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Here is a mock scenario I have to illustrate my confusion,the regular basic assumptions here apply ie smooth floor so there is no friction and air resistance, and a inextensible string . There is a force of 60 N acting on block 2, the tension, however, will be a lot lot lower than 60 N ie the force pulling block 1 would be a lot lower than 60 N, and i only know this purely from common sense but I don't know the intuition behind it, i don't think F=ma isn't really an intuitive explanation for it either,more so a mathematical proof behind the intuition.

I initially thought, well the tension is low because only a low force is required to get block 1 to the overall acceleration of the system, but that's basically another way of saying $F=ma$ and I'm also kind of suggesting that the tension is due to the acceleration ,almost like the tension is caused by the acceleration ie acceleration came first and tension is trying to find the right force to cause the block to match the same rate of acceleration of the whole system , but that's obviously not the right chain of events.

My second thought of intuition was the tension force is distributing the force on both objects based on mass such that the ratio of the mass to resultant force on both objects should be equal. But I'm not sure if this is a good intuitive explanation as to why the tension is very low I see the formula F=ma as proof to the intuition of why the tension is very low , my issue is I don't fully understand the intuition ie its not obvious to me , if someone could provide a better explanation id be grateful.

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    $\begingroup$ This is obviously a conceptual question. Please do not vote to close it as a "homework-like question or a check-my-work question focused on a specific computation instead of the underlying physics concepts". $\endgroup$
    – PM 2Ring
    Commented Mar 2 at 17:08

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Be careful with "acceleration X is due to force Y" or "force Y is due to force X." In many cases you can get away with it, but not always. I recommend remembering your F=ma equation with one additional symbol: $\Sigma F=ma$. The sum of the forces on an object is proportional to the acceleration of that object (the constant of proportionality being $m$).

This summation is important. In your example block 1 has exactly one force applied to it, which means you kinda-sorta get away with saying either its acceleration is due to the force of tension or that the force of tension is due to its acceleration. However, in the example of block 2, there are two forces. You have the tension force and the 60N pulling force in the opposite direction. In this case you cannot say the tension is due to the acceleration or the acceleration is due to the tension (or the acceleration is due to the pulling). If you really wanted to say something like that, you'd have to say "the acceleration is due to the sum of the effects of the 60N pulling force and the tension force."

This is really important in your example, because the best place to crack this intuition you have about the tensions is by looking at block 2, where the "due to" notation sort of gets in the way. It's better to think of it the way Newton thought of it, as "proportional to" rather than "due to." The accelerations and sum-of-forces are proportional. The accelerations and sum-of-forces go together. Where one goes, the other goes too, and vice versa.

So how do we see the intuition? Well, intuitively the blocks must be accelerating. We'll back up that intuition in a moment, but your wording implies you intuitively understand that there's acceleration here. $\Sigma F=ma$ has a very intuitive implication. If acceleration is non-zero, the forces must be unequal. If acceleration is non-zero, the sum of the forces must be non-zero too (numerically, they must sum to $\frac a m$).

And thus you have an intuitive argument for why the tension is not equal to 60N. For block 2 to be accelerating, the forces must be unequal. T cannot be 60N. And, in fact, since our intuition should tell us that the blocks are accelerating to the right, the sum of the forces must be to the right as well, so the tension must be less than 60N.

From there its probably a good idea to start really crunching the numbers with $\Sigma F=ma$ on both block 1 and block 2, and you can solve those to figure out what the actual tension force is. But the above argument indicates at the very least that it's less than 60N

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  • $\begingroup$ so would you say my comment about the ratio of the mass and resultant force of each object being equal , is wrong ? Additionally I just find using ΣF=ma as a form of intuition akin to "what came first the chicken or the egg", ie if I did not know about ΣF=ma how would you show me (not in a mathematical proof sense, more like "common sense" )that it makes sense that the tension in the string would be much lower than 60N ,I don't know like by paralleling into some in real life scenario? etc. $\endgroup$
    – j jose
    Commented Mar 2 at 15:40
  • $\begingroup$ I think I may have skimmed over that part, so sorry for not addressing it. If you define the "resultant force" as the equivalent to the sum of all forces on the object, then your statement is correct. I'm a bit hesistant on the notation only because if you replace "sum of forces" with the "resultant force," it also obscures the next step you need to come to the conclusion that the tension is lower than 60N. You need the fact that you are summing the forces to get to the next step. If your mind can comfortably make that leap then, "resultant force" works fine. $\endgroup$
    – Cort Ammon
    Commented Mar 2 at 15:44
  • $\begingroup$ Because right now the entire intuition behind why the tension is really low is based solely on ΣF=ma, but surely someone would know that the tension would be much lower even without knowing of the formula ΣF=ma?In my head ΣF=ma is proof of a already preexisting notion/intuition(which i dont know) like most things in physics $\endgroup$
    – j jose
    Commented Mar 2 at 15:46
  • $\begingroup$ And actually it is somewhat circular. If you look at how Newton constructed his laws of physics, he defined forces to be something with certain properties on objects (such as causing acceleration). He invented the concept (based on less formal constructions others had, of course). So it becomes a bit circular because you started with a pulling force defined to be 60N, and doubling back on its effect on the objects. $\endgroup$
    – Cort Ammon
    Commented Mar 2 at 15:47
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    $\begingroup$ @SolomonSlow Agreed, although here that gets in the way, and I've seen it get in the way before. It's a bit counterintuitive to combine all of the forces into one net force in one step, and then the only way to proceed in the next step is to immediately break them back apart, to get back to exactly where we started. $\endgroup$
    – Cort Ammon
    Commented Mar 2 at 16:44
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I'm not really answering the question that you asked—Cort Ammon did a great job of that—but here is, maybe, a different approach to solving the original problem.

The two blocks must move together like one object, and their total mass is 101kg. I would calculate the acceleration, $a$, that would be imparted to the 101kg mass by a 60N force. Then, I would calculate, what force, $f_t$, would be required to impart that same acceleration, $a$, to just the 1kg mass.

$f_t$ must be the tension in the string because the tension in the string is the only force that accelerates the smaller object.

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Your initial thought is correct. It is based on the fact that the mass of block 2 is much less than block 1 so it requires much less force to give it the same acceleration as block 1, as they both must have the same acceleration since they are connected together by the string.

But you should verify this by applying Newton’s second law first to the combination of the two blocks and then to block 1 only where the only horizontal force acting on it is the tension in the string.

Hope this helps.

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