Forces involved in a 'tug of war'

[A.T.]

where does it come from?

The vector sum of the lateral forces is zero. A zero force doesn't have to "come from somewhere".

can the result of a force be grater than 100%?

The lateral forces can be greater than the vertical one. Instead of an arc, consider a simple roof:

/\

What will be the ratio of horizontal to vertical forces when the slopes are less than 45°?

 

[CWatters]

If I got it right you are saying that if I apply a downward force of 300N on the arc, all of 300N end up on the ground, but some force is exerted on the walls horizontally, suppose the arc is a perfect semicircle, what is the intensity of this lateral force, and where does it come from?

If it's a perfect semi circle the ends meet the ground at 90 degrees so there is no horizontal force on the ground. If it's only an arc (as per my diagram) then yes there is a horizontal force. The intensity is Fgnd*sin(theta)

can the result of a force be grater than 100%?

Yes. It can be infinite.

What is the tension in a washing line or telephone wire? How much tension is needed to make the wire perfectly horizontal not hanging in a curve... Try hanging a brick from the middle of some string. Then lift the brick by pulling the ends of the string apart horizontally. How much force is required to make the string perfectly straight and horizontal? Are you strong enough :-)

 

[bobie]

If it's a perfect semi circle the ends meet the ground at 90 degrees so there is no horizontal force on the ground.

Suppose the semicircle (50Kg) is on a weighing platform, if you put on top of it a weight exerting 100N, all force goes laterally, no force on the platform, the recorded weight is still 50Kg?

 

[CWatters]

That diagram is precious, can you add gravity (man's weight 100Kg) and insert it in your post so that I can copy, using the quote button. Or , is g already included in the diagram and that's why we get 424N?

I could but it's better to do that on another diagram showing the torques acting on the man.

Lets assume the man isn't falling over. eg he is not rotating about his feet. (Technically I should say that he's not subject to rotational acceleration). To meet that condition the sum of the torques about his feet must also add to zero. So the anticlockwise torque due to his weight must equal the clockwise torque due to the tension in the rope.

I also have to make some assumptions about how high up the man the rope is attached and where his centre of mass is. Let's say that the rope is attached to the man at height h_r when he is stood up. So it's at h_r*sin(theta) above his feet when he leans at theta. Likewise his centre of mass is at h_m when he is stood up straight and h_m*sin(theta) when he is leaning.

Note that _if_ m,g,h_r and h_m are constants then in order to apply 300N he must lean at the correct angle which might not be 45 degrees. I'll let you substitute values and solve for theta.

If you insist the man must lean at 45 degrees then he would have to change h_r to make the torques sum to zero. Otherwise he will either fall or be pulled upright. (edit: or he cannot apply 300N to the rope, something must change to ensure the torques sum to zero).

In short if he's not accelerating (be it linear or rotational acceleration) then

The vertical forces must sum to zero
The horizontal forces must sum to zero
The torques must sum to zero.

 

[CWatters]

Suppose the semicircle (50Kg) is on a weighing platform, if you put on top of it a weight exerting 100N, all force goes laterally, no force on the platform, the recorded weight is still 50Kg?

If it's a perfect semi circle no force goes laterally it all goes vertically!

You also made some errors in the units. 50kg = 500N approx. So total force is..

500+100=600N

This would be split 300N on each side of the bridge.

If the scale is under both sides at the same time it will be subject to 600N vertically and read 60kg.

 

[CWatters]

If the semi circle was not perfect/full then there would be a lateral force that tried to stretch the scale horizontally. vertical scales do not measure horizontal forces. So the scale would still read 60kg (edit: because the vertical forces must still sum to zero, the scale isn't accelerating).

 

[CWatters]

Do try the experiment with a brick on a string (see post #37 above). Better still try it while standing on some scales. No matter how hard you have to pull on the string to try and get it horizontal the reading on the scales won't change.

 

[bobie]

Note that _if_ m,g,h_r and h_m are constants then in order to apply 300N he must lean at the correct angle which might not be 45 degrees. I'll let you substitute values and solve for theta.

If you insist the man must lean at 45 degrees then he would have to change h_r to make the torques sum to zero. Otherwise he will either fall or be pulled upright. (edit: or he cannot apply 300N to the rope, something must change to ensure the torques sum to zero).
.

The diagram is great but vanishes when I quote. Can you place one somewhere on the web so that I can pick it up?
There is something I still do not understand:
there is no horizontal force on the ground in this sketch
if the man's weight is 100 kg , then the force of g is 1000N right?
so the force of 300N is unrealistic? any ideas what pulling force a strong man of 100Kg can exert?
now there is a downward force of 1000N applied roughly at man's stomach, shouldn' t it be split in 2 parts and one shouldn't be tangential, as it is a torque, and the other along the body??
isn't the force he exerting on the groung along the same line of the body, and shouldn'it be split, again, by a parallelogram?
why can't he stand at 45° as long as the truck doesn't move??

Thanks Cwatters, you are very patient and helpful. I regret the system doesn't allow me to send any more thanks.

 

[Dale]

No, I was not thinking that. I have little experience but I was thinking of an arc, a vault. When you exert a vertical force on a bridge, an arc most of it is discharged horizontally, that is why a fortress has flying buttresses. Now, when you pull on the rope and on the truck you have to prop up to the ground, and , I thought, you must discharge some force on the ground.

Is that a wrong deduction?

If you were not thinking about conservation of force then you should definitely watch your wording. Even here, your wording strongly carries the impression that you think forces are conserved.

You talk about "discharging" forces. Charge is conserved. Once you have discharged a capacitor there is no more charge. Use of the word "discharge" is confusing if you do not intend to convey conservation.

I would recommend using the standard terms "exert" or "apply". To calculate how much force is exerted on the ground, I would start with the free-body diagram provided by sophiecentaur, except that the reaction force on the ground is at an angle θ instead of horizontal, or the free-body diagram provided by CWatters, except with the weight.

 

[Dale]

But you said literally if those forces add up to zero you are not exerting any force on the truck.

Despite your protestations to the contrary, to me it sounds like you really think that forces are a conserved quantity. I cannot see how to make sense of this comment without assuming you are thinking of conservation of force.

 

[Dale]

There are three forces acting on the man. Gravity is exerting a force of mg vertically downwards. The tension is exerting a force T horizontally. The ground is exerting a force R at an angle of θ from the ground.

Assuming that the man is not accelerating then by Newton's 2nd law we can write algebraic expressions relating the vertical and horizontal components of R to T and mg. Using those two equations you can solve for θ and R in terms of T and mg.

Why don't you try that?

 

[Dale]

The torques must sum to zero.

That is interesting. There must be some extra degree of freedom then, because otherwise the system is overdetermined.

 

[bobie]

Despite your protestations to the contrary, to me it sounds like you really think that forces are a conserved quantity. I cannot see how to make sense of this comment without assuming you are thinking of conservation of force.

Yes and no, Dalespam. I have little technical knowledge, write in a foreign language, therefore do not make myself understood.
No, I use discharge as when you discharge a shot, a blow. You apply a force vertically, but all or some is deviated in an other direction: 70% down and 30% sideways.
Yes I think of a force as coming from energy and producing energy. If this means 'conserved' I believe so. I cannot imagine, bar form 'tricks' of some kinds, like a lever etc. that you apply a 300N F and you in the end find yourself with 290 or 310.
If I am in space between to rocks and I stretch my body pushing on both and You say I exert a force of 300 N I take it that my overall 'power' is 300N, so if the rocks move (or not) I expect them to move at different speeds but according to 3rd law and working out in reverse the values I can find that I exerted 60% on one and 40% on the other, but the total force I applied must be 100%

What is wrong with this, it is rational, perhaps it doesn't follow the valid definitions?

That is why I cannot follow you when you say I exert 300N, 300 on the rope and 300 on the ground.
I'd appreciate if at last, you make me understand.

 

[A.T.]

Yes I think of a force as coming from energy and producing energy.

No, it isn't. When a book rests on the table, it can exert a force on the table indefinitely, without any input of energy.

my overall 'power' is 300N

Power is not measured in Newtons.

What is wrong with this,

You have basic misunderstandings about what forces are.

 

[bobie]

No, it isn't. When a book rests on the table, it can exert a force on the table indefinitely, without any input of energy.

Are you sure g is not working fulltime?

 

[Nugatory]

Are you sure g is not working fulltime?

Quite sure. There's no energy entering or leaving the system.

 

[olivermsun]

g is working full time, but the physical quantity of "work," which is basically transfer of mechanical energy to the table, is zero. The book is not increasing the kinetic or potential energy of the table just by sitting on top of it.

 

[Dale]

Yes and no, Dalespam. I have little technical knowledge, write in a foreign language, therefore do not make myself understood.
No, I use discharge as when you discharge a shot, a blow. You apply a force vertically, but all or some is deviated in an other direction: 70% down and 30% sideways.

Again, there is no conservation of forces so there is no reason that the "down" and "sideways" forces should add up to 100%. I am sorry, but everything you say confirms that deep down you have this idea that forces should be conserved.

There certainly could be a language barrier, but I think you have a concept wrong in addition to the language barrier.

Yes I think of a force as coming from energy and producing energy. If this means 'conserved' I believe so. I cannot imagine, bar form 'tricks' of some kinds, like a lever etc. that you apply a 300N F and you in the end find yourself with 290 or 310.

This is simply wrong. First, if your concept of force has to consider a lever to be a "trick" which operates outside the normal laws of forces then it is clear that your concept of force is wrong. With a lever I can apply 300 N and get 3000 N, quite easily. Even a simple inclined plane can change a 300 N horizontal force into a 3000 N vertical force. Others have mentioned how cables at shallow angles can exert immense horizontal forces in response to small vertical forces.

If I am in space between to rocks and I stretch my body pushing on both and You say I exert a force of 300 N I take it that my overall 'power' is 300N, so if the rocks move (or not) I expect them to move at different speeds but according to 3rd law and working out in reverse the values I can find that I exerted 60% on one and 40% on the other, but the total force I applied must be 100%
What is wrong with this, it is rational, perhaps it doesn't follow the valid definitions?

The same as above. Forces are simply not conserved. It doesn't make any sense to think of them in terms of percentages and so forth. You must simply let go of that idea.

Forces follow Newton's 3 laws. The proper way to analyze and understand forces is to draw free-body diagrams, and apply Newton's laws to them, as well as the usual rules of vector addition and decomposition.

As I said above, the free body diagram for the man has 3 forces acting on it, the force of gravity of magnitude mg pointing down, the tension force of magnitude T pointing horizontally, and the ground reaction force of magnitude R pointing diagonally upwards at an angle of θ from the ground.

The vector sum of those three forces is equal to ma as stated by Newton's 2nd law. From that, if you know those three forces then you can determine the acceleration, or if you know the acceleration you can determine those forces.

That is how you analyze forces, you use Newton's laws. No percentages enter in anywhere.

 

[bobie]

Again, there is no conservation of forces so there is no reason that the "down" and "sideways" forces should add up to 100%.
This is simply wrong. First, if your concept of force has to consider a lever to be a "trick"

Thanks for your explanations, before I understand that I need to clarify what is probably misleading me:
1) isn't impulse a force? (I thought the only difference is that a force acts indefinitely and an impulse a short time) if that it is true can we substitute one with other?

2) I can't quote CWatter's picture, I can't yet post a diagram, so look at this:

- Roughly where his left fist is, gravity is acting on A's barycenter C. If he weighs 100 kg on C is applied a force of 980 N which produces a torque like in a pendulum, right?. If θ = 45°, the actual $F_t$ acting on A will be 693 N in the tangential direction to C, at 45° with the ground.

I called this the 'discharged' force which is 71% of the original force. That is wrong! right, what is the correct terminology or is all that completely wrong?

- The point P where his right foot rests (in the picture here the centre O) is the point of support, the fulcrum. When he's not pulling and only g is in play, what is the force acting on P? Is parallelogram again useful here? if it is
then $F_p$ is = F * (siθ=cosθ), if he is pulling/leaning on the rope then $F_p$ should be equal to F (I said 100)%

Is this all completely wrong?

 

[CWatters]

Sorry for delay in replying. I'm in the UK.

The diagram is great but vanishes when I quote. Can you place one somewhere on the web so that I can pick it up?

http://i586.photobucket.com/albums/ss303/CWattersUK/TruckpullTorque.png

There is something I still do not understand:
there is no horizontal force on the ground in this sketch

I summed the torques about the mans feet. Torque = force * distance but the force you refer to is at the feet so the distance is zero. Likewise there is no vertical force shown at the mans feet.

if the man's weight is 100 kg , then the force of g is 1000N right? so the force of 300N is unrealistic? any ideas what pulling force a strong man of 100Kg can exert?

Consider what happens if the rope is attached to the man's centre of mass so that h_r = h_m= h and then he leans at 45 degrees.

Let the Tension in rope be T.

The man isn't accelerating/rotating so the torques sum to zero..

mg*h*cos(45) + T*sin(45) = 0

Solve for T

T = -mg*h*cos(45)/sin(45)

cos(45) = sin(45) so that cancels

T = -mgh which could easily be 1000n if the man weighs 100kg

Note: This explains why it's best for the man to hold the rope low down close to the ground. Perhaps even below his centre of mass. You get the leverage effect of h_m/h_r if they are different.

why can't he stand at 45° as long as the truck doesn't move??

He can but the tension in the rope might not be 300N.

 

[bobie]

Sorry for delay in replying. I'm in the UK.
http://i586.photobucket.com/albums/ss303/CWattersUK/TruckpullTorque.png
.

That's great!

I'll read the text and reply anon, in the meanwhile

could you be so kind as to scale it down a little and modify it putting the rope along the CoM (man's CoM is roughly at the sternum so they should coincide in a normall pull) and add the values for g 1000N (it's rougly 100 kg) have you read my answer to dalespam? isn't it necessary to draw the circle along which is applied the torque by g?

If it is not complicated can you tell me what program you used and if I can use it free? I am just a student.

 

[CWatters]

I used Paint.net which is free.

 

[CWatters]

That is interesting. There must be some extra degree of freedom then, because otherwise the system is overdetermined.

Consider this statement..

If a "ideal man" stands perfectly upright then it's not possible for him to exert any force on the rope.

By ideal man I mean one with feet of zero length so that his centre of mass is directly above his point of contact with the ground.

In this situation there is nothing to balance the torque produced by the rope. He mustlean back, or do something similar such as putting one foot behind the other which amounts to the same thing. Otherwise even the slightest tension in the rope would pull him over.

In short, even if there is sufficient friction with the ground, the limit to how much force he can apply on the rope is controlled by his mass and the geometry.

Aside: Find a strong man and have him stand with his heels and back to a wall. Tell him he can't move his feet. A skinny kid could pull him over easily as the man cannot get his centre of mass more than about a foot behind his point of contact with the ground.

 

[bobie]

, or do something similar such as putting one foot behind the other which amounts to the same thing. Otherwise even the slightest tension in the rope would pull him over..

Sure, but when CoM is beween the feet gravity exerts no torque and all its force is applied to the ground and is to no avail, like in the picture I posted, there the rope is about the height of center of mass (sternum) which is inside the base of the feet. he cannot exploit gravity.

 

[CWatters]

That's not correct. He can "exploit gravity" right up to the point where all his weight is on his front foot.

Consider what happens if something pulls on the rope with increasing force. At some point the man starts to rotate about the front foot and his rear foot just lifts off the ground. That sets a limit on how much force he/it can apply. eg how much he can exploit gravity. All the while his centre of mass is between his feet.

 

[CWatters]

If he has both feet on the ground (as per your clipart) then that complicates how you account for all the torques when you sum them to zero, but the approach and method is this same. The torques must still sum to zero if he isn't falling (eg if he's not subject to rotational acceleration).

[STRIKE]All you need to do is add an extra term for the weight on the back foot multiplied by the distance between his feet.[/STRIKE] The vertical force on the front foot can be ignored as that's the point about which you are summing the torques.

Edit: Actually it's not quite as simple as I implied in the strike out, but I'm running out of energy trying to explain it. Sorry.

 

[bobie]

That's not correct. He can "exploit gravity" right up to the point where all his weight is on his front foot.

Consider what happens if something pulls on the rope with increasing force. At some point the man starts to rotate about the front foot and his rear foot just lifts off the ground. That sets a limit on how much force he/it can apply. eg how much he can exploit gravity. All the while his centre of mass is between his feet.

I can't visualize it, I thought that he feels a pull only when CM is behind the base. What about the torque about the circle? if g 1s 1000N is it 710? can you draw it?

 

[CWatters]

I can't visualize it, I thought that he feels a pull only when CM is behind the base.

The CM has to be behind the point of rotation not the "base".

Sorry but I've got to go do other stuff for awhile.

 

[bobie]

He can but the tension in the rope might not be 300N.

Let me try:
if CM is at distance h (1m) from point P (fulcrum) and the rope is at that height

the torque of g is 100kg*1m*9.8* 0.7 = 693 n , so in order to balance the pull must exert 980n, which multiplied by 0.7 gives 693. But if bends slightly forward he can exert 300n without problems, is that right?

Now if g did not exist , or we consider only the pull, what force is applied to P on the ground 300n or 300/.7?

 

[jbriggs444]

Let me try:
if CM is at distance h (1m) from point P (fulcrum) and the rope is at that height

So the CM is 1 meter away from the fulcrum at an angle of 45 degrees above the horizontal. That is, it is about 0.7 meters back and 0.7 meters up, right? And the rope is at 1 meter height above the fulcrum.

EDIT: This appears to be a language difficulty. By "that height" you apparently meant "the height of the CM", not "1 meter". The intent is that the rope is about 0.7 meters above ground.

the torque of g is 100kg*1m*9.8* 0.7 = 693 n

693 Newton-meters. A torque is not a force. It has different units than a force.

so in order to balance the pull must exert 980n, which multiplied by 0.7 gives 693

Why are you multiplying by 0.7? The pull is 980 Newtons. The associated torque is 980 Newton-meters. Remember you put the rope 1 meter above the fulcrum, not 70 centimeters.

Edit: With the language issue corrected, the torque from the rope is indeed 693 Newton-meters.

But if bends slightly forward he can exert 300n without problems, is that right?

Not sure what you are trying to say here.

Edit: So by bending forward, the center of mass is imagined to move forward and up and the rope height is increased accordingly. Yes, if he leans forward far enough a torque of 300 Newtons times vertical distance from rope to fulcrum can balance a torque of 980 Newtons times horizontal distance from CG to fulcrum.

Now if g did not exist , or we consider only the pull, what force is applied to P on the ground 300n or 300/.7?

Are we still assuming that an equilibrium exists? If so, it is an impossible situation.

Edit: You are attempting to ask about the horizontal component of the force P applied on the ground. That's 300 Newtons. Conservation of momentum assures us of this.

 

[bobie]

So the CM is 1 meter away from the fulcrum at an angle of 45 degrees above the horizontal. That is, it is about 0.7 meters back and 0.7 meters up, right? And the rope is at 1 meter height above the fulcrum.
Why are you multiplying by 0.7? The pull is 980 Newtons. The associated torque is 980 Newton-meters. Remember you put the rope 1 meter above the fulcrum, not 70 centimeters.
Not sure what you are trying to say here.

Edit: So by bending forward, the center of mass is imagined to move forward and up and the rope height is increased accordingly. Yes, if he leans forward far enough a torque of 300 Newtons times vertical distance from rope to fulcrum can balance a torque of 980 Newtons times horizontal distance from CG to fulcrum.


Are we still assuming that an equilibrium exists? If so, it is an impossible situation.

Edit: You are attempting to ask about the horizontal component of the force P applied on the ground. That's 300 Newtons. Conservation of momentum assures us of this.

P is the point on the ground where are the feet: the fulcrum, right?
I wrote 1m distance from P (hm=h), as angle is 45° the height from the ground is .7m, right?
I am multiplying because the diagram says hmcosθ, which is .7, right. But I think the pull is not horizontal, as in the diagram, but tangential at hm, like in a pendulum, I got no reply to this.

I am not asking about the horizontal but the force of the pull acting from hm toward P at 45%, I said 300*cosθ, and was corrected to 300/cosθ, what is true and why? If I pull horizontally and prop up at 45° I thought the force is applied at 45° on the ground, than its horizontal component would be 300*cosθ

Thanks a lot

 

[A.T.]

But I think the pull is not horizontal, as in the diagram,

You are already confused by the geometry of this, and now you want to make it even more complicated, by introducing a non-horizontal rope?

 

[bobie]

You are already confused by the geometry of this, and now you want to make it even more complicated, by introducing a non-horizontal rope?

I was talking about the torque by g, isn't it like in a pendulum, tangential?

 

[sophiecentaur]

Thanks for your explanations, before I understand that I need to clarify what is probably misleading me:
1) isn't impulse a force? (I thought the only difference is that a force acts indefinitely and an impulse a short time) if that it is true can we substitute one with other?

2) I can't quote CWatter's picture, I can't yet post a diagram, so look at this:


- Roughly where his left fist is, gravity is acting on A's barycenter C. If he weighs 100 kg on C is applied a force of 980 N which produces a torque like in a pendulum, right?. If θ = 45°, the actual $F_t$ acting on A will be 693 N in the tangential direction to C, at 45° with the ground.

I called this the 'discharged' force which is 71% of the original force. That is wrong! right, what is the correct terminology or is all that completely wrong?



- The point P where his right foot rests (in the picture here the centre O) is the point of support, the fulcrum. When he's not pulling and only g is in play, what is the force acting on P? Is parallelogram again useful here? if it is
then $F_p$ is = F * (siθ=cosθ), if he is pulling/leaning on the rope then $F_p$ should be equal to F (I said 100)%

Is this all completely wrong?

It strikes me that you are not approaching this problem in the spirit, likely to help you in the long run. You need to start with the very basics of what was called 'statics' when I was at School. The basic definitions and the (simple) equations involved (involving basic Trig) with this sort of problem will always give you the right answer. If you don't do this 'right' then you cannot be sure of any conclusions you may come to because your predictions will not be bomb-proof. There is no quick fix for this sort of topic.

 

[Dale]

I see that there have been a bunch of replies, so some of this may be redundant.

1) isn't impulse a force? (I thought the only difference is that a force acts indefinitely and an impulse a short time) if that it is true can we substitute one with other?

No. Impulse is force times time, or more specifically the integral of force with respect to time. Impulse does not need to be a short time, but often in collision you are interested in forces which have a finite impulse over a very short time and don't care about the actual force. Impulse has units of momentum, it is definitely not force.

- Roughly where his left fist is, gravity is acting on A's barycenter C. If he weighs 100 kg on C is applied a force of 980 N which produces a torque like in a pendulum, right?. If θ = 45°, the actual $F_t$ acting on A will be 693 N in the tangential direction to C, at 45° with the ground.

I called this the 'discharged' force which is 71% of the original force. That is wrong! right, what is the correct terminology or is all that completely wrong?

Just call it an "exerted" force or an "applied" force. The percentage of the original force is irrelevant, since there is no conservation of force and therefore may not be any 29% force.

- The point P where his right foot rests (in the picture here the centre O) is the point of support, the fulcrum. When he's not pulling and only g is in play, what is the force acting on P? Is parallelogram again useful here? if it is
then $F_p$ is = F * (siθ=cosθ), if he is pulling/leaning on the rope then $F_p$ should be equal to F (I said 100)%

Is this all completely wrong?

I don't know, I would have to work out the forces but I cannot do that now. Yes, the parallelogram rule and all of the other standard rules of vector addition are important here. The fundamental relationship is Newton's laws, especially his 2nd law.

Personally, I would recommend simplifying the secenario and not worrying about torques. Consider a one-legged man so that you can have a single force on the ground, and consider both the force on the ground and the tension from the rope to be acting in line with the center of mass so that there is no torque, and consider the tension to be horizontal. Thus you have only 3 forces (horizontal tension (T), vertical weight (mg), and ground reaction (R) at an angle of theta with the ground) and 0 torques.

Do you know how to solve this simplified problem?