I was trying to make it in my head.
posted by voidcontext at 8:19 AM on May 18, 2007
posted by voidcontext at 8:19 AM on May 18, 2007
OK, although it's less about the strength and more about the quality of the footwear.
posted by edd at 8:21 AM on May 18, 2007
posted by edd at 8:21 AM on May 18, 2007
1) If two people are pulling on the rope with a force of, say, 10 Newtons, then
F1 - F2 = mass x accelleration, and since the forces are the same, you know the rope is stationary (accelleration = 0).
2) When the rope is attached to a tree, the rope is still stationary: accelleration = 0.
so Fperson - Ftree = 0. In other words, Fperson=Ftree, which is the same result that we got in situation 1.
posted by muddgirl at 8:21 AM on May 18, 2007
F1 - F2 = mass x accelleration, and since the forces are the same, you know the rope is stationary (accelleration = 0).
2) When the rope is attached to a tree, the rope is still stationary: accelleration = 0.
so Fperson - Ftree = 0. In other words, Fperson=Ftree, which is the same result that we got in situation 1.
posted by muddgirl at 8:21 AM on May 18, 2007
Another way to look at it: the absolute maximum force you can exert on a rope is the maximum force you can exert with your feet against the ground. That doesn't change, so two people pulling on a rope will never exert MORE force than one of them against a tree; at absolute best it will be equal. Most likely it will be a little less, since once of them is likely to be weaker/have less traction.
posted by Malor at 8:23 AM on May 18, 2007
posted by Malor at 8:23 AM on May 18, 2007
Response by poster: I'm really embarrassed about how much of this I've forgotten. I guess I need to hit the books.
posted by voidcontext at 8:39 AM on May 18, 2007
posted by voidcontext at 8:39 AM on May 18, 2007
maybe, the question really should be if equal force is applied by the two people AND the person vs. tree which rope is under more stress... ?
posted by sxtxixtxcxh at 8:53 AM on May 18, 2007
posted by sxtxixtxcxh at 8:53 AM on May 18, 2007
I didn't really pay attention to this answer until it got marked as 'best', but:
This is the same question as whether there is more force involved in a collision between a car going 50 mph and a wall or two cars head on, each going 50 mph. It's the same amount of force.
How can that be true? If both cars are going 50mph, you have all the kinetic energy of BOTH to dissipate, not just one.
posted by Malor at 9:03 AM on May 18, 2007
This is the same question as whether there is more force involved in a collision between a car going 50 mph and a wall or two cars head on, each going 50 mph. It's the same amount of force.
How can that be true? If both cars are going 50mph, you have all the kinetic energy of BOTH to dissipate, not just one.
posted by Malor at 9:03 AM on May 18, 2007
This is the same question as whether there is more force involved in a collision between a car going 50 mph and a wall or two cars head on, each going 50 mph. It's the same amount of force.
Wait, how would the collision between two cars not be a 100mph collision? If you're in one car, that other car is coming at you at 100mph. The stationary wall is still only coming at you at 50.
posted by LionIndex at 9:05 AM on May 18, 2007
Wait, how would the collision between two cars not be a 100mph collision? If you're in one car, that other car is coming at you at 100mph. The stationary wall is still only coming at you at 50.
posted by LionIndex at 9:05 AM on May 18, 2007
Response by poster: I unmarked it as best answer since it looks like it wasn't correct after all, although later ones were. I didn't want to mark all the answers as best, so I was just going to mark the first.
posted by voidcontext at 9:13 AM on May 18, 2007
posted by voidcontext at 9:13 AM on May 18, 2007
Response by poster: sxtxixtxcxh, that is something we also discussed. What is the result in that situation?
posted by voidcontext at 9:14 AM on May 18, 2007
posted by voidcontext at 9:14 AM on May 18, 2007
As long as when there are two people niether of them is moving (Tension involes weight and acceleration) then the force on the rope is the same.
posted by DJWeezy at 9:20 AM on May 18, 2007
posted by DJWeezy at 9:20 AM on May 18, 2007
Actually, no, BigSky is wrong. (Relativity.)
When we imagine watching two cars collide (or anything else), we arbitrarily choose a stationary point of perspective. Say we delegate the wall to occupy the role of stationary point. We are the wall and we stare at the oncoming car. It comes at us at 50 mph relative to us, and hits us with the momentum of an oncoming car at 50 mph.
If we delegate a spot midway between two cars to be our stationary point and we watch that spot as the cars rush toward it at 50 mph each relative to the midpoint, we might find that a little complicated to calculate.
We simplify our calculations by delegating one of the cars to be our stationary point. So. We are sitting in Car A which is perfectly motionless. The earth is moving under us at a rate of 50 mph relative to us. Car B is moving 50 mph relative to the earth but 100 mph relative to us. We are sitting perfectly still (just like we were when we were the wall) and Car B hits us with the momentum of an oncoming car at 100 mph.
Note that Car B could just as easily be chosen as the stationary point of reference; in that case, it experiences the momentum of Car A at 100 mph. It doesn’t matter who is hitting who; the effect is the same for both.
Anyway, if I am a designated stationary point and a car is moving toward me at a speed relative to me of 50 mph I will absorb the momentum of a car at 50 mph. This relative motion can happen in different ways - we could both be moving relative to the earth at 25 mph, or one of us could be motionless with respect to the earth and the other moving at 50 mph with respect to the earth, or some combination of the above. Any calculation needs to take into account that the situations above are indistinguishable.
posted by kika at 9:23 AM on May 18, 2007
When we imagine watching two cars collide (or anything else), we arbitrarily choose a stationary point of perspective. Say we delegate the wall to occupy the role of stationary point. We are the wall and we stare at the oncoming car. It comes at us at 50 mph relative to us, and hits us with the momentum of an oncoming car at 50 mph.
If we delegate a spot midway between two cars to be our stationary point and we watch that spot as the cars rush toward it at 50 mph each relative to the midpoint, we might find that a little complicated to calculate.
We simplify our calculations by delegating one of the cars to be our stationary point. So. We are sitting in Car A which is perfectly motionless. The earth is moving under us at a rate of 50 mph relative to us. Car B is moving 50 mph relative to the earth but 100 mph relative to us. We are sitting perfectly still (just like we were when we were the wall) and Car B hits us with the momentum of an oncoming car at 100 mph.
Note that Car B could just as easily be chosen as the stationary point of reference; in that case, it experiences the momentum of Car A at 100 mph. It doesn’t matter who is hitting who; the effect is the same for both.
Anyway, if I am a designated stationary point and a car is moving toward me at a speed relative to me of 50 mph I will absorb the momentum of a car at 50 mph. This relative motion can happen in different ways - we could both be moving relative to the earth at 25 mph, or one of us could be motionless with respect to the earth and the other moving at 50 mph with respect to the earth, or some combination of the above. Any calculation needs to take into account that the situations above are indistinguishable.
posted by kika at 9:23 AM on May 18, 2007
Response by poster: Just to clarify what I think I understand:
1. Two people, at either end of the rope, securely in place (say they are super glued to the floor), pulling on the rope.
2. One person attached to the floor, pulling on a rope secured to a fixed object.
3. Two people, secured to the floor, pulling from the same end of the rope, attached to a fixed object.
#1 and #2 are equivalent from the rope's perspective, and #3 should be about twice of either #1 or #2?
The force countering the people pulling on the rope in #2 and #3 is the inertia of the earth?
posted by voidcontext at 9:31 AM on May 18, 2007
1. Two people, at either end of the rope, securely in place (say they are super glued to the floor), pulling on the rope.
2. One person attached to the floor, pulling on a rope secured to a fixed object.
3. Two people, secured to the floor, pulling from the same end of the rope, attached to a fixed object.
#1 and #2 are equivalent from the rope's perspective, and #3 should be about twice of either #1 or #2?
The force countering the people pulling on the rope in #2 and #3 is the inertia of the earth?
posted by voidcontext at 9:31 AM on May 18, 2007
This is an interesting problem and a bit counter intuitive.
Car 1 hits car 2 and both cars stop, car 1 hits the wall and it stops.
Car 1 exerts a force on car 2, car 2 exerts an equal and opposite force on car 1.
Car 1 exerts a force on the wall, the wall exerts a force on car 1.
Which of these is bigger? They're the same it's the force of car 1 in either case. This is assuming that car 2 is the same size and speed as car 1. If car 2 is tiny, say a tricycle, then the momentum of car 1 is much greater and it will blow past the tricycle. But when hitting a car of equal size they both come to a stop just like when it hits the wall.
How is this similar to the original question? Two people pull on a rope. Person A pulls 50N to the right, the equal and opposite force here is provided by the opposing person, who pulls 50N to the left. With the tree, Person A pulls 50N to the right, the rope isn't going anywhere, the equal and opposite force from the tree is 50N to the left. Same stress on the rope.
This question from the book is available on the web. Take a look here: http://www.farmingdale.edu/~kramerpr/Home-page/Teasers/head-on.html
posted by BigSky at 9:32 AM on May 18, 2007
Car 1 hits car 2 and both cars stop, car 1 hits the wall and it stops.
Car 1 exerts a force on car 2, car 2 exerts an equal and opposite force on car 1.
Car 1 exerts a force on the wall, the wall exerts a force on car 1.
Which of these is bigger? They're the same it's the force of car 1 in either case. This is assuming that car 2 is the same size and speed as car 1. If car 2 is tiny, say a tricycle, then the momentum of car 1 is much greater and it will blow past the tricycle. But when hitting a car of equal size they both come to a stop just like when it hits the wall.
How is this similar to the original question? Two people pull on a rope. Person A pulls 50N to the right, the equal and opposite force here is provided by the opposing person, who pulls 50N to the left. With the tree, Person A pulls 50N to the right, the rope isn't going anywhere, the equal and opposite force from the tree is 50N to the left. Same stress on the rope.
This question from the book is available on the web. Take a look here: http://www.farmingdale.edu/~kramerpr/Home-page/Teasers/head-on.html
posted by BigSky at 9:32 AM on May 18, 2007
Err, Bigsky's answer is essentially correct, so if you marked and unmarked it, you should mark it again. (he was sloppy about the use of the word "force", but the gist of his statement, that the damage is the same in the two collisions, is correct.)
In automobile crashes, what counts is deceleration, not force. Compare the situations:
a) automobile going 50 mph encounters a brick wall, and in a distance of 3 feet, comes to a halt (the center of mass travels 3 feet forward after the bumper encounters the wall, as the front of the car crushes). 50 mph is 75 feet per second, so this takes about 1/10th of a second to occur.
b) automobile going 50mph encounters another automobile of equal weight also going 50mph in the opposite direction. In this case, the opposing automobile delivers the same amount of force as the brick wall, the auto again stops in 3 feet, and it again takes 1/10th of a second to occur. The deceleration is the same, and the damage is the same.
c) automobile going 50 mph encounters a brick wall of infinite mass traveling 50mph in the opposite direction. In this case, the automobile has to decelerate from 50 mph forward to 50 mph backward in a very short period of time (since the wall doesn't give at all). This is the equivalent of hitting an unmoving brick wall at 100 mph. Twice the deceleration, twice the damage as the above two situations. Situations like this occur when a passenger car crosses the middle line of a roadway and encounters a tractor-trailer going the other way. Since the truck is of very large mass compared to the car, the car suffers most of the deceleration and takes most of the damage, and the collision is considerably worse (for the car) than hitting an equivalent-weight car or a stationary object. For the truck, the collision is considerably better than hitting an equivalent-weight truck or stationary object.
As far as ONE AUTOMOBILE is concerned, the two collisions a) and b) are identical. In the second example, there are of course two automobiles, so people think there's twice as much energy involved (true) and therefore twice as much damage to the ONE AUTOMOBILE (false). The second car absorbs half the energy!
posted by jellicle at 9:35 AM on May 18, 2007
In automobile crashes, what counts is deceleration, not force. Compare the situations:
a) automobile going 50 mph encounters a brick wall, and in a distance of 3 feet, comes to a halt (the center of mass travels 3 feet forward after the bumper encounters the wall, as the front of the car crushes). 50 mph is 75 feet per second, so this takes about 1/10th of a second to occur.
b) automobile going 50mph encounters another automobile of equal weight also going 50mph in the opposite direction. In this case, the opposing automobile delivers the same amount of force as the brick wall, the auto again stops in 3 feet, and it again takes 1/10th of a second to occur. The deceleration is the same, and the damage is the same.
c) automobile going 50 mph encounters a brick wall of infinite mass traveling 50mph in the opposite direction. In this case, the automobile has to decelerate from 50 mph forward to 50 mph backward in a very short period of time (since the wall doesn't give at all). This is the equivalent of hitting an unmoving brick wall at 100 mph. Twice the deceleration, twice the damage as the above two situations. Situations like this occur when a passenger car crosses the middle line of a roadway and encounters a tractor-trailer going the other way. Since the truck is of very large mass compared to the car, the car suffers most of the deceleration and takes most of the damage, and the collision is considerably worse (for the car) than hitting an equivalent-weight car or a stationary object. For the truck, the collision is considerably better than hitting an equivalent-weight truck or stationary object.
As far as ONE AUTOMOBILE is concerned, the two collisions a) and b) are identical. In the second example, there are of course two automobiles, so people think there's twice as much energy involved (true) and therefore twice as much damage to the ONE AUTOMOBILE (false). The second car absorbs half the energy!
posted by jellicle at 9:35 AM on May 18, 2007
voidcontext, if you want to think of another case similar to your original problem it could be as follows. The internal force on a column supporting a 100kg barrel is 10.2 Newtons. The ground exerts that same force. If the weight of the barrel was 200kg then the ground would exert double the force. The ground is the same but it only exerts the force opposite of that applied. The internal force is also always that applied externally.
posted by JJ86 at 9:42 AM on May 18, 2007
posted by JJ86 at 9:42 AM on May 18, 2007
Response by poster: OK, that makes a lot of sense JJ86. And BigSky, I am sorry for taking away the best answer check mark, looks like what you said does parallel my question.
posted by voidcontext at 9:46 AM on May 18, 2007
posted by voidcontext at 9:46 AM on May 18, 2007
As far as ONE AUTOMOBILE is concerned, the two collisions a) and b) are identical. In the second example, there are of course two automobiles, so people think there's twice as much energy involved (true) and therefore twice as much damage to the ONE AUTOMOBILE (false). The second car absorbs half the energy!
That explains it--thanks. I wasn't thinking of the car damage as much as the kinetic energy involved, much as Malor indicates.
posted by LionIndex at 9:51 AM on May 18, 2007
That explains it--thanks. I wasn't thinking of the car damage as much as the kinetic energy involved, much as Malor indicates.
posted by LionIndex at 9:51 AM on May 18, 2007
BigSky is essentially correct as jellicle points out. What kika's analysis fails to take into account is that in a reference frame which initially moves along with one of the cars, both cars don't stop but crash and then move as a unit backwards at 50 mph.
Here's an intuitive way of thinking about the initial rope problem. Let's say that two people decide to pull apart two super-strong magnets. They attach a rope to each magnet.
Case 1: Friend A says to B: "Ok, we're equally strong. Lets just each pull on each magnet with our ropes"
Case 2: Friend A says to B: "Don't pull on your rope just hold it and I'll pull the other rope. But make sure that the magnet doesn't move."
Whats the difference? none. B will have to pull equally as hard in both cases. In case its not clear, in the second case, Friend B is acting as the tree.
posted by vacapinta at 9:56 AM on May 18, 2007
Here's an intuitive way of thinking about the initial rope problem. Let's say that two people decide to pull apart two super-strong magnets. They attach a rope to each magnet.
Case 1: Friend A says to B: "Ok, we're equally strong. Lets just each pull on each magnet with our ropes"
Case 2: Friend A says to B: "Don't pull on your rope just hold it and I'll pull the other rope. But make sure that the magnet doesn't move."
Whats the difference? none. B will have to pull equally as hard in both cases. In case its not clear, in the second case, Friend B is acting as the tree.
posted by vacapinta at 9:56 AM on May 18, 2007
"The force countering the people pulling on the rope in #2 and #3 is the inertia of the earth?"
In the vast majority of movement (all?), the equal and opposite force is applied to the earth. When you take a step you push your body X Newtons forward and you push the earth X Newtons backwards. You and the earth are of disproportionate mass, and X newtons effects a perceptible change in your position, not so much on the earth.
The pulling on the rope and collision of two cars questions are equilibrium systems and JJ86's answer is an analogy. I mention the above so you know that in these equilibriums the horizontal force on the earth is not the important factor, it comes in when you describe movement.
jellicle, thanks for the assist.
posted by BigSky at 10:02 AM on May 18, 2007
In the vast majority of movement (all?), the equal and opposite force is applied to the earth. When you take a step you push your body X Newtons forward and you push the earth X Newtons backwards. You and the earth are of disproportionate mass, and X newtons effects a perceptible change in your position, not so much on the earth.
The pulling on the rope and collision of two cars questions are equilibrium systems and JJ86's answer is an analogy. I mention the above so you know that in these equilibriums the horizontal force on the earth is not the important factor, it comes in when you describe movement.
jellicle, thanks for the assist.
posted by BigSky at 10:02 AM on May 18, 2007
Response by poster: Another question:
#1: Two people
#2: One person + tree
Is the fact that some of the energy in #2 going to moving the entire earth (some ridiculously slight amount) taken away from the strain put on the rope? IE: the earth moves slightly, thus relieving some of the strain?
In #1, the earth is not moved, since both people pulling are equally and oppositely dug in, and all the strain remains on the rope.
That's sort of like...
#1 Two people in space, with jet packs, pulling away from each other, attached by rope, vs:
#2 One person with jet pack, pulling away while attached with a rope to a small spaceship that is not actively expending energy on moving (only inertia involved).
Is there less strain on the rope in #2?
I guess it would be equivalent, now that I have more of a handle on this.
posted by voidcontext at 10:05 AM on May 18, 2007
#1: Two people
#2: One person + tree
Is the fact that some of the energy in #2 going to moving the entire earth (some ridiculously slight amount) taken away from the strain put on the rope? IE: the earth moves slightly, thus relieving some of the strain?
In #1, the earth is not moved, since both people pulling are equally and oppositely dug in, and all the strain remains on the rope.
That's sort of like...
#1 Two people in space, with jet packs, pulling away from each other, attached by rope, vs:
#2 One person with jet pack, pulling away while attached with a rope to a small spaceship that is not actively expending energy on moving (only inertia involved).
Is there less strain on the rope in #2?
I guess it would be equivalent, now that I have more of a handle on this.
posted by voidcontext at 10:05 AM on May 18, 2007
Response by poster: The strain would be the same, but the rope would move in case #2, some amount dependent on the mass of the counterweight.
I think I've got it now.
posted by voidcontext at 10:07 AM on May 18, 2007
I think I've got it now.
posted by voidcontext at 10:07 AM on May 18, 2007
Response by poster: Or but no, because if the whatever the counterweight is, is like the mass of a ping-pong ball, the rope would be under hardly any strain at all?
posted by voidcontext at 10:16 AM on May 18, 2007
posted by voidcontext at 10:16 AM on May 18, 2007
Is the fact that some of the energy in #2 going to moving the entire earth (some ridiculously slight amount) taken away from the strain put on the rope? IE: the earth moves slightly, thus relieving some of the strain?
No, there's no energy going to moving the entire Earth. If that were true you'd have reactionless drive and you'd be violating the conservation of momentum.
Noone's moving in either situation.
posted by edd at 10:39 AM on May 18, 2007
No, there's no energy going to moving the entire Earth. If that were true you'd have reactionless drive and you'd be violating the conservation of momentum.
Noone's moving in either situation.
posted by edd at 10:39 AM on May 18, 2007
"Is the fact that some of the energy in #2 going to moving the entire earth (some ridiculously slight amount) taken away from the strain put on the rope? IE: the earth moves slightly, thus relieving some of the strain?"
Perhaps the tree moves some theoretical amount but the key to the problem is that they are in equilibrium. The measurable strain on the rope is the same. All of the force that the person is exerting is opposed by the tree. The horizontal force on the earth is not important. Whether his feet were attached to the ground or not would make no difference.
The mass of the counterweight to the tree is not important to the strain put on the rope. A difference in mass is relevant in describing why one force causes one object to move and the other one not, at least not a perceptible amount. But here, no one is moving.
I think mentioning that about the earth moving when you step confused you. It wasn't very relevant but I wanted you to understand why equal and opposite force was not the earth when you have equilibrium between the tree and the puller.
posted by BigSky at 10:41 AM on May 18, 2007
Perhaps the tree moves some theoretical amount but the key to the problem is that they are in equilibrium. The measurable strain on the rope is the same. All of the force that the person is exerting is opposed by the tree. The horizontal force on the earth is not important. Whether his feet were attached to the ground or not would make no difference.
The mass of the counterweight to the tree is not important to the strain put on the rope. A difference in mass is relevant in describing why one force causes one object to move and the other one not, at least not a perceptible amount. But here, no one is moving.
I think mentioning that about the earth moving when you step confused you. It wasn't very relevant but I wanted you to understand why equal and opposite force was not the earth when you have equilibrium between the tree and the puller.
posted by BigSky at 10:41 AM on May 18, 2007
Response by poster: OK, thanks. I'm just over thinking it again.
posted by voidcontext at 10:46 AM on May 18, 2007
posted by voidcontext at 10:46 AM on May 18, 2007
The original question's been answered, but a word on terminology. "Strain" is the amount something deforms because a force is applied to it. In this case, the rope will stretch some amount under tension. If I understand some comments here correctly, where some are using 'strain' you'd be more correct to say "stress", or maybe avoid both and say "tension". Stress is the force applied to something divided by the cross-sectional area of that thing, and measured perpendicular to the force. In this case, the cross-sectional area of the rope.
Stress and strain are obviously related and if we know enough about the material in question, one can often be calculated from the other. In real-world situations and materials the relationship can be quite complex, however.
The original question, as stated, doesn't need an understanding of stress and strain, just forces. In the case of the force constrained by a rope, "tension" is probably the most useful word.
Sorry for the pedantry. Just a pet nit I'm picking. Outside of engineering and physics, stress and strain are some of the most commonly mis-applied terms.
posted by normy at 11:09 AM on May 18, 2007 [1 favorite]
Stress and strain are obviously related and if we know enough about the material in question, one can often be calculated from the other. In real-world situations and materials the relationship can be quite complex, however.
The original question, as stated, doesn't need an understanding of stress and strain, just forces. In the case of the force constrained by a rope, "tension" is probably the most useful word.
Sorry for the pedantry. Just a pet nit I'm picking. Outside of engineering and physics, stress and strain are some of the most commonly mis-applied terms.
posted by normy at 11:09 AM on May 18, 2007 [1 favorite]
Response by poster: Normy, that's all stuff I need to know in order to understand the distinctions involved, and I do plan on reading up more and fixing these gaps in my understanding.
Pedantry is welcome and encouraged :)
posted by voidcontext at 11:38 AM on May 18, 2007
Pedantry is welcome and encouraged :)
posted by voidcontext at 11:38 AM on May 18, 2007
The amount of tension a puller can put on a rope is the lesser of either his/her A) physical strength, or B) the amount of traction in his or her shoes.
If I'm pulling against a rope tied to a tree, I can exert force up to those limits; the rope will 'feel' my exertion as tension. The opposing force is the inertia of the tree's roots in the ground; they're hard to pull out. If I'm immensely strong and have gigantic traction (and the rope is strong enough), I can pull the tree out of the ground. This generally takes a tractor, but at least in theory could be done by a very strong person with very, very large shoes. :)
If I'm pulling against someone else who is precisely matched to me, the tension on the rope is the same; it's just that I'm being opposed by strength instead of inertia. If one of us is weaker or has less traction, our strength or feet will give out before the other. Thus, the strain on the rope is the least of four possible forces: My strength, his strength, my traction, his traction. This can NEVER exert more force than I can exert alone.
The real answer to your question is this: two people pulling on a rope will always exert the tension that the weaker/smaller person can manage. If you're pulling against a tree, it's all just one person. Adding another person can ONLY exert the same or less tension, never more. This part I'm really sure about.
As far as head-on collisions go, I'm less sure, but I think the answers so far are wrong. Kinetic energy is a square function on speed. It's what does damage and must be dissipated. The formula is 1/2mv2. A 2000lb car hitting a wall at 50mph has to dissipate (0.5 * 2000) * (50 ^ 2) = 250,000 units of energy. Two cars striking each other at 50mph can be treated as one stationary and one moving at 100mph, so the force involved becomes (0.5 * 2000) * (100 ^ 2) = 10,000,000 units of energy.... 40 times more damage. Even if you get a perfect division of damage, that's still 20 times as much per car.
Note also that cars are usually unequal in size, and the smaller car will almost always take the brunt of the damage.
This is why head-on crashes are so wildly lethal compared to rear-enders.
posted by Malor at 4:36 PM on May 18, 2007
If I'm pulling against a rope tied to a tree, I can exert force up to those limits; the rope will 'feel' my exertion as tension. The opposing force is the inertia of the tree's roots in the ground; they're hard to pull out. If I'm immensely strong and have gigantic traction (and the rope is strong enough), I can pull the tree out of the ground. This generally takes a tractor, but at least in theory could be done by a very strong person with very, very large shoes. :)
If I'm pulling against someone else who is precisely matched to me, the tension on the rope is the same; it's just that I'm being opposed by strength instead of inertia. If one of us is weaker or has less traction, our strength or feet will give out before the other. Thus, the strain on the rope is the least of four possible forces: My strength, his strength, my traction, his traction. This can NEVER exert more force than I can exert alone.
The real answer to your question is this: two people pulling on a rope will always exert the tension that the weaker/smaller person can manage. If you're pulling against a tree, it's all just one person. Adding another person can ONLY exert the same or less tension, never more. This part I'm really sure about.
As far as head-on collisions go, I'm less sure, but I think the answers so far are wrong. Kinetic energy is a square function on speed. It's what does damage and must be dissipated. The formula is 1/2mv2. A 2000lb car hitting a wall at 50mph has to dissipate (0.5 * 2000) * (50 ^ 2) = 250,000 units of energy. Two cars striking each other at 50mph can be treated as one stationary and one moving at 100mph, so the force involved becomes (0.5 * 2000) * (100 ^ 2) = 10,000,000 units of energy.... 40 times more damage. Even if you get a perfect division of damage, that's still 20 times as much per car.
Note also that cars are usually unequal in size, and the smaller car will almost always take the brunt of the damage.
This is why head-on crashes are so wildly lethal compared to rear-enders.
posted by Malor at 4:36 PM on May 18, 2007
Malor, you are confusing energy and force in your analysis. Further, you have a couple of arithmetic errors that gave you to the incorrect 40x factor. (By the way, there is a mix of mph and pounds, so I'll just use pseudo energy units instead of joules to stick with your example.)
For the first case of a car into a wall, it has a total of 2,500,000 energy units, not 250,000 units. It dissipates that energy when it hits the wall.
For the second case, two cars hitting each other at 50 mph, each has 2,500,000 units of energy and assuming they are the same mass, they each dissipate 2,500,000 units of energy, the same as for a stationary wall and one car. The total energy dissipated is 5,000,000 units. You forgot that there are two cars dissipating energy in this case. The deceleration forces on each car are the same as for the car/wall case.
For your third case, one car hitting a stationary car at 100 mph, the total energy is 10,000,000 units. When they hit, again assuming equal mass, they each dissipate 2,500,000 units, the same as previously, for a total of 5,000,000 units. So what happens to the extra 5,000,000 units of energy? It remains kinetic energy as both cars continue moving in the same direction as the 100 mph car. This kinetic energy exactly equals what is left over from the crash. If you assume the cars stick together and work out the numbers, they continue moving together at 50 mph, which makes sense if you consider the equivalence of stationary or moving reference frames. Two cars hitting each other at 50 mph and remaining stationary is equivalent to one car hitting a stationary car at 100 mph but afterwards moving at together at 50 mph. The part you are forgetting in your case of hitting a stationary car is that the car does not remain stationary, unlike a wall. (This assumes that no braking is involved in all cases.)
So it turns out that for all three of those cases, the energy dissipated in the crash by each car is exactly the same -- 2,500,000 units. The deceleration forces felt by the passengers are also exactly the same in each case.
posted by JackFlash at 5:45 PM on May 18, 2007
For the first case of a car into a wall, it has a total of 2,500,000 energy units, not 250,000 units. It dissipates that energy when it hits the wall.
For the second case, two cars hitting each other at 50 mph, each has 2,500,000 units of energy and assuming they are the same mass, they each dissipate 2,500,000 units of energy, the same as for a stationary wall and one car. The total energy dissipated is 5,000,000 units. You forgot that there are two cars dissipating energy in this case. The deceleration forces on each car are the same as for the car/wall case.
For your third case, one car hitting a stationary car at 100 mph, the total energy is 10,000,000 units. When they hit, again assuming equal mass, they each dissipate 2,500,000 units, the same as previously, for a total of 5,000,000 units. So what happens to the extra 5,000,000 units of energy? It remains kinetic energy as both cars continue moving in the same direction as the 100 mph car. This kinetic energy exactly equals what is left over from the crash. If you assume the cars stick together and work out the numbers, they continue moving together at 50 mph, which makes sense if you consider the equivalence of stationary or moving reference frames. Two cars hitting each other at 50 mph and remaining stationary is equivalent to one car hitting a stationary car at 100 mph but afterwards moving at together at 50 mph. The part you are forgetting in your case of hitting a stationary car is that the car does not remain stationary, unlike a wall. (This assumes that no braking is involved in all cases.)
So it turns out that for all three of those cases, the energy dissipated in the crash by each car is exactly the same -- 2,500,000 units. The deceleration forces felt by the passengers are also exactly the same in each case.
posted by JackFlash at 5:45 PM on May 18, 2007
This thread is closed to new comments.
This is the same question as whether there is more force involved in a collision between a car going 50 mph and a wall or two cars head on, each going 50 mph. It's the same amount of force.
This is explained well in Epstein's book Thinking Physics, which is very reader friendly.
posted by BigSky at 8:15 AM on May 18, 2007