Forces involved in a 'tug of war'

[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?

 

[CWatters]

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 693NM.

I believe this is correct!

Assumptions:

The 1m is measured along his body.
He's leaning at 45 degrees.

So the torque equation is...

m*g*h*cos(45) + T*h*sin(45) = 0

h cancels
cos(45) = sin(45) = 0.7 so that also cancels

gives

m*g + T = 0

T = -mg = 980 NM

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

Correct. If he doesn't lean at 45 degrees he will exert a lower force on the rope.

... what force is applied to P on the ground 300n or 300/.7?

Ok to answer the last part of that you have to look at the diagram showing the horizontal forces.

If the man is not accelerating then the horizontal forces must sum to zero. The only two horizontal forces acting on the man are:

The tension in the rope T = 300N
The friction force with the ground (call it Fgnd).

So..

T + Fgnd = 0

or

Fgng = -T

If T = 300N then Fgnd = -300N

The negative sign is because they are in opposite directions.

Edit: Note this is just the horizontal force the man applies to the ground not the total force he applies to the ground. If you want to work out that you have to calculate the angle that he has to lean at to generate 300N instead of 980N

 

[bobie]

Ok to answer the last part of that you have to look at the diagram showing the horizontal forces.
If the man is not accelerating then the horizontal forces must sum to zero. The only two horizontal forces acting on the man are:
The tension in the rope T = 300N
The friction force with the ground (call it Fgnd)
T + Fgnd = 0
Fgng = -T
If T = 300N then Fgnd = -300N
The negative sign is because they are in opposite directions.
Edit: Note this is just the horizontal force the man applies to the ground not the total force he applies to the ground. If you want to work out that you have to calculate the angle that he has to lean at to generate 300N instead of 980N

I do not follow you, he is exerting F =300 N, so the maximum force on the ground cannot exceed that value. You are saying, if I got it right, that all the force is deviated horizontally? there is no vertical component? is that possible?
I made a sketch according to the pallalelogram rule, (gravity exerts no torque and is fully applied to the ground):
http://s47.photobucket.com/user/lisa0rg/media/PullT_zps6c8c5be2.jpg.html?filters[user]=141333040&filters[recent]=1&sort=1&o=0
http://s47.photobucket.com/user/lisa0rg/media/PullT_zps6c8c5be2.jpg.html?filters[user]=141333040&filters[recent]=1&sort=1&o=0
I don't know why the image doesn't show properly, click on the link ,
but anyway are forces wrong?


Thanks a lot for your suggestion paint.net is great!

 

[A.T.]

I do not follow you, he is exerting F =300 N, so the maximum force on the ground cannot exceed that value.

If he is in equilibrium all horizontal forces on him must balance. If the rope pulls him at 300N to the left, then the ground pushes him at 300N to the right. Sum of horizontal forces on him is zero.

You are saying, if I got it righr, that all the force is deviated horizontally? there is no vertical component?

Sure there is a vertical component of the ground reaction force. In equilibrium something must balance his weight. If his weight is 1000N down, then the vertical ground reaction is 1000N up. Sum of vertical forces on him is zero.

 

[bobie]

If he is in equilibrium all horizontal forces on him must balance. If the rope pulls him at 300N to the left, then the ground pushes him at 300N to the right. Sum of horizontal forces on him is zero.

Sure there is a vertical component of the ground reaction force. In equilibrium something must balance his weight. If his weight is 1000N down, then the vertical ground reaction is 1000N up. Sum of vertical forces on him is zero.

- No I am not talking of reaction, AT, the reaction of course will be equal and contrary, but the ground can cope with any force, that is not the problem, but
if you exert a F at an angle , F becomes F cos y. Resolution-of-Forces Is it always valid or not? if he is pushing on the ground at 45° must there not be an x and a y component? how can the whole force go un the x axis?, according to my sketch, btw can you make it visible?, the ground is reacting to 980 N (G) and 210 N (push from rope) , for a total 1190N where is the problem. The ground is reacting horizontally to his right foot with a force of 210N (stiction)

Forces are not always in equilibrium, when the truck moves it means that the reaction is <299N, right?

 

[A.T.]

210 N (push from rope)

Rope tension is 300N in your diagram.

Forces are not always in equilibrium, when the truck moves it means that the reaction is <299N, right?

If it moves at constant speed the forces are in equilibrium. If it accelerates then you are likely accelerating too, so the ground force on you greater than the rope force on you.

 

[bobie]

Rope tension is 300N in your diagram.
.

Yes rope tension is 300N on the truck and the right foot, and on the ground it becomes 210 x-axis and 210 x-axis
what is wrong?

 

[A.T.]

Yes rope tension is 300N on the truck and the right foot, and on the ground it becomes 210 x-axis and 210 x-axis
what is wrong?

You are losing against the truck, and accelerating towards it.

 

[bobie]

You are losing against the truck, and accelerating towards it.

No, the forces balance,
here: http://s47.photobucket.com/user/lisa0rg/media/PullT_zps6c8c5be2.jpg.html?filters[user]=141333040&filters[recent]=1&sort=1&o=0
The truck is budging

http://s47.photobucket.com/user/lisa0rg/media/PullT_zps6c8c5be2.jpg.html?filters[user]=141333040&filters[recent]=1&sort=1&o=0
Thanks, anyway, I'll see for what dalespam says (tonight) of the reason why the parallelogram must not be applied here, or tomorrow for CWatters

 

[A.T.]

No, the forces balance,

No they don't. The friction of 210N is not sufficient to balance the opposed 300N rope pull.

The truck is budging

Not only budging, but plowing the field with you.

why the parallelogram must not be applied here, or tomorrow for CWatters

CWatters explained to you how to apply the parallelogram on page 2:
https://www.physicsforums.com/showpost.php?p=4840187&postcount=29

 

[bobie]

CWatters explained to you how to apply the parallelogram on page 2:
https://www.physicsforums.com/showpost.php?p=4840187&postcount=29

If the force on right foot is 424N, that means that the forces on x and y-axis sum up to 424*cos 45 * 2 = 600N.
You are doubling the force applied by the man, that seems rather unrealistic.
The forces in the sketch are perfectly balanced as the big red arrows show, so 3rd law is respected , and also 2nd law , because the reaction of the truck is slaller and the truck gives way a few cm and the rope is accelerating.

 

[A.T.]

If the force on right foot is 424N, that means that the forces on x and y-axis sum up to 424*cos 45 * 2 = 600N.

Adding x to y components is meaningless nonsense. This is not how vector addition works.

The forces in the sketch are perfectly balanced

Nope. All the x-components must add to zero. But they don't.

 

[bobie]

Adding x to y components is meaningless nonsense. This is not how vector addition works.
Nope. All the x-components must add to zero. But they don't.

Have you looked at the sketch? can you make it visible in your post?
the y and x components match perfectly 210 -210, 300 -300, if 300 -290, they do not match and the puller won his trophy.

Be kind and give me a couple of links where to learn how to add vectors
These are not good:
Resolution-of-Forces
http://www.physicsclassroom.com/class/vectors/Lesson-3/Resolution-of-Forces

http://www.physicsclassroom.com/class/vectors/Lesson-3/Resolution-of-Forces
Vector-Addition
http://mathworld.wolfram.com/VectorAddition.html
vector

 

[A.T.]

the y and x components match perfectly

Nope. The x components of the forces acting on the man do not add up to zero.

F_rope + F_friction = -300N + 210N = -90N ≠ 0

 

[Dale]

Thanks, anyway, I'll see for what dalespam says (tonight) of the reason why the parallelogram must not be applied here

Sorry, there has been a lot of conversation. I am not sure what it is that you would like me to respond to.

 

[bobie]

Sorry, there has been a lot of conversation. I am not sure what it is that you would like me to respond to.

Please say a final word on the value of the pull force on the left foot and on th x and y-axis and the use of parallelogram.
Pull on rope/truck
http://s47.photobucket.com/user/lisa0rg/media/PullT_zps6c8c5be2.jpg.html?filters[user]=141333040&filters[recent]=1&sort=1&o=0

I think that will end the controversy. I'll start a thread on impulse / force if you want to clarify those concept.
Thanks Dalespam, you are exquisite as ever

 

[A.T.]

I'll start a thread on impulse / force if you want to clarify those concept.

You should make yourself familiar with the concept of vectors. I don't see how you can even claim that the forces in your diagram balance, without understanding how forces are added.

 

[bobie]

Nope. The x components of the forces acting on the man do not add up to zero.

F_rope + F_friction = -300N + 210N = -90N ≠ 0

Where did you get those numbers?
firction is 210 equal to Px component -210
pull of rope is 300 equal to -300 reaction of truck (stiction + inertia); when they do not match you have + 300 -299, ... -290 = +1...+10 and the track has moved

 

[A.T.]

Nope. The x components of the forces acting on the man do not add up to zero.

F_rope + F_friction = -300N + 210N = -90N ≠ 0

Where did you get those numbers?

Read again.

firction is 210 equal to Px component -210
pull of rope is 300 equal to -300 reaction of truck

It makes no sense to add 3rd Law paired forces, which act on different object each. You have to add all the forces acting on a single object: the man.

 

[bobie]

Read again.


It makes no sense to add 3rd Law paired forces, which act on different object each. You have to add all the forces acting on a single object: the man.

Who says that? reference please!
That does not make sense as the three reactions act all on the man whereas his actions are exertted on three different obiects.

 

[Dale]

Please say a final word on the value of the pull force on the left foot and on th x and y-axis and the use of parallelogram.

The diagram is pretty cluttered. When you do a true free-body diagram one of the important rules is to only draw the forces acting ON the free-body. If you have multiple objects that you want to analyze then you draw a separate free-body diagram for each object and on each diagram place only the forces acting ON that object. Specifically, on this diagram you should not include the force of the leg on the ground, only the force of the ground on the leg. Also, you should not include the force of the arm on the rope, but only the force of the rope on the arm.

If you delete all of the forces acting on other objects then you are left with the attached drawing. Note how it is much clearer to analyze this diagram than the previous one. By removing the clutter you immediately see the following:

1) The vertical forces do balance out, meaning that the person is not sinking into the ground or jumping off the ground.

2) The horizontal forces do not balance out, meaning that the person is being accelerated towards the truck at a rate of (290-210) N/ 100 kg = .8 m/s². Either the rope tension must be less or the friction force must be greater for the person to not accelerate.

3) The force on the front leg is purely horizontal, which is probably not what was intended. I suspect that you want the force on the front leg to be at a 45° angle, but that is not what is drawn here.