   Forces in supporting rope
  Hi everyone, Ive been wrestling with this problem for ages and I either just cant figure out the write answer, or have figured it out but dont understand how its possible.IMAGE1:IMAGE2:I have included the two diagrams above, I belive that it is IMAGE one that correctly anylises the model.The problem is if a mass(m) is droped, which is under the influence of gravity(g), what tension will it produce in the rope (which is inextendable) it is supported by which is attached to a fixed surface. My belief being that the imidiate tension in the rope produced by a "Jerk" will far exceed the tension in the rope thereafter where the tension in the rope will equal the gravitational force.My resoning for this comes from IMAGE1.fig1 shows the mass falling with acceleration g and the rope os slack. fig2 shows the point when the mass has a velocity of 0 m/s. I belive that at this point I am still required to include the force producing acceleration. The tension in the rope R must therefore be equal to the force of gravity on the mass plus the force producing acceleration giving these equations.R = mg + ma  = m9.8 + m9.8  = 2mgSo from this the initial force in the rope would be twice the force of gravity on the mass. If this is true I fail to understand why. If the velocity of the mass due to the acceleration was large I would have thought that the initial tention in the string would be greater then that if the mass had a lower velocity thus the tension in the string would be dependent on the displacment of the mass.I do belive that this is wrong, and I welcome as much detail to a solution as anyone can give, I so want to understand the nature of this problem, thanks![Edited by - Genesiiis on April 6, 2010 7:26:14 PM]
  If I didn't misunderstand you, your problem is that you assume the rope as completely inextendable.Thus, the mass falling with velocity v is stopped immediately by the rope.Now the force coming from stopping the mass equals m*a = m*(dv/dt).As dt->0, F diverges to infinity.So you can't actually compute the force under the assumption that the rope is completely inelastic as it would be infinite.;  Extendablity in ropes is what produces tension (which is a force in the opposite direction of the extending direction due to the "regaining of position force" of the inner molecules of the rope), so if you are considering a very stiff inextendable rope, then you have converted it to a cable, which has no effect when the objects don't exceed its length, but holds the objects from exceeding its length, not by forces (because in real life this situation doesn't happen, because cables aren't extended anyway), but by just editing their positions.The other case (which I think is what you are looking for) is that if you have a rope that can't be elongated but has some extendabilty to allow tension. In this case, if you take the equilibrium state at which the mass is suspended with no additional forces and it's velocity is equal to zero. The tension will have a magnitude of mg which is the mass of the object times the acceleration due to gravity but its direction is in the opposite direction of gravity. This tension will not move the object upwards but will help the object just to remain in air (as it's just opposing gravity). Any other additional forces that you apply to the body in the same direction of the gravity will be added to the magnitude of the tension. That builds up the equation to be mg+ma (magnitude only ofcourse). But just note that the applied force will not be exactly mirrored in the tension, only a fraction will be mirrored due to some losses of energy in the rope (in the form of heat and kinetic energy) so just a fraction determined by a constant will be added (this constant could be the elasticity constant which characterizes a string), thus the final equation will be: mg(to hold the body in air) + maK(where k is from 0->1 and determines how much of the applied energy is lost).Note: if you substitute with zero in k then all the applied force will be lost and the object will just stand in air (which is the case of a cable).HTH,Atef Hashem;Quote:Original post by AHashemExtendablity in ropes is what produces tension (which is a force in the opposite direction of the extending direction due to the "regaining of position force" of the inner molecules of the rope), so if you are considering a very stiff inextendable rope, then you have converted it to a cable, which has no effect when the objects don't exceed its length, but holds the objects from exceeding its length, not by forces (because in real life this situation doesn't happen, because cables aren't extended anyway), but by just editing their positions.Theres nothing magical about objects which cannot be extended (cables as you called them), they don't mysteriously edit the position of anything. The force applied when the ball falls to the end of a rope/cable which can't be extended will simply be a constant multiple of the dirac delta function in the opposite direction to that the ball was falling. In particular if an object with mass m is being acted on by gravity g and is falling at velocity v when it reaches the end of the rope/cable at time t0 then the force on the object as a result of the rope/cable will be F_br = -v * m * dirac(t - t0) - g * m * H(t - t0) and the force on the cable will be the negation of this. Hence the total force on the ball will be F_br plus the force from gravity or F = F_br + g * m.Note: H(t) is the heaviside function.Quote:Now the force coming from stopping the mass equals m*a = m*(dv/dt).As dt->0, F diverges to infinity.So you can't actually compute the force under the assumption that the rope is completely inelastic as it would be infinite.This is almost right, the force m*(dv/dt) does approach infinity at the point in time where the ball comes to a stop, however, this doesn't mean we can't work with the force. If we assume that the force is applied constantly over some time small time interval and that the ball comes to a stop at t = 0 then we get...F = { m * dv / dt    if -dt/2 < t < dt/2    { 0    otherwiseAnd tacking the limit as dt -> 0 this becomes m * dv * dirac(t) which although we still can't calculate the force does give us a nice representation for it and allows us to manipulate it and use it to calculate other quantities like the velocity1.1 This is actually a small lie, to compute the velocity requires integrating the dirac delta, however, the integral of the dirac delta in the traditional sense doesn't exist, we can however take the generalized integral which may or may not be acceptable depending on the problem domain.[Edited by - Julian90 on December 27, 2007 8:22:02 PM];  I was just assuming that he's doing this for a game, so the simplification of the problem was just to edit the positions of the object to avoid them from exceeding the hard constraint. But pure physics, I think you are right :)Atef Hashem   ;  Thanks guys for the replys. No i am actually doing this to furture my mathematical knowledge, but thanks AHashem, its just been puzzling to me for some time but Julian90 explaination clears things up for me thanks again!   