20.12.2013

## Man pulling back,how to get a girl to kiss you under the mistletoe,what to do if you want your ex boyfriend back song,i got back with my ex but its not the same youtube - Plans On 2016

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Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. 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 ? You can stand (like in this picture but holding a rope in both hands) with bent knees and keep your arms stretched.
If the rope is horizontal, then no part of the pulling force results in vertical force on the ground.
Olin although you are right, you can say exactly the same thing without mentioning that it is freshman physics. If someone were interested, in the analysis below I try to confute the 'misconception' that a pull on the rope of 300N will produce a lateral pull on the ground of the same amount.
I have tried to do my homework and I am posting an answer only to show how I expected the issue to be explained to a layman. Although the title has a wide scope, my post (from the original version), was mainly asking about the distribution of force(s), and, secondly, on the best way to pull a heavy weight. The sketches show the so-called 'one-leg problem' which applies to any rod on which two contrasting torques are applied. Since the rope is horizontal the two torques (180°, 90°) act at a right angle and the angles $\lambda,\theta$ are complementary, therefore $cos\lambda = sin\theta$. The angle of 45° is almost the ideal one, since the natural position and length of the forearms allow one to keep the rope horizontal and aligned with the CoM, that in a man is about is sternum. Let's consider now the case presented in the question: I set the force of the pull at 300 N, let's assume that his weight is ca. The pull from the rope (-300) is transmitted as compression to the front leg )and to the ground) at 45°, the total compression force on the front foot is 300 + 124 = 424N. Edit: that is rather confused, I have shown that with both feet at the fulcrum when 'you are about to tip' the horizontal force is 300N, but that is 'limit case' that never occurs when the legs are astride.
Sketch B shows the opposite case, when all weight is on RL: if there is no tension on the rope (T = 0) the reaction of the ground takes all 600N and he can lift his front leg. If a weight is lying on the floor it is not spending energy because gravity is spending energy for it. The mistake is that in order to balance torques, the majority of the weight will end up on the front foot.
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.
I was going to object to your model of the resistance, since the train (presumably!) has wheels. 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? 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.
Well, you're asking about two different things: the distribution of force, and the distribution of energy.
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). 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! Since all of the energy goes into the train, which technique you use is pretty much irrelevant to energy expenditure.

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. My impression was that the efficiency of a muscle performing some action is the ratio of mechanical work performed to actual energy burned. TorqueWhen 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. 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.
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. Note that in both case (1) and (3), your hands are in effect "carrying" (part of) the weight of the rope.
WikiArt.org allows unlimited copying, distributing and displaying of the images of public domain artworks solely. Suppose this last problem is solved 100% (like in the second picture) preventing the feet from slipping.
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. 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!. Just correct the misconception, explain as you did, then just say that if you want to read about it, find a physics book about forces or even better include a reference. In this position one cannot exploit gravity, but almost all is pull is transmitted to the train. The tangential torques balance when $T * sin\lambda (h) = W * cos\lambda (b = sin \theta)$. It is important that the rope be not above the CM, to avoid additional torque, lower might be better, but it is an unnatural position that does not allow the best effort. Because of the prop on the block, he can easily shift is rear foot forward or backward changing the distribution of the weight on the legs and the force exerted on the block, to remain in equilibrium in any situation. In this case, all the weight is on the front leg, the rear leg (in red in two positions to show it is not influent and free to move) can be lifted, or even join the front leg on the block: in this case the pull can be exerted bending the knees and stretching. The comment mentions mass, but I never mentioned mass and, as g is considered =10, 60 Kg = 600N. I think it is superfluous to remark that it seems foolish to do anything wasting more energy than the least necessary. I could obtain the same result with less energy if I pulled along the track, that, also, seems obvious. Laying parallel to yhe ground, the train and the rope he will exert his force at 0° , wasting no energy on perpendicular pull. In the limit where the person is about to tip over, the back foot has no weight at all on 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.
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. I am on the weaker end, so I would probably benefit from the first option, where my weight can help pulling; a stronger person should be able to produce more power adopting a more appropriate position. 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. By one of Newton's laws, the net force exerted on an object is equal to its mass times its acceleration. 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. 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.

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.
Suppose you're standing perfectly straight, and you pull on the rope while bracing with your legs. I will define it as the technique that allows the greatest force to be applied to moving the train along the track. 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. 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). Images of that type of artworks are prohibited for copying, printing, or any kind of reproducing and communicating to public since these activities may be considered copyright infringement. Is it possible to improve the percentage of energy which is actually transmitted to the train? There is no point mentioning that this is freshman physics, this is why some people are saying that your tone was condescending. It has been rightly noted that is better to avoid a rope, a light cable of nylon would be ideal: 5 m of such cable are not heavy, and we could avoid hanging it higher on the train as Floris suggested. When we join the two sketches and sum up all the forces from tension and the forces from weight there is equilibrium: $\Sigma f=0$ and $\Sigma \tau = 0$.
But this position is dangerous, because if the train moves (or he loses his grip=, he may fall back to the ground and break his neck. The sketch added in the other answer does not consider the angle and the forces acting on each leg. At that point the entire weight is on the front foot and you will see that the horizontal force is 300N. 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.
A clever combination of arms and legs may help you get that extra power at the beginning of the experiment. Since you're not accelerating at a significant rate, the net force exerted on you must be approximately zero. 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.
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. Since these two forces are applied to different locations on your body, the effect is that you'll rotate and fall forwards! The man in picture 1 creates a torque along his back - and the back muscles have to provide the counter torque.
A handle might give a better grip, and a block before the front foot might avoid kinetic friction and slipping.
When the object budges he will experience a clockwise torque and accelerate; as he is pulled back, he can re-adjust his foot to prop up against excessive acceleration. 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.
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.

## Comments to Man pulling back

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