newton's third law

Experimental Rope Logarithms

rallain — Tue, 01 Jun 2010 16:28:17 +0000

Experimental Rope Logarithms

Sue from Math Mama Writes... sent me an email about wrapping a rope around a pole. In that post, Sue thinks about rope looped around a post. When you wrap a rope around a post, the friction between the rope and the post can help you hold something (like a horse) that is much stronger than you.

The first case she thinks about is using several posts. What if you wrap a rope around one post and pull? What if you use 2, or three posts? The idea is that if one posts 'multiplies' the force by 10, two posts would have an effect of multiplying by 100 and so forth. That seems reasonable.

The next thing, what about one post with just half a turn? What about multiple turns around the same post? This is what I looked at first since it is the easiest (you don't need multiples of the same post).

Setup

Here is a basic diagram of my experimental setup.

Where these force probes are Vernier probes using Logger Pro. One of the probes is clamped to the table and the other held by me. First, let me test one probe connected to another by a string (no wrapping). Here is a plot of force 1 vs. force 2:

Fitting a linear function, I should get a slope of 1, but I get 1.152. This means that the two probes are not perfectly calibrated (I guess I should have checked this before hand). Now what about half of a turn around the post? (I calibrated the probes - they match a little better now)

So, for 1/2 turn, there is a force difference with the relationship:

Now, what about one complete turn?

This gives a relationship of:

I tried two loops, but I just could get good data. How about multiple posts? I already have data for one loop around one post. What about two posts?

Can I get three posts? Yes, I can.

I am not going to try 4 posts. I know my force probes would be unhappy.

Back to the theory

For multiple posts, the idea is that if 1 posts multiplies the force by a factor of a, then two posts should have a total multiplicative factor of a² and three would have a factor of a³. In general, if one posts produces a force multiplication of a and the number of posts is n, then:

Let me take my three whole turn data points. I already have the ratio of F₁ to F₂ (the slope of the graphs). So, re-writing the above expression, I get:

If I plot the natural log of the ratio of forces vs. the number of wraps, I should get a constant. Here is that plot.

So, the natural log of a is the slope of this line - .8825. This would give a value for a as 2.41. This should be the same as the slope for one turn, but it is not. Oh well, it isn't too far off.

What about n = 1/2? Does that work? The natural log of the slope for half a turn is 0.385 which is fairly close to half the value of ln(a).

The physics

Physically, how does this work? The normal model for friction says that the frictional force is proportional only to the force the two surfaces are pushing against each other. Not sure if that works here.

Energy? This seems like it is cheating. Do you get more energy than you put in? No, because you don't even put in any energy if the rope doesn't move. This is not a simple machine. What about forces? This seems like it makes more force and thus violates Newton's third law. Well, it doesn't. If I pull with 1 Newton and a horse on the other end pulls with 10 Newtons, these are not "equal and opposite forces". You could think of it this way. The horse pulls on the rope with a magnitude of 10 Newtons. So, looking at the post, the horse pulls one way with 10 N, me the other way with 1 N. This leaves 9 Newtons. The ground exerts 9 Newton force on the post. If the horse were strong enough, it could pull the post out of the ground.

rallain Tue, 06/01/2010 - 12:28

MythBusters' energy explanation

rallain — Thu, 06 May 2010 16:32:57 +0000

MythBusters' energy explanation

I already mentioned the MythBusters' crashing two cars episode where they correctly doubled the speed of a pendulum type object. Overall, this was a very visual (although expensive) demo. There was one part that left a sour taste in my mouth - the final explanation from the narrator. First, they showed this.

And then they had an explanation that went something very similar to to this (after restating what the sign above said)

"Although the two-car crash doubles the speed, the energy the crash is transferred to twice the mass resulting in a crash that looks like just one car hitting a wall at 50 mph."

Here is the graphic that went with that.

I had to re-listen to this narration a couple of times because something seemed not quite right. First, let me comment on the last diagram. Really, maybe it would have been better to leave this off. It doesn't really add any useful explanation other than to point out that the kinetic energy is dependent on the square of the velocity. And what about the narrative? I think what bothered me is that they said the two-car crash doubles the speed, but what they probably should have said is "the two car crash doubles the kinetic energy and this energy is spread out over 2 cars." Using their statement, you would say "oh, double speed means 4 times as much energy."

Here is the real question: what concept are you trying to get across? Newton's third law? The idea of kinetic energy? Conservation of momentum? I would just pick one and stick with it. Otherwise, you are kinda implying that "action and reaction" have something to do with kinetic energy. Oh, have I ever mentioned how much I hate "action and reaction" explanation of Newton's 3rd law? Action? Reaction? How about this for Newton's third law:

Forces are an interaction between two objects. Forces come in pairs. Or, if you must: For every force there is an equal and opposite force.

Oh sure - you can make the action reaction thing work, but it can also cause problems. Either way, I would suggest sticking with the energy explanation. Here, I want to help. I am going to give an energy explanation that the MythBusters could use and a force explanation for the two-car collision.

Two-car crash, energy explanation

Why are two cars crashing into each other not the same as one car going into a wall at twice the speed?

Explanation: In terms of energy, the energy of motion is called kinetic energy. Kinetic energy depends on the square of the velocity. This means a car moving at twice the speed with have 4 times the kinetic energy.

So 10 + 10 is not the same as 40.

Two-car crash, force explanation

This is a little more complex, but I will try to make it simple. First, two key points:

Forces are an interaction between two objects. Object 1 pushes on object 2 the same as object 2 pushes on object 1 (same interaction).
A force on an object changes the object's momentum where momentum is mass times velocity.

Suppose a car crashes into a wall with a velocity v. While it is interacting with the wall, the wall exerts a force (F) on the car and the car exert a force F on the wall.

Where the force the wall exerts on the car and the car on the wall have the same magnitude. Now, what if I replace the wall with another identical car traveling at the same speed?

Since the initial momentums are the same and the forces are same, the effects are the same on the two cars. So, two cars are the same as one car into a wall. If I now double the speed, I will have different initial momentum, so it will not be the same.

Other stuff

I actually forgot that I have talked about the MythBusters colliding two cars before. I wrote this post the first time they did this myth. That is a little more detail than I have here. Also, a similar thing came up when the MythBusters tried to pull two phone books apart. Here is my discussion of forces in that situation.

One more thing. I would like to emphasize how awesome this demonstration was. You hear people discussing things just like this all the time, but no one actually does it. For many physicists, the actual experiment doesn't mean much. However, to many people this experiment is important. They just need to fix their final explanation (call me next time and I will be glad to help).

rallain Thu, 05/06/2010 - 12:32

Tags

I can't wrap my brain around how Newton's 3rd law logically proves Jamie's hypothesis wrong.

Doubling the mass (action) results in what type of reaction? (Sorry for using those terms.)

@C. Felix,

I am not sure I proved he was wrong, maybe instead I just showed that two cars colliding is the same as 1 into a wall at the same speed. If that works, then I guess it follows that double the speed would not be the same.

I started to think about this in terms of reference frames and this really makes me believe that energy is the wrong way to explain this problem to the average person.

If you look at the proper reference frame of the orange car, because the event is happening to it. The person in the car would see a yellow car with E=40 hitting them or a wall with E=40 hitting them.

Am I thinking about the reference frame incorrectly?

Got this idea from Al Guenther: to teach newton's third, cut a plastic pear in half and put a force vector in each half. HA! Forces come in pears. Thought you might like that if you hadn't seen it before.

@Matt,

Great idea. How about this modification. Make a plastic arrow for a vector. Take a real pear and embed this plastic arrow inside. Then when you cut it open in class you can say vectors come in pears.

Ummm... Isn't the velocity in the kinetic energy equation suppose to the relative velocity between two objects, not relative to a fixed point in space or the point of the collision. Isn't the relative velocity between two cars traveling at 50 mph the same as one car travelling toward a wall at 100 mph.

Me thinks a few mistakes were made.

@Evan,

Kinetic energy is not independent of reference frame. If you move to the reference frame of one of the cars, you would see a different total KE before the collision. What you would agree on is the changes in energy during the collision.

My statement was that the v in kinetic energy equation is based on relative velocity. Relative velocity is just that, Relative. Using the earth as a universal reference is a very common mistake. Each object's kinetic energy relative to the earth is irrelevant. We only care about the kinetic energy relative to the objects in the experiment. We're forced to choose one of the objects as a frame of reference. With the wall the choice is obvious. We have to use the wall as a frame of reference. With two cars travelling towards each other we have to pick one of the cars as our frame of reference and consider that car stationary. Thus, the relative velocity is the same between the two experiments. If the mass of each car is the same and the relative velocity is the same, then the kinetic energy is the same. It's like saying 1/2m(100)^2 = 1/2m(100)^2.

Basically, I'm saying your explanation using kinetic energy is wrong because you calculated kinetic energy relative to the earth and not relative the objects in the experiment.

I'm also saying that a car hitting the wall at 100 mph hit that wall with the same kinetic energy as two cars hitting head on at 50 mph. Maybe not relative to the earth, but relative to one of the cars.

Because kinetic energy is dependent on frame of reference, it's absolutely critical that you choose a valid frame of reference.

But nobody has addressed deformation, which Adam and Jamie used *as their measurement* of impact (along with G-forces i.e. deceleration). Of course the clay between the weights on two moving hammers squished only as much as the clay in one hammer did... there was twice as much total clay to absorb the impact. Likewise, when one moving car hits the wall, the wall shows no deformation - only the car is squished. When two cars hit with the same relative velocity as in car vs wall, *both* cars get deformed - so each individual car deforms less.

@Pteryxx

I agree 100%. The difference in damage had more to do with elasticity then it did with force. Elasticity was not constant as the wall was far more elastic then the cars. The amount of force due to relative kinetic energy was the same for the 2 cars traveling at 50 and the one car traveling at 100. The difference was the one car traveling at 100 was the only object in the collision that could absorb energy through deformation. Thus it was damaged far more. The energy had to go somewhere.

I agree with Rhett that the action-is-minus-reaction is a particularly awkward way to formulate that law of motion.

Whatever Newton's reasons were to formulate the third law in that form, a far superior form is present in the Principia. In the Principia, the three laws are followed by corrollaries, and the fourth corrolary states the third law in dynamical form:

"The common center of mass of two or more objects does not alter its state of motion or rest by the actions of the bodies among themselves."

In retrospect we would have been in a far better situation if Newton would have asserted the third law in that dynamical form. (In fact many textbooks do present that dynamical form as the third law.)

I can't resist fantasizing what things would be like if that dynamical form would be the canonical form. Then that whacky "action-is-minus-reaction" thing would not be there to confuse the guys who write the Mythbuster narrator's script.

Going back once more to the dynamical form:
"The common center of mass of two or more objects does not alter its state of motion or rest by the actions of the bodies among themselves."

This form emphasizes the mutualness, and it is uncommitted as to what force is. It's a statement about motion, not a statement about force. For better education that "action-is-minus-reaction" thing ought to be phased out, and replaced with the dynamical form.

You write: "The two-car crash doubles the kinetic energy and this energy is spread out over 2 cars." I think this is the clearest explanation for why 2 cars hitting each other @ 50 mph might do the same amount of damage (to Car #1) as would Car #1 hitting a stationary wall @ 50 mph.

For me, the tricky part is convincing myself why, in the wall case, we don't need to worry about some of that initial KE getting transferred to the wall instead of going into damaging Car #1. One model that seems to work is if you think of the wall as a spring with spring constant k >> F_max/L, where F_max is the max force w/ which the car pushes back while being crushed, and where L is a characteristic length scale in the process of crushing the car (F_max*L ~ energy that goes into damaging Car #1 during crash).

Then, in the two-car collision, where each car has initial speed v, all the initial KE is shared equally between Car #1 and Car #2 during the crash:

2 * 1/2 m v^2 = 2 * (Energy that goes into damaging Car #1),

=> Energy that goes into damaging Car #1 = 1/2 m v^2.

But in the Car#1-hitting-wall collision, where Car #1 again has initial speed v, the initial KE is shared unequally between Car #1 and wall during the crash:

1/2 m v^2 = E_wall + (Energy that goes into damaging Car #1)

Now, IF the wall acts like an ideal spring w/ k >> F_max/L, and if we can assume the wall/spring ends up compressed by a (tiny) distance x, which is how far it had to compress to push back on the car with the required crushing force F_max, we have

E_wall = 1/2 k x^2
= 1/2 k (F_max/k)^2
= F_max^2 / (2k)
= F_max*L / (2 kappa)
= O(1/kappa)

where kappa = kL/F_max >> 1 is the dimensionless large parameter in our approximation. And recall that, by definition,

Energy that goes into damaging Car#1 ~ F_max*L
= O(1);

that is, in the limit kappa->infinity, E_wall is negligible in comparison with the energy that goes into damaging Car #1. Going back to our energy-transfer eqn for car-hitting-wall, and taking the 1st term on the RHS to zero, we thus have

Energy that goes into damaging Car #1 ~= 1/2 m v^2,

which is the same amount of energy that went into damaging Car #1 in the two-car collision when *both* cars had initial speed v.

But is it accurate to treat the wall like a large-k spring in this manner? I don't know!

Re the comments about reference frames: Remember that the COM reference frames of the cars are NONinertial. (As the cars get crushed, they decelerate.) So the physics can't be expected to work properly in those frames. From his own COM frame, Car #1 thinks there's twice as much energy to go around than there actually is (1/2 m*(2v)^2 = 2 mv^2 instead of 2* 1/2 mv^2 = mv^2), and he'll therefore be confused about why there isn't more damage to the two vehicles.

@Carol,

I am not ignoring your comments, I am just taking time to think of a response. Sorry for the delay.

1. One almost always has to make simplifying assumptions when "modeling".

2. One must "model" in order to keep the math reasonable.

3. Sometimes the real world answer is lost in the over simplifications of the models.

4. In this case, I think the key is to ignore the crumple factor, focus on the choosing the inertial reference frame for each case and let the kinetic energy tell the tale.

They should have made a wall traveling at 50mph and a car traveling at 50mph hit each other.

The energy argument is wrong, by the way. The momentum one is correct. The problem with the energy argument is that you have all assumed that there is such a thing as "absolute kinetic energy". Unfortunatley, that does not exist. Only changes in energy are meaningful - this is 1st year physics. I'll give you an example. Say you are in a train moving at speed v (not accelerating) and the windows are blacked out. You would say that you have zero kinetic energy. Now, and observer on the ground says that you have 1/2 m v^2 kinetic energy. Who is right? It's meaningless since the notion of absolute KE is just as meaningless as absolute potential energy (PE). If you considered changes in energy, then things would be more meaningful.

Finally, this is an in-elastic collision, so what you can apply is conservation of momentum. Mechanical energy is not conserved in this collision.

Summary: Newton's third law, and the cons. of momentum explain this result. That is all you need. This whole issue of different frames of reference (cars frame, ground frame) is a red herring.

I have a queston if there were to eople in the wouds how would last longer the skiny one or the fat one ??????

Basics: Free Body Diagrams

rallain — Tue, 30 Sep 2008 09:55:52 +0000

Basics: Free Body Diagrams

**Pre Reqs:** [Intro to Forces](http://scienceblogs.com/dotphysics/2008/09/basics-what-is-a-force.php), [Vectors](http://scienceblogs.com/dotphysics/2008/09/basics-vectors-and-vector-ad…)

Hopefully now you have an idea of what a force is and what it isn't. What do you do with them? The useful thing to do with forces is to determine the total force acting on an object. At the beginning of the introductory physics course, you will likely look at cases where the total force is the zero vector. This is called equilibrium. Even if you are looking at cases where the forces don't add up to the zero vector (I say that instead of just "zero" to remind you that the total force is still a vector). Physicists like to represent forces on an object by drawing a Free Body Diagram. This is simply a representation of an object and a graphical representation of all the forces acting on that object.

Simply put, in a free body diagram, all the forces acting on the given object are represented as arrows. Let me start with a simple case, a box sitting on a table.

There are only two forces acting on this box (essentially). The table pushing up on the box and the gravitational force of the Earth pulling down on the box. The free body diagram for this box would look like this:

![Screenshot 02](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/screensho…)

Note that I have used proper vector notation on my force vectors. The force of the table pushing up on the box is labeled as N because these types of forces are called "normal forces" - maybe I will talk about that more later. Another useful thing is to include the labels "table-box" and "Earth-box" to indicate that each force is an interaction between two objects. A final note on this first example is the length of the arrows representing the forces. They are the same length indicating that they are the same magnitude of force. Since these forces are the same magnitude, but different directions, the total force on this box is zero vector.

One final note. I put a dot in the middle of the box. That is where I started all the forces from. It doesn't *really* matter where the force is, but this might make it a little easier.

**A more complicated Example**

Now suppose that I have two blocks, block A sitting on top of block B that is sitting on a table. In this case, I can draw a free body diagram for both block A and block B:

![Screenshot 03](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/screensho…)

Here you can see the advantage of extreme-labeling of the forces. I know it is a pain to keep writing "the force of block B on block A" but you can see something. All the forces on block A end in "block A" and all the forces on B end in "B". This notation can really help you keep track of which forces are on which block. A common mistake is to include the gravitational force of the Earth pulling on block A on the block B diagram. The thinking is that gravity is pulling block A down onto block B - which is true. However, the gravitational interaction is between Earth and A and Earth and B.

**Newton's 3rd Law**

Here you might notice something else. I have left the force of B on A and A on B as both red vectors and they are both the same length. This is a fundamental property of forces. If Newton were around today, he would state this property as:

*Forces come in pairs. For every force there is another force on a different object that has the same magnitude, but opposite direction*.

So, in a sense, the two forces are the same thing. They are a representation of the interaction between block A and B.

Finally, notice that the force of the table pushing on block B is much larger than the other forces. Why is this? Well, block B also has gravity pulling down (the Earth pulling on block B) and block A is pushing down. In order to make the total force zero vector, the table has to push up with a greater magnitude. Notice that when I have two forces acting on the same object in the same direction, I can just put the forces in a line. This is useful in that it looks like one force of a longer length.

**One more example**

Here is a slightly more complicated example for a block sitting at rest on an inclined plane.

![Screenshot 05](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/screensho…)

In this case, there are three forces on the block. I imagine that everyone is ok with the gravitational force of the Earth pulling on block A - right? Here you see why the force of the plane pushing on the block is called the normal force. It is because that force is perpendicular to the surface (normal). There is another force between the block and the plane that is NOT normal. It is the friction force and it is parallel to the surface.

**Adding Vectors on the inclined plane**

Suppose you want to calculate the frictional force or something using the assumption that the forces all add up to zero vector. Here you can use a small trick. Since N and the friction force are perpendicular, you can put the x-y axis tilted so that those two forces are ONLY in x or y direction:

![Screenshot 06](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/screensho…)

This would give the equation for the x-direction as (I am going to call the normal force N, friction F and the gravitational force G):

![Screenshot 08](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/screensho…)

Note that these are not vectors, here the symbols stand for the magnitudes of the vectors. Also, I will leave it as a geometry exercise for you to show that the angle between the gravitational force and the y-axis is the same as the angle of the incline.

**Identifying Forces**

I understand it can be difficult to determine what forces are acting on an object. All forces that you will see can be in one of two groups:

Long range forces: These are forces between two objects where the objects do not have to touch (thus long range). Really, there are only two interactions you will see that do this. The gravitational interaction (between objects with mass) and the electromagnetic interaction between things with electric charges.
Contact forces: Secretly, there is no such thing as contact forces (see this post) but we will pretend for simplicity. Contact forces are from things that are touching that object. Examples: friction, normal force, tension from a rope, hand pushing on something, air resistance.
When you are identifying forces, first look for long range. In the first semester physics this will probably JUST be gravity. All the other forces on that object are from things touching it.

In your journey to create free body diagrams, I encourage you to properly label your forces. This will help you find forces that really shouldn't be there.

rallain Tue, 09/30/2008 - 05:55