physics question - do forces propogate through a crowd?

[TuxDave]

Are you trying to solve it from a real life perspective or a "spherical cow" perspective?

If everyone had their arms out and locked and were pushing then larger crowd = more force, but in real life there's going to be a lot of elastic collisions so as RaistlinZ said there's probably a non-linear relationship with a upper threshold.

Don't we all live in a vacuum with no friction where everyone is a point mass?

 

[deadlyapp]

What your engineering friend said reminds me of something I read in the math section of Childcraft. Now I need to go dig up that book to see what it said.

Edit: Ok, work with me here.

1) Static model so nothing is moving
2) No one except the last person in the line is pushing with their feet (so no force is getting pushed against the friction of the ground). Everyone except the guy in the back is standing upright or whatever.

So if the guy in the back pushing with his feet is asserting force A, the guy he's pushing on is pushing back with force A and thus had to push forward with force A, etc.... all the way to the front.

Wouldn't that cause the guy in the front to only feel force A?

If there is only a single contact with the ground, only the force that can be made without the feet slipping on the surface can be transmitted. Ultimately the final person will not feel the force A because there is a certain amount of damping that the human body absorbs, therefore its likely that they will feel nothing.

This problem is relatively easy to envision. Each person in the crowd has their feet on the floor and can exert a maximum horizontal force equal to the force of friction on their feet, given by their weight and the coefficient of friction between their soles and the ground. If each person was solid and absorbed nothing then that force would simply add up and the first person would transmit all that force. The issue is that each person is also going to instinctively counteract the rear force, therefore diminishing their forward force.

Ultimately I bet you would find in a group situation that the person in the very back exerts all their force forward, and as you move forward in the group, less and less of their friction force is exerted forward, and is rather exerted backwards to a certain extent. Eventually, depending on the bottleneck, you may find that there is a point that there is no force in the forward direction by a person, and all their force is trying to counteract backwards.

Its a rather complex problem since each persons tolerance is different. I've been in mosh crowds where people will go limp and do nothing, people will push forward and not counteract at all, or people will use all their force to counteract the rearward force.

 

[gophins72]

Say a crowd of people is pushing on a large glass door (like a mall entrance), does the overall strength of the crowd increase as the crowd gets larger?

depends on if the crowd of people are zombies or not

 

[Kev]

depends on if the crowd of people are zombies or not

that's actually how we started this argument... talking about zombies

 

[OCGuy]

that's actually how we started this argument... talking about zombies

Zombies and physics is what you talk about with co-workers?

Please don't ever work in my office.

 

[zinfamous]

I'm not sure, but considering all of those dangerous free-floating microwaves running through that crowd, that amount of people pushing on one door would probably trigger the microwaves into starting a massive explosion.

 

[gorcorps]

Your friend will not be an engineer long.

 

[torpid]

Your friend will not be an engineer long.

Yup; he has politician or upper management written all over him.

 

[DisgruntledVirus]

What OP is trying to ask, is if the engine in a MMF 3-way train gets pounded twice as hard?

My gf would say no, since it's usually more of an eiffel tower in which case she acts more of a pendulumn suspended in the middle of the tower banging into the sides of it.

 

[IronWing]

Let's find out! When is The Who touring again?

 

[DisgruntledVirus]

OP, the engineer (which how he got a degree with this type of thinking scares me, but another topic) is wrong.

As others have said the force would not be linear and would be dampened by the bodies absorbing some of the force, but there would be a net gain from each person pushing in the same direction. If he doesn't believe this is true, a simple experiment can be done without any gauges even (although not scientifically accurate). Have him push against a wall just by himself. Then you, and add other people to really show the force added, push against him. If you have a force meter that he could push upon though that would be the only way to prove it to him.

 

[Kev]

Zombies and physics is what you talk about with co-workers?

Please don't ever work in my office.

did I say it was a coworker? Yay reading comprehension.

 

[Caveman]

Friend is wrong. There are losses of course but think about it this way... Next time you run out of gas and have to push your car, do you want to do it yourself or get some help? This "system" is more rigid than the one you propose but the principle is similar...

 

[DrPizza]

Haha I did! Then him and another guy ganged up on me in agreement.

Well, as a physics teacher, I suggest that you do a little experiment with your coworker to prove him wrong: have him hold up a book in front of his face. Extend your arms. Have the other coworker shove you from behind.

Alternatively, when a train is stopped and about to reverse, have him stand behind the last car (in the direction the train is going to go.) He should be able to stop the train, because there's no way the locomotive could exert a force through all the separate cars in the train.

So, either you're phrasing something wrong and they're correct on a technicality (somewhat likely), else your coworkers are too stupid to be engineers.

 

[thraashman]

As someone else said F=ma (force=mass x acceleration). Now, obviously that formula doesn't work directly as the entire mass isn't focused upon the same single point, but the idea is there. The mass is gigantic in a crowd, but the acceleration is less. Click on the link someone posted I think it has the correct application of that formula in it.

 

[gophins72]

i find that in engineering discussions, those who speak loudest are usually the most correct...


...or something

 

[torpid]

Football and Rugby regularly disprove this "engineer" as well.

 

[TecHNooB]

It's really important to understand what he means by 'front' of the crowd.

 

[CLite]

Well how the strength of the crowd gets transmitted depends on if the people are squashed against the doors or still pushing with their hands.

If the people are still pushing with their arms then the only force being transmitted is what is being transmitted through their arms, so the entire force against the door could be considered the front people's exerted strength. They are able to exert more force because instead of depending on friction from their feet to anchor their push they have people pushing on their backs to help stabilize them, but ultimately the front is what is transmitting the force so technically it is what they are exerting. If they are unable to handle this force and achieve an equilibrium with arm strength alone then the next scenario arrives (squashed body).

In squashed body scenario the front people's bodies are just pressed flat against the wall and they clearly aren't "exerting" any force with their muscles, they are just acting like a medium for the force from the crowd.

So your friend is correct in an arm pushing situation and incorrect in squashed body situation.

 

[eLiu]

Q: Why did ancient infantry stand many-ranks deep? Ancient = greek hoplites, roman legions, various medieval era units, etc.

Do you think they did that so that it would look epic in our video games...????

The people in the back supported the people in the front. And no, it wasn't just a matter of getting more spears within poking range of the enemy--these formations were much deeper than the length of a spear/halberd/whatever. If you watched '300', they "demonstrated" the value of deep ranks; by "demonstrated", I mean 'severely exaggerated' but the idea is there.

edit:
This is the same idea as how solids/liquids/gasses transmit force through them. The behavior of a densely packed crowd of people is somewhere in the solids/liquids range, so force will definitely propagate through.

 

[CLite]

As someone else said F=ma (force=mass x acceleration). Now, obviously that formula doesn't work directly as the entire mass isn't focused upon the same single point, but the idea is there. The mass is gigantic in a crowd, but the acceleration is less. Click on the link someone posted I think it has the correct application of that formula in it.

You would want to use F=-kx and Ff=u*Fn, as F=ma doesn't really apply here (Technically F=MA would apply in this case as it does for any solid mechanics problem, although not in the way you described it, but it's easier to start at higher level equations).

The crowd can be locked rigidly and still transmit force to the door. The total force would depend on their individual muscles which can be modeled as a linear springs, their weight, and the coefficient of friction of their footwear on the ground.

 

[Hacp]

This question really got me thinking since it seems like a simple problem to solve if you make enough assumptions. So I made a few outrageous assumptions and here is what I got.

A few assumptions is that there are no "viscous" effects, that the act of a person pushing someone in front will not affect the person to his right or his left. Another assumption is that everyone has the same strength, thus can push with the same force. This makes the problem pretty straightforward.

Edit: Someone pointed out to me that the summation was a geometric series which could be approximated in an even simpler way. Of course I had forgot all about geometric series but I uploaded the more accurate solution.

As you can see, your engineering friend has a point. When the number of people gets really large, the nudge of the guy in the back won't be felt at all, practically speaking.

 

[TecHNooB]

This question really got me thinking since it seems like a simple problem to solve if you make enough assumptions. So I made a few outrageous assumptions and here is what I got.

A few assumptions is that there are no "viscous" effects, that the act of a person pushing someone in front will not affect the person to his right or his left. Another assumption is that everyone has the same strength, thus can push with the same force. This makes the problem pretty straightforward.

sum( F(a,b,c,d) * u(e,f,g,h,i,j,....,z)^n )

fixed!

 

[DrPizza]

As someone else said F=ma (force=mass x acceleration). Now, obviously that formula doesn't work directly as the entire mass isn't focused upon the same single point, but the idea is there. The mass is gigantic in a crowd, but the acceleration is less. Click on the link someone posted I think it has the correct application of that formula in it.

That formula is f_net =ma. If the doors don't accelerate, then the net force acting on them is zero. That doesn't mean the crowd's force = zero though; it means that the force exerted on the doors by the door frames, building, etc. = the force of the crowd acting on the door.

 

[zinfamous]

Football and Rugby regularly disprove this "engineer" as well.

I also think about the strength in the Spartan Phalanx

OP should tell the engineer to watch 300 b/c that's like...well, totally real.

:hmm: