Forces involved in a 'tug of war'

[A.T.]

What is kinetic friction?

http://en.wikipedia.org/wiki/Friction#Kinetic_friction

But when you think about bearings, they actually use rolling, so it's more rolling resistance due to deformation, rather than kinetic friction.

does any of the force go into KE at all?

When the truck accelerates at v>0, yes.

 

[bobie]

http://en.wikipedia.org/wiki/Friction#Kinetic_friction
When the truck accelerates at v>0, yes.

If memory serves, in another thread wasnt't it you or as it another who maintained that if there is no friction I can move anything with 'negligible' energy?

 

[A.T.]

If memory serves, in another thread wasnt't it you or as it another who maintained that if there is no friction I can move anything with 'negligible' energy?

If there is no friction (or other resistance), moving something at constant speed doesn't require any energy input.

 

[CWatters]

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.

As stated, in horizontal equilibrium the horizontal components sum to zero. The only two horizontal components are the 300N force at the rope and friction. So these must be equal and opposite. If they aren't the same then there would be a net horizontal force on the man and he wouldn't be in equilibrium.

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...

For the same reason as above. If they were different there would be a net horizontal force on the man and he would start accelerating! Clearly that cannot happen if the only horizontal forces are due to friction (Friction can't be a source of energy).

 

[A.T.]

Friction can't be a source of energy.

Friction with a static object can't be a source of energy. In a reference frame where the ground moves, friction with the ground can do positive work.

 

[jbriggs444]

Friction with a static object can't be a source of energy. In a reference frame where the ground moves, friction with the ground can do positive work.

The above is certainly correct.

If one examines the sum of the work done by each object on the other in a pair that are interacting only with friction then the total [center-of-mass] work done across the interface can only be zero or negative.

I think that this is the sense of "can't be a source of energy" that CWatters had in mind.

 

[Dale]

http://s47.photobucket.com/user/lisa0rg/media/Double_zps8dcf5a18.jpg.html?filters[user]=141333040&filters[recent]=1&sort=1&o=0

These seem fine to me. What they show is that a one legged child (weighing 300 N) has to lean back much further than a one legged man (weighing 750 N) if they each want to exert 300 N tension in the rope. That makes sense and goes along with what you would intuitively expect.

The only potential suggestion I would make is to somehow indicate that the 300 N horizontal force (black arrow) is a component of the ground reaction force (green arrow) and not a separate force. Usually, when I was doing free-body diagrams for homework, I would indicate that by placing the back of both arrows at the same point and making a dotted line from the tip of one arrow to the tip of the other arrow (forming a little triangle with the two vectors). That is not essential, as long as it is understood, but just a suggestion to keep in the back of your mind for future free-body diagrams.

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.

That is guaranteed by the equilibrium condition that ∑f=0. It is not a general rule of nature, but an assumption of the problem.

 

[bobie]

Usually, when I was doing free-body diagrams for homework, I would indicate that by placing the back of both arrows at the same point and making a dotted line from the t.

You mean something like this:
"Force of inertia" is not a commonly used term. Furthermore, it is unclear if there is some physics concept you are actually asking

If you think some sketch are useful for future viewers, please attach them to relative posts, as I am going do delete my account at bucketshop, and I do not know how to do it. When I clik 'image' they give me no choice to apload from my PC.

 

[Dale]

You mean something like this

No, I mean on the drawing, indicating that one arrow is not a separate force, but just an identified component of another force by something like this.

 

[bobie]

No, I mean on the drawing, indicating that one arrow is not a separate force, but just an identified component of another force by something like this.

I posted wrong link , I mean something like this (this is the final sketch)
http://s47.photobucket.com/user/lis...[user]=141333040&filters[recent]=1&sort=1&o=0

 

[bobie]

Friction with a static object can't be a source of energy. In a reference frame where the ground moves, friction with the ground can do positive work.

I made a sketch examining the 3 possible positions. Can you give it a look?

three positions

I am not sure what happens in sketch C:
if the weight 600N is all on the right leg, there is no component on the front foot, both diagonally and horizontally. Of course there is not equilibrium. But what really happen?

I imagine that the pull on the rope will generate a torque and the man will move up/left in a circle, but immediately his weight will bring him down. Will he oscillate or what? it's really intriguing.

How do I attach a thumbnail?

Thanks

 

[A.T.]

I made a sketch examining the 3 possible positions. Can you give it a look?

Try putting less arrows into one diagram.

if the weight 600N is all on the right leg, there is no component on the front foot, both diagonally and horizontally. Of course there is not equilibrium. But what really happen?

I see no horizontal component on either leg, so the rope will pull him left, while he apparently glides on ice.

 

[bobie]

Try putting less arrows into one diagram.I see no horizontal component on either leg, so the rope will pull him left, while he apparently glides on ice.

There is no ice, there is a block before his foot, he will rotate anticlockwise, but as soon as his rear foot is not touching the ground any more won't he be pulled back?
The horizontal component shold be -212, as in the sketch below

 

[A.T.]

There is no ice, there is a block

If the block doesn't exert any horizontal force, then it's same same as if it wasn't there, just ice.

 

[bobie]

If the block doesn't exert any horizontal force, then it's same same as if it wasn't there, just ice.

I didn't put a figure because the block exerts any force up to 100000...N., reacting in proportion to any F-x force

 

[A.T.]

I didn't put a figure because the block exerts any force

That is one arrow that should be there, unlike many of the others.

 

[Dale]

All of these recent diagrams are wrong. They all contain forces acting on other objects. These diagrams should only contain four forces: the weight, the pull of the rope, and two ground reaction forces (one for each foot). You could break the ground reaction forces into horizontal and vertical components if desired, so there should be a maximum of 8 arrows.

I have told you this many times. It is extremely frustrating.

 

[bobie]

You could break the ground reaction forces into horizontal and vertical components if desired, so there should be a maximum of 8 arrows.

I have told you this many times. It is extremely frustrating.

I hope you can check last sketch, I followed your instructions and an example by Sophiecentaur:

Forces 4-8 ...

In the sketch on the left I drew forces from 1 to 4 and on the right: from 5 to 8, I numbered all other forces so that you can tell me quickly which I am allowed to include besides the 8.

As the rear leg is under the CM I could not put a figure to 7 and 8, I suppose those components are 0?. Are the other numbers right?

So 1-4 are obligatory, 5-8 optional.
I am sorry to appear dumb, but, can you explain why is it forbidden to add more forces,( breaking also the forces that go toward the ground in horizontal and vertical components?) Force number 9 is precious, it helps visualize the components of the reaction force.

Here there is no balance, the weight is all on the rear leg and exceeds the pull by 300N, but there is no torque because the pull in excess is absorbed by the rear leg. Right?

If I want the train to move, what shall I do, shift the rear leg to the left, so that the CM is out of the body? But too much would be dangerous?

Thanks for your patience if I made it well this time, I'll not ask other questions.
Thanks

 

[sophiecentaur]

eight+ forces

I hope you can check last sketch, I followed your instructions.
I have numbered the forces from 1 to 4 and from 5 to 8, and numbered all other forces. As the rear leg is under the CM I could not put a figure to 7 and 8.

I am sorry to appear dumb, but can you explain why is it forbidden to add more forces, breaking also the forces that go toward the ground? They help me calculate the whole reaction. Here for example , as there is not equilibrium, I cannot use the equations you gave me, neither can I use pythagoras √300²+600²

Here it seems that there is no balance, the weight is all on the rear leg and exceeds the pull by 300N, what happens?

I would like to think that you have looked elsewhere at how this sort of problem is tackled. If you have, you will notice that forces are not drawn out in the middle of nowhere - as in your picture. A force is normally drawn, acting on or through an object.
On your picture, the horizontal forces do not appear to add up to zero. If that were the case then the system would not be in equilibrium. You are imposing values on forces when you should, in fact, be calculating them, starting with the forces and distances that are actually known. Start with the weight of the man and the tension in the rope. Everything else comes from those two. If you don't want to use Maths to solve this then you are really going to get nowhere and, as we can see, you just end up with a verbal discussion which doesn't help towards a solution.
Have a look at the thousands of examples of free body problems and how they are solved. This is an example.

 

[bobie]

, you will notice that forces are not drawn out in the middle of nowhere -ich doesn't help towards a solution.
Have a look at the thousands of examples of free body problems and how they are solved. This is an example.

I did not put the forces where they belong, I grouped them in forces from the pull (blue) and weight (red). When I put them in proper place they said it was too crowded. That is a skecth just to check that I got right the 4 obligatory forces and the 4 optional ones, but here I cannot place the components for the reaction of the ground to weight.

 

[sophiecentaur]

I did not put the forces where they belong, I grouped them in forces from the pull (blue) and weight (red). When I put them in proper place they said it was too crowded. That is a skecth just to check that I got right the 4 obligatory forces and the 4 optional ones, but here I cannot place the components for the reaction of the ground to weight.

Just look at the diagrams in the link I gave you. That is the classic way to draw such diagrams. Narrow lines with arrows for forces, for instance. The part of the object where the force acts is very relevant (moments and all that) so it should be made clear where a force acts. The simple situation you are dealing with should never involve a diagram that's "too crowded" if you use the convention thin lines and single character labels.

Why not? the force on the ground (or the components) all act at his front foot in the simplest situation. If you want to include the back foot (just for the purpose of representing the body more accurately) you can assume that, at equilibrium, the vertical force on that foot will be zero - or it will be acting to subtract from the useful force he is applying to the rope. If he chooses to lean back far enough so that he needs support, then he will be using more force than necessary with his front foot. This would increase the required friction force on the ground and put more stress on his leg. Balanced is best value - as the Maths will show when you write out the equation with all possible forces included. Many problems like this can be made easier by a bit of pre-thought and some variables can often be omitted from the start. You have to be careful, though, when doing this and not make wrong assumptions or you can miss out something important.

 

[bobie]

I have modified the drawing according to your example, the also use arrows, smaller than mine.

 

[Dale]

I hope you can check last sketch, I followed your instructions

No, you didn't. You just went off and did whatever you felt like doing, as always.

I am sorry to appear dumb, but, can you explain why is it forbidden to add more forces

On a free body diagram you only draw the external forces acting on the body. There are only 4 external forces acting on the man. Any additional forces you might add are either internal forces or they act on other objects. Those forces do not belong on a free body diagram.

bobie, I am done for this round. I wish you luck, but I simply cannot accept working with someone who makes me repeat the same instructions as many times as you do. It is beyond irritating. I have given you essentially the same message from post 90 to post 157 which you have consistently ignored. I will leave with one last repetition of the instructions (which I am sure is in vain) and a free-body diagram showing the correct setup.

On a free body diagram you draw only the external forces acting on the body, no internal forces and no forces acting on other bodies. Additionally, if you break a force up into components then it is important to indicate on the diagram that the components are not additional forces, but merely components of an already indicated force.

I will post a diagram after I work out.

 

[bobie]

Thanks, I hope you believe me I am sincerely sorry, if I irritate you, but, honestly, I yhought that in the drawing on the left I did what you asked.

 

[Dale]

Here is the diagram. This is the maximum number of forces (4) and the maximum number of arrows (8) that should be on any of these diagrams. For a one-legged problem the R force and its corresponding components are removed (leaving 3 forces and a maximum of 5 arrows).

bobie, it doesn't matter if you honestly believed that the drawings you posted were correct or not. The fact that they were not after so many pages of posts is a serious problem whether it is intentional or not. I am done. Best of luck.

 

[sophiecentaur]

Thanks, I hope you believe me I am sincerely sorry, if I irritate you, but, honestly, I yhought that in the drawing on the left I did what you asked.

I think your problem is that you want to get the whole of 'Mechanics' by just this one example, rather than learning each step that's needed. You have to appreciate the effect of multiple forces, acting at a point and then you need to grasp the principle of moments. I am not too surprised that you are not making progress with this calculation because you do not seem prepared to go through the process of learning the subject from scratch. Physics doesn't work like that; it needs rigour. That's a point that many people seem not to realize.
If you had spent the time working at the subject rather than causing the generation of more than a hundred posts, you would have been home and dry by now.