Forces involved in a 'tug of war'

[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.

 

[Dale]

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

I agree. I suggested a simplified scenario several pages ago. I think it is better to start simple and add complications later.

 

[CWatters]

Ok this is my final word on this problem. Got to fix a leaking WC! I have produced the following diagrams.

1) The first shows some assumptions about the geometry such as him leaning at 45 degrees and his COM being over his rear foot etc. I also assume he is not accelerating (vertically, horizontally or in rotation).

2) The second shows the how the torques must sum to zero. The distances from the pivot are all the same due to the geometry. Let's call that distance h (see assumptions drawing). If we take anti clockwise as +ve then..

-980*h + 300*h + 680*h = 0 √

3) The third diagram shows the consequences for the vertical components. Let's take vertical as positive then

-980 + 300 + 680 = 0 √

4) The fourth diagram shows the horizontal components. I have made the reasonable assumption that the 300N friction force is shared in proportion to the vertical load on each foot. That calculation is done on the diagram. Let's assume right is positive then..

92 + 208 + (-300) = 0 √

5) The final diagram combines the horizontal and vertical components at each foot to give the total force at each foot (using Pythagoras and basic trig.). Note that the total force acting on each foot acts at about 73 degrees to the horizontal. There is no reason why this should be 45 degrees.

That's all folks.

 

[CWatters]

PS: Some may question my assumption about the way the friction force is shared between the two feet but I think it's perfectly reasonable for a flat uniform smooth floor like a Gym. You also have to remember that in this problem the man is NOT trying to pull as hard as he can (980N at 45 degrees). He is only applying 300N. Most of his weight is carried by his rear foot.

 

[Dale]

I have made the reasonable assumption that the 300N friction force is shared in proportion to the vertical load on each foot.

I like that assumption also, I think I would have made the same one. It is interesting that one consequence of that assumption is that the forces from the ground are parallel to each other.

The other assumption I might consider would be that the forces are whatever is required to minimize the torque about the hips. But that seems excessively complicated.

 

[bobie]

the 300N friction force is shared by both feet.
... some of the excess torque generated by his mass must be supported by his rear foot
...(which is under his COM)
... friction force is shared in proportion to the load on each foot then the friction force on each foot is
...Rear..300* 680/980 = 208N
..So now we have horizontal and vertical components for the force on each foot...
Rear..Vertical = 680N...Horizontal = 208N

1)...and his COM being over his rear foot etc.
4) The fourth diagram shows the horizontal components. I have made the reasonable assumption that the 300N friction force is shared in proportion to the vertical load on each foot. That
92 + 208 + (-300) = 0 √
... Note that the total force acting on each foot acts at about 73 degrees to the horizontal. There is no reason why this should be 45 degrees. .

PS: Some may question my assumption about the way the friction force is shared by each feet

You may be right, but your basic assumption that LF is bang under the COM and consequently g is transmitted entirely on the back foot.

This situation is unrealistic because if you are pulling (and being pulled, let's not forget that the truck is in itself a simplification of the original problem:tug-og-war') CM is continuously moving , adjusting itself to the torque (sometimes instinctively or authomatically).

If one foot is under CM you can transmit all your weight on your left leg if no other force is acting on you, you can lift your right foot from the ground and then adjust your posture so that CM is aligned with the center of the left foot. (Position B)
gravity A-B

I have followed DaleSpam advice and started from the simplest position: no rope. Here is my sketch:
Position B
If the rear foot is at B the whole weight is on the LF, and no g is acting on RF, you may lift it and nothing happens.But, if it is beyond B (at A, or further) the weight is shared in growing proportions by RF.
When you are pulling the rope, the appropriate position of CM/lF is found authomatically, pushing against the block at the required force. Here are the forces necessary to compensate a pull of 300 N:
Position A

As you can see there is a horizontal F-x on RF (-88N) which is opposed by the Block pinned to the ground (do you call this 'friction'?) and a positive Fx from gravity on LF (DaleSpam disagrees but any civil engineer will confirm that a weight on sloping supports (and the legs are sloping here) generates a horizontal normal push/ force). In the sketch it's 148N but it can vary and adjust itself to needs) .This x-force will anyway compensate the -x force you supposed will produced when you pull on the rope.

I figured an angle of 10°-15° for LF, but I am not sure when 148 is right. What we know for sure is that the R-leg angle must be 45°, at least as a starting point, else we would be introducing too many variables.

If G on LF 856N, the left foot will gain the necessary 124 N missing to reach the balance (-88-212) the horizontal F-x which is necessary to compensate the torque fom the rope +300N.

If you agree on this A diagram, I'll show you the next step, when there is tension on the rope.

If you had enough, I thank you for your invaluable help that has guided me in this tangled issue.

 

[A.T.]

your basic assumption is that g is transmitted entirely on the back foot.

Not what he said. Read again.

This situation is unrealistic

Place the feet differently, assume higher tension, and you get a different distribution.

As you can see there is Fx from gravity on LF

You are confusing friction with gravity.

any civil engineer will confirm that a weight on sloping supports generates a horizontal force

There is no slope here.

 

[Dale]

bobie, this is incredibly frustrating. You are completely ignoring large portions of what we actually do say and completely inventing things that none of us ever said and pretending that is what we said. Please read what we write and don't invent what you want to hear but actually learn what was said. If you have questions about what we say then ask, but don't put words in our mouths.

You may be right, but your basic assumption that LF is bang under the COM and consequently g is transmitted entirely on the back foot.

No, he never said that. g acts at the center of gravity and is not transmitted at all. Furthermore, the normal force on the rear leg is only 680 N, so even if you want to refer to the ground reaction as a "transmitted" g force it is not entirely on the back foot.

I have followed DaleSpam advice and started from the simplest position: no rope.

The scenario that I suggested as being the simplest was to have only one leg. It is not a tug of war without a rope, but you certainly could have a one-legged tug of war.

DaleSpam disagrees but any civil engineer will confirm that a weight on sloping supports (and the legs are sloping here) generates a horizontal normal push/ force)[/I].

This is not true, and it is also not what I said. Your putting words in my mouth is really starting to irritate me.

Ask any civil engineer you like to draw a free-body diagram for a tug of war man on flat ground. I guarantee that not one of them will break g into vertical and horizontal components. At the ground there may or may not be a horizontal component of the reaction force, depending on the assumptions they make, but I guarantee that none of them will put a horizontal component of gravity.

 

[bobie]

Ask any civil engineer you like to draw a free-body diagram for a tug of war man on flat ground. I guarantee that not one of them will break g into vertical and horizontal components. At the ground there may or may not be a horizontal component of the reaction force, depending on the assumptions they make, but I guarantee that none of them will put a horizontal component of gravity.

I am awfully sorry if I irritated you, but I was not referring to Tug-of-war.
I am saying that just standing on ice or on a waxed floor and with your legs astride at an angle of less than 45/35°, you will see that there is a horizontal pull on your foot.
In picture A only gravity is acting on the man, and friction from the ground and the block offers a reaction to that, and that +F on rear foot compensate the -F suggested by CWatters.

I realize the discussion has been lengthy so I thank you all again

 

[Dale]

I am saying that just standing on ice or on a waxed floor and with your legs astride at an angle of less than 45/35°, you will see that there is a horizontal pull on your foot.

No, there isn't. What there is is a bending moment about each of your hip joints.

If you are on a frictionless floor then by definition there is no horizontal force at the ground, and there is never any horizontal component of gravity. Because there is no horizontal force at the ground, there must be an outward bending moment about each hip, which you compensate for by using your hip adductor muscles.

The hip adductor muscles are the same muscles that you would use to oppose a genuine horizontal pull at the floor. So perceptually it feels very similar, but there is in fact no horizontal pull and the bending moment you are opposing with those muscles is generated by a purely vertical force.

Again, ask any engineer you like to draw a free body diagram of a man on flat ice with legs astride and I guarantee that they will not break gravity into horizontal components.

 

[bobie]

No, there isn't. What there is is a net torque about each of your hip joints.
.

Pardon my ignorance, DaleSpam, but a torque on hip does not mean that your foot is pushed outward? if so, I poorly expressed my self, else I'll learn something completely new.

Is this what you meant 1 foot?
http://s47.photobucket.com/user/lisa0rg/media/Man2F_zps48a11680.jpg.html?filters[user]=141333040&filters[recent]=1&sort=1&o=0

What is the force of gravity acting on the two legs? (and on the block)?

 

[jbriggs444]

Pardon my ignorance, DaleSpam, but a torque on hip does not mean that your foot is pushed outward?

The torque at your hip is your muscles pulling your legs toward one another. Because of this torque, your legs are supporting an inward-and-down sheer stress in addition to an outward-and-down compression stress. If you standing in an equilibrium on frictionless ice there can be no net inward or outward force on either foot. The combination of the sheer force and the compression force must be purely vertical.

Your feet are not pushed outward by the ice. They are pushed upward. As Dale has pointed out, this feels the same as if they are being pushed outward because both forces cause torques in the same direction.

What is the force of gravity acting on the two legs? (and on the block)?

Gravity acts on the body. A force from the body (at the hips) acts on the legs. A force from the legs (at the ankles) acts on the feet. A force from the feet (at their bottom) acts on the block.

The force of gravity does not act on the block, except, of course, to the extent that the block has mass.

 

[bobie]

1) Your feet are not pushed outward by the ice. They are pushed upward.

2)
The force of gravity does not act on the block, except, of course, to the extent that the block has mass.

1) I could not understand as that is absolutely new to me. When I was on a waxed floor once, I thought my feet were pulle apart.
Besides , even now, when I climb theseon one of these to change a bulb, if I forget to properly spread the sides,as soon as I mount they are immediately pushed apart.
I hope you confirm that it happens to you, too. And that is why this tool has rubber shoes.

Now if this is true, can you explain to me what is the difference between a ladder and a leg? Why doesn't a leg behave like a ladder?

2)Are you referring to the image? G does not push on the block when exerting the torque on the body? what is the value of G-torque (red on the circle)?

 

[CWatters]

You may be right, but your basic assumption that LF is bang under the COM and consequently g is transmitted entirely on the back foot.

This situation is unrealistic...

I totally agree but then I'm not trying to make the problem realistic. I'm trying to show you how to go about solving this type of problem.

 

[TumblingDice]

I am saying that just standing on ice or on a waxed floor and with your legs astride at an angle of less than 45/35°, you will see that there is a horizontal pull on your foot.

That's silly. Your foot is pushed out because the angle of your leg decreases while staying the same length and being attached to you. As DaleSpam said, all related to your COG and vertical gravity vector from there. You should be able to tell the difference between that and pulling on your foot.

 

[A.T.]

When I was on a waxed floor once, I thought my feet were pulle apart.

The legs are rotated apart because the vertical forces on your feet create a troque around the hip. There are no horizontal forces pulling the feet apart.

G does not push on the block when exerting the torque on the body?

No, the body pushes on the bloc.