The best way in which a man can pull a train

Question

I have chosen this example as a paradoxical limit, but my question really regards the optimization, distribution and outcome of force when you try to pull a weight: how can you optimize your effort/energies spent?

In order to pull a weight you need a rope and a firm grip on the ground. Suppose this last problem is solved 100% (like in the second picture) preventing the feet from slipping. The rope is perfectly horizontal, suppose the man's weight is 100kg. The angle may be between 0° and 45, at your choice. Now, you can decide you want to keep:

1. your leg(s) firm and stretched and apply your force to the rope pulling it like in this picture.

   Suppose you are exerting a force (on the train) of 300 N, how much force is going to the train and how much is going to the ground ? and, how much energy is being wasted? Is it possible to improve the percentage of energy which is actually transmitted to the train? (without tricks or external factors)

2. You can stand (like in this picture but holding a rope in both hands) with bent knees and keep your arms stretched. Then you stretch your legs and apply a push on the ground, in this case you are exerting 300N directly on the ground and indirectly on the train , as the displacement of your trunk will pull the train.

   Of course this position is most favourable a): because the legs are usually stronger then the arms [probably, force is more than double here] b): because the angle is near 0°. That's why this man can pull a train!:

   

   but all other factors being equal, how does push-on-the ground affect the distribution of forces as compared to pull-on-the rope?

   Which technique is more effective/favourable, which makes the train move with less energy spent? Is the outcome exactly the same as in the previous case? if not, why? what physical principle makes the difference? Can you draw two sketches and show how forces interact and are distributed?

EDIT:

Supposing you are exerting a force... how much of it is going to the train and how much to the ground?

This is difficult to answer since you seem to have a major misconception about forces. The best answer is to go back your freshman, or even high school, physics book and read the section on forces...Olin Lathrop

I am doing my homework, but this site is all about clarifying, explaining misconceptions. I regularly help students (like: here to dispel theirs and, lo their misconceptions do not seem major to me, nor trivial, everyone got his own, sometime or other!.

Answer 1

This is difficult to answer, since you seem to have a major misconception about forces. The best answer is to go back your freshman, or even high school, physics book and read the section on forces. However, briefly:

The pulling force isn't somehow split between the train and the ground. The rope will pull with the same force on whatever is holding each end. If the rope is pulling on the train with 300 N, then the man is pulling on the rope with 300 N, and the man must be pushing on the ground with a force of 300 N laterally. The vertical component of the force the man is pushing on the ground with will be his weight plus some fraction of the pulling force depending on the angle of the rope.

If the rope is horizontal, then no part of the pulling force results in vertical force on the ground. In this case, which is pretty close to your bottom picture, the horizontal component of the force on the ground is equal and opposite of the force the rope is being pulled with, and the vertical component is simply the man's weight.

Answer 2

Well, one sub-question at a time.

Suppose you are exerting a force of 300 N, how much KE is going to the train and how much is wasted to the ground?

Applying a force to an object doesn't necessarily involve energy expenditure. If there is a weight lying on the ground, then the weight is exerting force on the ground, but the weight isn't expending any energy by doing so.

Energy is spent by pushing an object in the direction that it is moving, and the amount of energy spent is equal to the force exerted on the object times the distance that it moves. (If the force and the motion are in slightly different directions, ignore the part of the force that's perpendicular to the direction of movement, and just pay attention to the part of the force that's going in the same direction.)

Since the ground doesn't move when you push on it, no energy is lost to the ground. All of the energy goes into the train.

All other factors being equal, how does push-on-the ground affect the distribution of force/KE as compared to pull-on-the rope?

Well, you're asking about two different things: the distribution of force, and the distribution of energy. As I said above, all energy goes to the train, not to the ground. How about the force?

It's probably safe to assume that while you're pulling a train, you're not accelerating at a significant rate (unless you're extremely strong or the train is extremely light). By one of Newton's laws, the net force exerted on an object is equal to its mass times its acceleration. Since you're not accelerating at a significant rate, the net force exerted on you must be approximately zero.

Of course, you are, in fact, exerting a lot of force, both on the train and on the ground. The reason that these forces can add up to zero is that the two forces are in opposite directions: if you're pulling eastward on the train, you must be pushing westward on the ground.

This explains why, if the forces suddenly become unequal (perhaps because the rope breaks, or your feet slip, or the train starts moving under its own power), you suddenly accelerate!

Which technique is more effective/favourable, which makes the train moves with less energy spent?

Since all of the energy goes into the train, which technique you use is pretty much irrelevant to energy expenditure. The kinetic energy of the train is determined completely by how fast it is moving. This means that, if no energy is lost to friction or other effects, then the amount of energy you must expend is determined completely by the speed of the train. (So if you want to move the train using as little energy as possible, the solution is to move it slowly!)

Now, there's kind of an implicit question here:

Which technique exerts the greatest amount of force on the train?

You'll definitely want to use your legs instead of your arms. In theory, the maximum amount of force you can exert on the train is equal to the maximum amount of force your legs are capable of producing.

There's a difficulty, here, though, which is that if a force is borne by a chain of components, all of the components are subjected to all the force. If you have a rope in each hand and you're pulling with a force of 300 N, then your legs are exerting a force of 300 N and your arms are exerting a force of 300 N, and your torso is exerting a force of 300 N as well. Your arms will handle it all right, since they aren't moving; human muscles are better at holding still than they are at pulling something.

Still, it takes some effort just to hold the ropes with your arms. So it'll be best not to use your arms and wear a harness.

And there's one final problem. Suppose you're standing perfectly straight, and you pull on the rope while bracing with your legs. Since these two forces are applied to different locations on your body, the effect is that you'll rotate and fall forwards! In order to exert a large amount of force without falling over, you'll need to lean way back, exactly as the people in your pictures are doing.

Answer 3

You have two subproblems here. One is the motion of the train and the other is how to grab the rope.

Let's focus on the train. It has a mass $m$ and there is some friction with the ground (assumed low because it is on rolls) $\mu N$ , being $ \mu << 1$. You apply a force $F$, assumed modulo constant. So, the horizontal acceleration is:

$$\sum F = m a$$ $$F \cos \theta - \mu m g\ \sin \theta = m a$$

You want to find the angle that maximises that expression. That, of course, depends on the numerical values. Doing the math, it comes to be:

$$ \theta = 2 \left[\arctan \left( {\sqrt{F^2 + \mu^2 m^2 g^2} - F}\over{\mu m g} \right)\right]$$

That is a mess of an equation.

Here you are splitting your force into a component lifting the train, and thus reducing friction (setting $\mu=0$ makes $\theta=0$), and another component actually moving it (if you were Superman strong you could lift the whole train upwards and then move it horizontally almost effortlessly). Note that the train doesn't know how you are holding yourself, it just sees the angle and the force of the pull.

Now we know the optimal angle (and I will tell you, it is very small), you just need to find a way to exert it. As you said, you do need a good grip both on the ground and on the rope, and use your muscles to do lever. The optimal way would depend on the strength of the person, but legs, or a combination of legs and arms, are probably the ones capable of getting most force. Note that if you are going to do this on a regular basis (professional train puller) you would need to find a way of doing it without crushing your back.

I have left out a detail: there is a difference between static and dynamic friction. When two objects are statically in contact with each other, the friction is bigger than when they are moving, so it will be more difficult to get the train moving that to keep it going. A clever combination of arms and legs may help you get that extra power at the beginning of the experiment.

Disclaimer: do not attempt this at home or the station.

Answer 4

We have to define "most efficient". I will define it as the technique that allows the greatest force to be applied to moving the train along the track. Then we have three considerations relative to the position of the person pulling: torque on the body, transmission of force to the rope, and force on the train.

Torque
When you pull a train with a rope, and the rope is higher than the support of your feet, then you will experience a torque that tries to pull you over. You can lower the point from which you are pulling to reduce the torque - at the same time moving your center of gravity further away from the support point (horizontally), thus increasing the torque that keeps you from tipping forward. This is why the strong man in picture 2 is practically horizontal.

Transmission
Notice that the man in picture 2 is practically horizontal: this means that the force of his legs is transmitted to the shoulders through compression of the spine, rather than by tension in the back muscles. In the latter case, you probably know from experience (and could demonstrate with a simple diagram) that tremendous forces appear on the back muscles when your back is not straight when you lift - up to 7x the weight lifted (because the muscles run close to the spine, so the lever is very short). The man in picture 1 creates a torque along his back - and the back muscles have to provide the counter torque.

Angle
Third - you want the rope to be perfectly horizontal: in that way, the full tension is applied directly in the direction that the train can move. Now a real rope has finite mass, so it will hang in a catenary. Now what is the best strategy is for hand height when the rope is not straight? Consider three options:

1. Hands at same height as point of attachment on train
2. Hands lower: rope horizontal at hands
3. Hands higher: rope horizontal at train

Note that in both case (1) and (3), your hands are in effect "carrying" (part of) the weight of the rope. Now let's assume that the angle between rope and horizontal is $\theta$ on each side for case 1, then with weight of rope $W$ and tension $T$, we know that

$$W = 2 T \sin \theta$$

because the weight has to be carried by the sum of the vertical components of tension. In this case, the horizontal component of force is

$$F = T \cos \theta\\ = T \sqrt{1- \left(\frac{W}{2 T}\right)^2}$$

When one side is horizontal, we get angle $\phi$ where

$$W = T \sin \phi$$

Balance of forces says that the horizontal forces on the rope must be equal on both sides, so if the side of the rope where you pull is perfectly horizontal, then all your force will transmit as horizontal force to the train - the train won't be bothered at all by "carrying" the vertical component of the weight of the rope.

So that's my "final answer": the most efficient technique is the one that

a) Has your hands low (to minimize the torque that tries to pull your body over)
b) Ensures that your center of gravity is far behind your support (to maximize the counter torque)
c) Ensures that your back is under direct compression - so the back muscles don't have to work
d) Has the rope horizontal at your hands - if the rope is heavy, this might involve fixing the rope at the train at a higher point

In all this, making sure that all components are stiff (rope doesn't stretch, feet don't slip, foot support has no "give") so none of your energy is stored in them (even if they would "give it back" when you release the force - that's too late for the train...). Remember that

$$E = \frac12 k x^2\\ F = k x \implies \\ E = \frac{F^2}{2k}$$

So when $k$ is as big as possible, the least energy is stored where you don't want it...

Looking again at your picture 2 - it is clear that the man in the picture "knows his physics"; even if he might have come to the above conclusion empirically, you see almost all the elements of my solution in his posture (except that in order to use his mouth, he has to arch his back somewhat. I think he could pull more if he used his hands. Just sayin').

Just to clarify the point of what force, and what torque, is applied at what point in the system, here is a little sketch (specifically addressing the force on the front foot). Assuming that the person is "pulling as hard as possible without falling over", we can draw:

Here the red arrows represent the tension in the rope T, and the weight of the person W. With all the weight on the front foot, we can easily write down the balance of forces (vertical force balance: green force = -W'; horizontal force balance: -T'). Balance of torques similarly follows from

$$W\cdot d = T \cdot h$$

The important thing to note is that it is not necessary for the resultant of the two green forces to point along the leg. Instead, it points to the point where $W$ and $T$ appear to meet. This is how you get the balance of torques. And when you do the analysis this way you will get lateral force on tracks == force of rope.

Answer 5

There are two misconseptions that lay at the bottom of your question:

• that force is being wasted by being applied to the wrong object (or that there is a wrong object for the force to be applied to).
• that the best way to do something is necesarily the one that wastes the least energy.

I will start by looking at the case when a weight hanging from a pulley is pulling the train. Here no force is being exerted on the ground by the weight, (meaning that the ground exerts no force on the weight, that the weight accelerates). So some energy is in fact being "wasted" in this case. However, we cannot have this setup without the weight accelerating, because we are exchanging gravitational potential for kinetic energy, and the weight needs to move. So, it's not correct to say that energy is wasted. We can limit this waste by different schemes: essentially, we need the train to move further than the weight. However, this reduces the amount of force a given weight will apply to the train and so this is probably a worse way to move the train. (Surely we want to do the opposite, have the weight move further than the train. This will make more energy go to the weight, but a given weight will allow us to exert more force)

Also, it's not correct to say that no force is being applied to the ground since the rope exerts a force on the pulley, which is why the force the weight exerts on the rope can be different (it's rotated by 90 degrees) from the force the rope exerts on the train. The pulley is not acellerating so the ground is exerting a force on the pulley. Therefore there are forces being exerted on the ground.

In the case of a person pulling on a train, we are no longer looking at exchanging gravitational potential for kinetic energy. Instead we are exchanging chemical energy for kinetic energy.

We are not accelerating the earth to a noticable degree, hence no energy goes into the ground. If we are not accelerating, we are not getting any kinetic energy. In this case we are pulling the train towards us, rather than pulling the train along with us. We will lose some energy to heat (the same happens in the case of the weight, but not to the same degree).

If we are not acellerating, the forces balance. If we are pulling the train along with us we will exert more force on the ground than on the train as long as we are accelerating, and when we are moving at a constant speed the forces will balance. No force can be said to be wasted, because the force we exert on the ground is determined by the acceleration that we are achieving.

This does not mean that there are no good or bad ways to pull the train. But these boils down to: where the forces are coming from. We are quite good at pushing with our legs, so the method used in the picture is efficient.

For people that are not so strong ,we might get more force by grabbing the rope and leaning away from the train. In this case we are using the body as a lever with the legs as the fulcrum. If the parts of the system is acelerating at the same speed we are still exerting the same forces, but now part of these forces are exerted by our skeleton resistance to compression.

TL;DR: "All other factors beeing equal" reduces too much the problem. The best way to pull a train is about leveraging the parts of the body in the bestway which is able to produce the needed force. - If you posit that, this does not matter. Then the answer is that the way in which you pull the train does not matter.

Answer 6

I believe that a simple drawing can clarify some of the misconceptions. In dynamics, you must analize the forces acting on the individual objects separately. I have drawn the forces on the train and on the person:

You can see that if the person is not moving the floor must do a force equal to the tension. The internal forces inside the person do not matter here. they are internal. What is the best angle for the rope? one which allows the person to make the largest internal force. Also important, the angle on the rope will change the horizontal component of the tension to make it smaller, but it will also will diminish the friction force on the train because it will reduce the normal force N. What is the perfect angle depends on the coeficcient of friction, that is, if there is an angle such that the net horizontal force on the train, $F_x=T_x-F_r$, is maximal. That is, find the angle $alpha$ that makes maximal the horizontal force on the train:

$F_x=T(\cos \alpha + \sin \alpha)-\mu m_{train}g$