![]() ![]() When you pull your friend forward although the forces between you and your friend are equal in magnitude what about the force that you bother experience due to the Earth?Īssuming that you had equal masses if net force on your friend is greater than the net force on you then your friend will undergo a greater acceleration. The motion of you and your friend depends on all the forces which are acting on you you and your masses. Obviously, there will be no tension force in case that the second point is not given. While the first point is always true, the second one doesn't have to be true, say, if the rope and the weight are in a free fall accelerating towards the earth, rather than in an equilibrium situation. Here we are looking at the forces on one object exercised by all other objects it interacts with. The weight and the rope (and thus each segment of the rope) are in a force equilibrium, so the total force on each segment exercised by both adjacent segments totals to zero.Here we are looking at forces acting on different objects interacting with each other. Each segment of the rope acts on any of the adjacent segments with the same force as the adjacent segment acts on it.In the rope example there are two things playing a role and they should be kept apart: But the parts which come due to you two interacting with each other will still be equal. ![]() If further forces like friction are present, then the total forces acting on each of you might differ. ![]() If both of you are in vacuum and no further forces are present, the change in momentum will be equal for both of you. If you and your friend are interacting, when he pulls you, he will feel that you apply to him the same force as he applies to you. If this is correct then I can see the uniform tension around the rope as an application of newton's third law but the fact that this resulted from the weight negates the third law and also this will mean that m1=m2 where m1 and m2 are the masses at each segment People say this weight is the reason for the counterforce in that segment, and that this force will travel all the way through the rope to the other segment resulting in a uniform tension. That segment will experience a force $m1g$ from the weight. Is tension the counterforce to weight? Consider a segment of rope from which a mass is suspended in Atwood's Machine. I have solved various pulley mass problems but I have not thought about applying Newton's Third Law of Motion to it. My second question is about the tension in the rope in Atwood's Machine (two unequal masses connected by a rope on either side of a friction-less pulley). Isn't this a case where Newton's third law of motion fails? Or, does this happen because of the difference in our masses? Suppose I get a pull from my friend with some force, then I get pulled forward by a large distance as compared to the distance covered by my friend. I have always struggled with how to apply this law to problems and real life. According to Newton's third law, whenever objects A and B interact with each other, they exert equal and opposite forces upon each other. ![]()
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