Forces involved in a 'tug of war'

[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.

 

[CWatters]

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.

The effect you refer to ONLY exists because the human body is NOT RIGID. There is a hinge at the hips. If the man was rigid there would be no apparent horizontal force at the feet.

It is difficult to account for this effect. How rigid is he? The hinge at the hips cannot be totally free to move or he would fall.

Edit: I agree with this reply by A.T...

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.

 

[Dale]

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

A bending moment on the hip means that the leg is rotated outwards. In this case the outward rotation is caused by an upwards force at the foot, not a horizontal force.

Is this what you meant 1 foot?

Yes, but again you have all sorts of extra forces that don't belong on the diagram (forces that the man exerts on other things) and again you have given gravity a horizontal component that it does not have.

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

For at least the 3rd time, gravity acts at the center of gravity by definition, and it acts vertically downwards at that point.

 

[Dale]

if I forget to properly spread the sides,as soon as I mount they are immediately pushed apart.

They are not pushed outward by a horizontal force at the floor, they are rotated out by the bending moment from the vertical force at the floor.

In fact, under normal conditions the horizontal friction force at the floor will cause an inward torque, not an outward one. In the case of a frictionless floor the bending moment is entirely due to the vertical normal force.

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?

A leg behaves exactly like a ladder for this purpose. You are misunderstanding the forces acting on a ladder in the same way you are misunderstanding the forces acting on a leg. Specifically, for both you are mistakenly assuming that an outward bending moment at the hip/hinge implies a horizontal force at the foot.

 

[bobie]

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.
.

Is this what you meant?
http://s47.photobucket.com/user/lisa0rg/media/T2_zps18695702.jpg.html?filters[user]=141333040&filters[recent]=1&sort=1&o=0
I considered 300 and 750 N to simplify it.

If these are torques , like in a pendulum?, they act tangentially or perpendicularly?
Please , anyone give me the correct values for the three notes ((1),(2),3)) this would help me most.

If the forces were equal 300 and 300 we would have equilibrium, and applying the parallelogram I would find the force on the pivot = 424. But if I use it here instead of a square I get a 2.5:1 rectangle and the resulting force is at 68°, how do I find the force acting on the pivot, here?
Can you, also, tell me how to use signs, shall I follow the cartesian axes or what?

Thanks for your comprehension

 

[A.T.]

If these are torques , like in a pendulum?, they act tangentially or perpendicularly?

For a 2D problem the torques act clockwise or counter-clockwise depending on their sign. Technically they are vectors perpendicular to the 2D plane you are analyzing:

http://en.wikipedia.org/wiki/Torque

 

[bobie]

For a 2D problem the torques act clockwise or counter-clockwise depending on their sign. Technically they are vectors perpendicular to the 2D plane you are analyzing:http://en.wikipedia.org/wiki/Torque

In the diagram the CM is the hip and is subject to a torque, if there where no pivot the vertical g would pull the (leg) to the left but since ther is a pivot, it must follow a circle. Now,

- Are the tangential values in the image exact?
- And, what are the correct values for the forces at the notes (1,2,3) ?

These answer can hel me more than any link. I read those articles, but I cannot apply the principles to a particular example (yet)

 

[Dale]

Is this what you meant?
http://s47.photobucket.com/user/lisa0rg/media/T2_zps18695702.jpg.html?filters[user]=141333040&filters[recent]=1&sort=1&o=0
I considered 300 and 750 N to simplify it.

This is very close, but it is even simpler than what you drew. There should only be three arrows on the diagram. The red vertical arrow is the force of gravity on the man, the blue horizontal arrow is the rope tension pulling on the man, and the green diagonal arrow is the ground reaction force on the man (it should be pointing up and to the right since the ground is pushing up and to the right). The extra blue and red arrows are unnecessary.

If these are torques , like in a pendulum?, they act tangentially or perpendicularly?
Please , anyone give me the correct values for the three notes ((1),(2),3)) this would help me most.

One of the reasons to simplify the diagram is that it makes calculating the torques much easier. If the system is in static equilibrium then the torque is 0 about any axis. So, to simplify things we can take the axis to be the center of gravity. All three forces pass through the center of gravity so their torque about the center of gravity is 0.

I how do I find the force acting on the pivot, here?

Since the man is in static equilibrium, if you know the tension, T=300, and you know the weight, W=750, then you can use Newton's second law, ∑F=ma with a=0 to find the reaction force, R, at the ground. So $W+T+R=0$ by Newton's second law. If you break that up into components then:

Vertical: $|W| \sin(-90)+|T|\sin(180)+|R|\sin(\theta)=0$
Horizontal: $|W|\cos(-90)+|T|\cos(180)+|R|\cos(\theta)=0$

That gives us two equations in two unknowns* $|R|$ and $\theta$. Solving you get $|R|=808 N$ and $\theta=68°$.

Can you, also, tell me how to use signs, shall I follow the cartesian axes or what?

As long as you are consistent in your convention it does not matter too much, everything will come out correctly. In the above I measured angles counter clockwise from the ground. So $\theta$ is a positive acute angle, gravity is acting at -90° and the tension is acting at 180°.

I could have picked any different convention for my angles, and as long as I used it consistently, I would have obtained the same results.

*note that you gave 45° as a known angle, but that makes the system overdetermined. If you specify the angle then the tension must be an unknown. You can only specify two parameters out of the set {θ, R, W, T}.

 

[CWatters]

I don't understand your diagram. Where does the -750 figure come from?

 

[CWatters]

Since the man is in static equilibrium, if you know the tension, T=300, and you know the weight, W=750...

If the 750 is the mans weight I don't think he can be in equilibrium.

The torques don't appear to sum to zero...

300*1*sin(45) + (-750)*1*cos(45) ≠ 0

PS: This is now the one leg problem isn't it? There is no rear leg to support part of his weight or have I misunderstood post #12 ?

 

[Dale]

If the 750 is the mans weight I don't think he can be in equilibrium.

The torques don't appear to sum to zero...

300*1*sin(45) + (-750)*1*cos(45) ≠ 0

PS: This is now the one leg problem isn't it? There is no rear leg to support part of his weight or have I misunderstood post #12 ?

Yes, this is the one-leg problem. bobie accidentally overspecified the problem, I assumed that the weight and the tension were known and the angle and ground reaction were unknown. Also, as drawn the torques are guaranteed to sum to 0, regardless of the angle and the reaction force since all three pass through the center of gravity.

 

[bobie]

Yes, this is the one-leg problem. bobie accidentally overspecified the problem, I assumed that the weight and the tension were known and the angle and ground reaction were unknown. Also, as drawn the torques are guaranteed to sum to 0, regardless of the angle and the reaction force since all three pass through the center of gravity.

Yes I considered the real weight of a man 75 kg.
I have put the right values for equilibrium.

T300

I am glad that we reached a first important conclusion. I added what is most important for me the x force on the foot :300N

Now this shows what I was not able to express clearly: that the force on the rope -300 cannot by itself for a force on the pivot of 424, which generates a counter force from the block of 300N,
The extra 124 must come from (300N) gravity .Therefore if we add the rear leg this must carry only 450N and the man system will be in equilibrium.
I hope I didn't get it wrong again. If that is right, can I add the rear leg? The problem is: where , at what distance /angle from CM must the rear foot be in order to carry 400N not 1N more not 1 less in order to be in equilibrium? I suppose the normal from CM must fall outside (behind) the foot in order to carry only 40Kg and excercise a net force of 300 on CM, but have no clue about which formula to use.

Thanks Dalespam, you know me, please tell me when you run out of patience!

 

[Dale]

I am glad that we reached a first important conclusion. I added what is most important for me the x force on the foot :300N

Now this shows what I was not able to express clearly: that the force on the rope -300 cannot by itself for a force on the pivot of 424, which generates a counter force from the block of 300N,
The extra 124 must come from (300N) gravity

I don't know where you got these numbers. The force at the ground is 808 N at 68°. The x component is indeed 808 cos(68) = 300 N and the y component is 808 sin(68) = 750 N. I don't know where 424 or 124 come from.

I showed you how to get the correct answer from Newton's 2nd law. Did you understand that? If not, please ask specific questions.

 

[bobie]

Is this what you meant?
If the forces were equal 300 and 300 we would have equilibrium, and applying the parallelogram I would find the force on the pivot = 424.

I returned to the case of equilibrium G = 300.
Specific question: the last image is fundamental, I 'll never change it if you confirm it's OK, now we need to know where to put the rear leg so that it carries exactly 450N, so that 300N go to compensate the pull.I supposed it must be on the left of the CM, so that the normal falls outside the body, is that right?

question: what is the formula to calculate the distribution of the weight on two legs? In which position of the rear leg /if gravity is 750 N) the system is again in equilibrium?

Thanks, consider that the lasr question.

 

[Dale]

I returned to the case of equilibrium G = 300.

I don't know what you mean by "returned to the case of equilibrium". Everything I did assumed equilibrium, so we never left the case of equilibrium. How are we returning to it when we never left it.

Equilibrium is defined by ∑f=0 and ∑τ=0. I solved for ∑f=0 above, as shown, and because all of the forces go through the same point ∑τ=0 is guaranteed.

If G is the weight then it seems like you are changing the weight from 750 N to 300 N, which is fine, but it would help if you would not just randomly change things. With a weight of 300 N then the force at the ground is 424 N at an angle of 45°. Is that what you had intended?

 

[sophiecentaur]

I don't know what you mean by "returned to the case of equilibrium". Everything I did assumed equilibrium, so we never left the case of equilibrium. How are we returning to it when we never left it.

Equilibrium is defined by ∑f=0 and ∑τ=0. I solved for ∑f=0 above, as shown, and because all of the forces go through the same point ∑τ=0 is guaranteed.

If G is the weight then it seems like you are changing the weight from 750 N to 300 N, which is fine, but it would help if you would not just randomly change things. With a weight of 300 N then the force at the ground is 424 N at an angle of 45°. Is that what you had intended?

Yes. I was a bit surprised by a reference to Newton 2, earlier on. There need be only vanishingly small acceleration. To get the 'just enough' condition- so you start to get motion, you just have to replace the Equation with an Inequality - giving the same numerical answer.

 

[CWatters]

Specific question: the last image is fundamental, I 'll never change it if you confirm it's OK.

I believe this the "last image"..

http://s47.photobucket.com/user/lis...[user]=141333040&filters[recent]=1&sort=1&o=0

This does not appear to be consistent.

The man cannot lean at 45 degrees and only produce a force of 300N.

Dale assumed that you had made a mistake on the diagram and in his post #127 he calculated the angle had to be 68 degrees.

Q: Do you agree with his calculation?

If not then why not?

It's not acceptable to keep changing the problem when you don't agree with the solution given for the previous version.

 

[CWatters]

question: what is the formula to calculate the distribution of the weight on two legs? In which position of the rear leg /if gravity is 750 N) the system is again in equilibrium?

I refer you to the basic procedure I outlined in post #107

I know you didn't like the assumptions I made (rear leg under COM) but the basic procedure is the same. Just modify it to suit your new version of the problem.

 

[bobie]

I refer you to the basic procedure I outlined in post #107
.

I have made a sketch of both the situation in equilibrium
http://s47.photobucket.com/user/lisa0rg/media/Double_zps8dcf5a18.jpg.html?filters[user]=141333040&filters[recent]=1&sort=1&o=0

The Fx (friction?) on the foot remains 300, should we conclude that whenever there is equilibrium, whatever the angle and whatever weight, it will always be 300? that is amazing.

I made also a sketch of the distribution of weight on each leg, and it seems that Friction is always the same (and opposite) at each foot, the weight here is 800N
http://s47.photobucket.com/user/lisa0rg/media/ManAstr_zps15a9bd01.jpg.html?filters[user]=141333040&filters[recent]=1&sort=1&o=1
Can you please answer these simple specific questions:

- what is called friction when the foot is not slipping but is firm against an obstacle.
- shall I call the friction that prevents the truck from moving stiction?
- the 300N required to move the truck are working only against stiction, since non-zero energy is required to move it is friction is 0

I hope I made it, at last.

Thanks, you have been really kind

 

[A.T.]

- what is called friction when the foot is not slipping but is firm against an obstacle.

The contact force component parallel to the contact surface is stiction (static friction).

- shall I call the friction that prevents the truck from moving stiction?

It's rather complex. Stiction in the bearings, rolling resistance due to deformation of tires.

- the 300N required to move the truck are working only against stiction, since non-zero energy is required to move it is friction is 0

When stiction breaks down, kinetic friction appears.

 

[bobie]

The contact force component parallel to the contact surface is stiction (static friction).
It's rather complex. Stiction in the bearings, rolling resistance due to deformation of tires.
When stiction breaks down, kinetic friction appears.

So 300N are necessary to overcome to stiction and rolling resistance. What is kinetic friction? does any of the force go into KE at all?
Are all three sketches all right, at last?