Third law of motion is so confusing?

Question

According to Newton's third law, whenever objects A and B interact with each other, they exert equal and opposite forces upon each other. I have always struggled with how to apply this law to problems and real life. >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. Isn't this a case where Newton's third law of motion fails? Or, does this happen because of the difference in our masses? 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). I have solved various pulley mass problems but I have not thought about applying Newton's Third Law of Motion to it. >Is tension the counterforce to weight? Consider a segment of rope from which a mass is suspended in Atwood's Machine. That segment will experience a force $m1g$ from the weight. 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. 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 ...can somebody explain this to me

Answer 1

Newton's 3rd law holds in both cases. In the first, when your friend exerts force A on you, you exert force B on them, where B=-A. But because the forces acting on you only include A and whatever force you apply to balance yourself, you will likely accelerate according to F=ma. Now your friend knows that he/she is about to exert a force, and having experienced Newton's laws, his/her body will naturally brace itself for the resulting force that acts on them (the reaction force). Thus the total force acting on their body is much less than the force acting on yours.

In the second question, the mass m1 experiences forces due to gravity (which m1 also exerts on the Earth) and the tension force from the rope. m1 exerts an equal force on the rope, but because we only care to consider forces acting on a particular object, we can set up our system: m1g-T=m1a; -m2g+T=m2a (notice I changed the signs to account for the different reference frame on the other side of the pulley). You could also find the forces acting on the rope, by each of the two masses, but that is not a very interesting result since the rope is assumed to be massless. Overall, the misunderstanding you seem to have is that you are considering the two forces as isolated and canceling each other out; however, the two forces are rarely isolated and act on two separate objects which causes both objects to accelerate according to Newton's 2nd law.

A good example to visualize this is to imagine a person laying on a flat bench and doing a bench press. Gravity exerts a force on the weight and the weight exerts the same force on Earth. The person exerts a force on the weight and the weight exerts the same force back to the person. The person exerts a force on the bench which exerts a force back on the person. And the bench exerts a force on the Earth which exerts a normal force back to the bench. So if you consider the one movement at constant speed, you have the weight feels gravity and the force of the lifter which has no net acceleration. The lifter feels the force of the weight pushing down (equal to the force exerted by himself), and the force of the bench which cancel each other out. The bench feels the weight of the lifter and the force he exerts on the weight and exerts that force onto the ground. The ground exerts a normal force which keeps the whole system balanced.