Forces involved in a 'tug of war'

[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.

 

[Dale]

Who says that? reference please!

A.T. is correct. If he does not post a reference then I am sure I can dig up several.

 

[bobie]

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.

Thanks, Dalespam
3) the force at 45° was in blue in the drawing 300N that has been split by the parallelogram. Is that procedure wrong (can I say 'split'?).
Now, the force any man can pull is easily measured (by instruments maybe, or ) in a simple way adding a couple of pulleys ans checking what is the maximum F. Suppose it is 2000N (is it a plausible figure) we'll make him do his outmost and consider that value.
If he is pulling 2000N the force on his foot (we excluded g, but we might solve the problem in this way shifting the CM so that on the right foot there is an extra F = 90).
But I am not looking for a cheap solution. I want to understand how you justify that exerted F is 2000 and on the foot becomes 2800. Can you suggest me a **formula to calculate how g is distributed between the two legs?

2) The reaction of the truck is nominal. If he were in a tug of war then 300N would be effective. Here we are dealing with passive resistance by stiction and inertia, the reaction immediately varies / adjusts itself with the action doesn't it? if it doesn't the man instinctively bend forward shifting (discharging) its weight on the fore leg, right?

Thanks
**do you think such formula exists?

P.S. Why I cannot upload my image from the web?

 

[Dale]

Thanks, Dalespam
3) the force at 45° was in blue in the drawing 300N that has been split by the parallelogram. Is that procedure wrong (can I say 'split'?).

That was the force on the ground FROM the leg, not the force from the ground ON the leg. Remember, in a free-body diagram we only include the forces that are acting ON our free body, not the forces FROM our free body which are acting on other objects. So, this should have been drawn as a 300 N force at 45° upwards acting on the leg.

Splitting the force into components is fine, but it just does not belong on this free-body diagram.

Can you suggest me a **formula to calculate how g is distributed between the two legs?

Yes, if the man is in static equilibrium then:
$\Sigma f=0$ and $\Sigma \tau = 0$.

2) The reaction of the truck is nominal. If he were in a tug of war then 300N would be effective. Here we are dealing with passive resistance by stiction and inertia, the reaction immediately varies / adjusts itself with the action doesn't it? if it doesn't the man instinctively bend forward shifting (discharging) its weight on the fore leg, right?

Whatever the force ON the man is, that is what you put on a free-body diagram. If the magnitude of that force is unknown then simply assign it a variable name (like "T") and then solve for its value. Whether the force is "passive resistance" or whatever else is not important.

For this scenario, if we have that the mass of the weight is 980 N and if we are given that the force on the front leg is 300 N at an angle of 45° upwards then the force on the other leg is 768 N upwards, and the tension in the rope is 212 N horizontal (assuming equilibrium).

 

[A.T.]

Who says that? reference please!

I gave you one on page 2 already. Did you read it?
https://www.physicsforums.com/showpost.php?p=4840189&postcount=30

That does not make sense as the three reactions act all on the man whereas his actions are exertted on three different obiects.

The terms "action" and "reaction" in the context of the 3rd law are assigned arbitrarily, and are interchangeable.

 

[CWatters]

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

No, I meant that I had only calculated the horizontal component. You can calculate the vertical component by looking at the vertical forces (which also sum to zero).

To work out the total force at the mans feet you need to add (vector add) the horizontal and vertical components.

 

[bobie]

No, I meant that I had only calculated the horizontal component. You can calculate the vertical component by looking at the vertical forces (which also sum to zero).

To work out the total force at the mans feet you need to add (vector add) the horizontal and vertical components.

Hi CWatters, I have corrected my sketch:
ManPullsT
http://s47.photobucket.com/user/lisa0rg/media/ManPullsT_zps61720dbc.jpg.html
Can you tell me if it is OK?
I am not sure if all forces on the body must be negative, as usually g is considered negative and the reaction positive. Shall I use + and - according to the axis x, an what about the forces at 45, which are minus?
Can you tell me hou to calculate the remaining g on the left leg as here I cannot use the parallelogram.

THanks again, you have been very kind to suggest paint.net and I used also your bucket shop.
Can you tell me why your image is stand alone and mine has all the ads etc?

 

[A.T.]

Hi CWatters, I have corrected my sketch: Can you tell me if it is OK?

It's incomprehensible. What are those G124, G88, R212? Make a legend defining the names. Why don't you take DaleSpam's advice and draw only the external forces acting on the man? There are only three of them. Spohiecentraur did that diagram for you on page 1, with just one small error (direction of ground reaction).

 

[Dale]

Hi CWatters, I have corrected my sketch:
ManPullsT
http://s47.photobucket.com/user/lisa0rg/media/ManPullsT_zps61720dbc.jpg.html
Can you tell me if it is OK?

None of the blue forces should be on this diagram. Also, gravity, by definition, has no horizontal component, so I don't know why you have the extra green arrows. Finally, on the red arrows at the front foot it is unclear if you are intending to break the 45° reaction force into components. If so then where is the vertical component, if not then why is there a horizontal component?

 

[bobie]

None of the blue forces should be on this diagram. Also, gravity, by definition, has no horizontal component, so I don't know why you have the extra green arrows. Finally, on the red arrows at the front foot it is unclear if you are intending to break the 45° reaction force into components. If so then where is the vertical component, if not then why is there a horizontal component?

The vertical component was missing it is -88 N. I'll post a new corrected diagram probably before you get back to this post.
The blue forces are in the diagram just for you to check if is at last correct.

As to gravity when you put yor legs astride (trying to do a split) when the angle exceeds 45° the weight is distributed on each leg and the dorce is 'split' /divided in 2 normal components, if the floor is waxed you slip and may dislocate your femur /hip. Moreover the blu force on the front foot cannot exceed 300N and the missing 124 N (if you want to get an x-force 300N) can come only from g.

** here is the image Pull-complete

 

[CWatters]

Can I persuade you to show that the horizontal, vertical and torques each sum to zero. That way you/we can see if it's all consistent.

For example do the vertical components of the forces acting on each foot add up to 980N?

 

[bobie]

Can I persuade you to show that the horizontal, vertical and torques each sum to zero. That way you/we can see if it's all consistent.

For example do the vertical components of the forces acting on each foot add up to 980N?

Why should the forces on the forefoot add up to 980N, since most of g is transmitted (is this term more acceptable than 'discharged') on the back foot? CM is almost above it. He can almost balance on it and lift the forefoot, when he is not pulling.
Is everything else OK?

 

[Dale]

The diagram is incorrect as long as any of the blue forces are on it.

As to gravity when you put yor legs astride (trying to do a split) when the angle exceeds 45° the weight is distributed on each leg and the dorce is 'split' /divided in 2 normal components, if the floor is waxed you slip and may dislocate your femur /hip.

No, weight acts at the center of gravity, by definition. What is distributed is the normal force. There, should only be one green force and it acts vertically downwards from the center of gravity. It has no horizontal component, and it is not distributed down each leg.

 

[CWatters]

Why should the forces on the forefoot add up to 980N..

I didn't say that the forces on the forefoot must add up to 980N.

 

[bobie]

I didn't say that the forces on the forefoot must add up to 980N.

Do you think the diagram is now OK? Are there any more problems?

 

[CWatters]

Personally I wish you hadn't complicated the situation by separating front and rear feet!

I don't believe you can calculate the total force on the front foot unless you calculate how the 300N friction force is shared between each foot. So I attempt that...

We have established that if the man was to lift his rear foot off the ground it appears he would apply 980N of force to the rope. If he only wants to pull with a force of 300N then some of the excess torque generated by his mass must be supported by his rear foot (which is under his COM)

If I make reasonable assumptions about the geometry of the drawing then the vertical component of the force on the rear foot must be 980-300=680N.

Then if the vertical components must add to 980N the vertical component of the force on the front foot must be 980-680=300N

Now look at the horizontal components...

The horizontal forces due to friction must sum to 300N but how much is carried by each foot? If we assume that the friction force is shared in proportion to the load on each foot then the friction force on each foot is

Front..

300* 300/980 = 92N

Rear..

300* 680/980 = 208N

So now we have horizontal and vertical components for the force on each foot...

Front...

Vertical = 300N
Horizontal = 92N

Rear..

Vertical = 680N
Horizontal = 208N

You can apply Pythagoras to work out the total force on each foot and some trig to work out the angle.

I'll just do the front foot...

Total^2 = 300^2 + 92^2

Total = 313N

Your diagram appears to show 424N but you don't explain how you get that figure.