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enter image description here

The reasoning for this question comes out of pure curiosity. say if there were, for example 4 bugatti veyron's ( represented by the rectangles on the diagram) all starting from the middle and all attached by rope, and they accelerate out at 90 degrees to each other, reaching a top speed of say 200mph (89.4m/s) before the rope becomes taught (diagram in top right corner) what kind of force will the centre point experience and also the force the driver. assumptions can be made regarding the distance and time, also assuming there are no material limits and ideal situations meaning in the end the centre point does not move and rope joints are infinitely strong. id just like to see how the newtons laws could be applied to a question like this.

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    $\begingroup$ Why specify the type of vehicle let alone the brand? It doesn't seem relevant. $\endgroup$
    – JMac
    Apr 3 '17 at 11:11
  • $\begingroup$ @Jmac May be that, OP is a big-fan of Bugatti ;) $\endgroup$ Apr 3 '17 at 13:10
  • $\begingroup$ @RaajaG Still doesn't show why it's relevant. Just seems like a lot of irrelevant details. I'm trying to see if they are relevant. $\endgroup$
    – JMac
    Apr 3 '17 at 13:22
  • $\begingroup$ @JMac use of this particular vehicle to specify the mass needed for the question also, why not :) $\endgroup$
    – Mcccccc
    Apr 3 '17 at 13:34
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    $\begingroup$ @Mcccccc Are you concerned about this exact scenario with 4 Bugatti Veyron's tugging on cables? In that case, we probably want more information on the surface conditions and such to determine the true output on the rope. If you're concerned about the general case of 4 ropes being tugged in differing directions, it doesn't matter what is tugging the ropes, just the applied force (which is perfectly clear as just a number, doesn't matter if it's a Bugatti or a winch). $\endgroup$
    – JMac
    Apr 3 '17 at 13:37
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The force on each car is purely longitudinal because the restraining force of necessity runs thru the central point (origin). For example, the East and West forces cancel so far as the origin is concerned, so no lateral force exists from the viewpoint of the North or South car.
The restraining force on each car, then, is the same as if it were tied to an unmoveable tiedown point-- the force thus is the torque currently being applied via each car's engine and drivetrain.

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  • $\begingroup$ I was thinking that the force will come from the momentum of the vehicles reaching the specified velocity and then the moment the rope becomes taught the rate of change of that velocity is what will cause the overall force? am i wrong making this assumption? @Carl Witthoft $\endgroup$
    – Mcccccc
    Apr 3 '17 at 14:03
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The question assumes no material limits, but specifies the very fine Bugatti Veyron, which is made of specific materials. Let's just take two cars driving apart at high speed, joined by an infinitely strong rope. Assuming 89.4m/s and a mass per car of 1,888kg, the total energy of 2*1/2 mv^2 = 15MJ.

If the cars were to come to a halt, this is the amount of energy which would have to be dissipated. This energy could be go into the elongation of the ropes, or more likely into the deformation of the cars. Depending upon what the rope was attached to, either the cars would be ripped apart until the requisite amount of energy has been absorbed by the structure of the car, or the ropes will detach themselves from the car due to local failure of the connection point.

This question is more about energy than balancing forces.

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With the information and assumptions you have given, I would say that the forces act in a hyperbolic fashion toward each other, creating a saddle region at the center point as in your figure. To have a rough look see the image below, in which the red arrows indicates the forces acting in a hyperbolic fashion around the center location creating the above-mentioned saddle region.enter image description here

Hope this helps!!

Edit 1: As J.Ari mentioned, only the net force at (0,0) will be 0N, all the others will behave hyperbolically.

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    $\begingroup$ Won't there be a net force of 0N at the center point at all times because each car's force is being cancelled out by the car going in the opposite direction generating equal force? $\endgroup$
    – J. Ari
    Apr 3 '17 at 12:05
  • $\begingroup$ @J.Ari It's a tricky argument :), the force will be 0N only at origin that is at (0,0), if and only if, all the cars are at rest i.e, when they aren't pulling, if not the net force will behave hyperbolically, whenever the cars are not at rest. I have already mentioned the (0,0) with a red-dot :) $\endgroup$ Apr 3 '17 at 13:06
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    $\begingroup$ There's no such thing as a "hyperbolic force" Force is a vector and can only point in one direction at a given time. $\endgroup$ Apr 3 '17 at 13:43
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    $\begingroup$ @Mcccccc That would require infinite force. $\endgroup$
    – JMac
    Apr 3 '17 at 14:26
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    $\begingroup$ @RaajaG It's very unclear what you mean. The ropes should all have tension in them between the car and whatever is connecting the ropes to each other. For rigid bodies and ideal ropes each rope should have a single value of tension acting in the direction the rope is being pulled. You seem to be drawing odd things and not explaining them clearly. $\endgroup$
    – JMac
    Apr 3 '17 at 14:55

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