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No. That equation says -10N is the force A is exerting on the ground. Your equations lead to A exerting a force -15N on the rope and B exerting a force +15N on it.kkps said:Ft-10N=1*a
No. That equation says -10N is the force A is exerting on the ground. Your equations lead to A exerting a force -15N on the rope and B exerting a force +15N on it.kkps said:Ft-10N=1*a
10N<-A---ROPE---B->20Nkuruman said:You realize that thread is 2 years old. That said, your free body diagram is incorrect. If the rope is the only entity that exerts a force on A and B, the direction of the force on A is from A to B and the direction of the force on B is from B to A. These two are equal and opposite according to Newton's third law.
please see my answer to kuruman. A is pulling on rope with 10N to left, B is pulling on rope with 20N to right. we do not have to add Ground in picture, A is somehow generating 10N, B 20N. Then we can translate the question to an Atwood machine problemharuspex said:No. That equation says -10N is the force A is exerting on the ground. Your equations lead to A exerting a force -15N on the rope and B exerting a force +15N on it.
My answer is treating the problem like an Atwood machine problem. I think I forgot to mention that I am treating the string as massless. I am working on a simple explanation for the "string with mass" case. I think this thread is a good interaction, though there is lot of unnecessary confusion in the discussion. But the original question is a doubt anyone could havePeroK said:You should open a new thread. Your answer makes little sense to me.
Which you've just added to! Your solution is conceptually wrong, as it violates Newton's third law, which I think has already been pointed out.kkps said:there is lot of unnecessary confusion in the discussion.
can you explain why this question cannot be simplified to an Atwood machine question as I pointed out in reply to kuruman? After that, can you explain why my solution is conceptually wrong and how it violates Netwon's third law?PeroK said:Which you've just added to! Your solution is conceptually wrong, as it violates Newton's third law, which I think has already been pointed out.
B cannot be pulling on the string with ##20N## and the string pulling back with ##15N##. That violates Newton's third law.kkps said:what force is not balanced violating third law?
That is the proposition in post #1. It appears to be a situation invented by the student, not a problem he or she was given.kkps said:A is pulling on rope with 10N to left, B is pulling on rope with 20N to right
In post #68 you described your equations as based on free body considerations for A and B. Necessarily that means your are considering all horizontal forces on those bodies, namely, the force from the ground and the force from the rope.kkps said:we do not have to add Ground in picture
to make it simple, translate the question to this:PeroK said:B cannot be pulling on the string with ##20N## and the string pulling back with ##15N##. That violates Newton's third law.
you are trying to prove that the question makes no sense, but the question makes a lot of sense to the original poster. I am trying to understand what the original poster meant, and answer the intent of the original poster.haruspex said:That is the proposition in post #1. It appears to be a situation invented by the student, not a problem he or she was given.
If the rope has zero mass then it is not physically possible, and if you start with an impossible combination of facts you can deduce anything.
please see post #80haruspex said:In post #68 you described your equations as based on free body considerations for A and B. Necessarily that means your are considering all horizontal forces on those bodies, namely, the force from the ground and the force from the rope.
If your FBD has the force of tension and the force of a person pulling, what is the body those two forces are acting on?
You seem not to understand the concept of an FBD.
If the original poster intends a massless rope then the question posits an impossible scenario according to Newton's second law. ##\sum F = ma##. ##m = 0## so ##\sum F = 0##.kkps said:you are trying to prove that the question makes no sense, but the question makes a lot of sense to the original poster. I am trying to understand what the original poster meant, and answer the intent of the original poster.
Please don't!kkps said:if you add mass, your answer is not going to be much different . I will post that answer soon
Indeed, it is not. Gravity pulls on B with 20N, but if B is accelerating at 5m/s2, B only pulls on the string with 15N, same as the tension.kkps said:B cannot be pulling on the string with 20N
Well, it would be interesting to see that, and take the limit as the rope's mass tends to zero.PeroK said:Please don't!
For ##m_1 = 1## kg and ##m_2 = 2## kg, the acceleration is $$a=\frac{m_2-m_1}{m_2+m_1}g=\frac{1}{3}g$$and the tension is 13##\frac{1}{3}## N as indicated in post #85.haruspex said:Indeed, it is not. Gravity pulls on B with 20N, but B is accelerating at 5m/s2, so B only pulls on the string with 15N, same as the tension.
Thanks. I made the mistake of assuming @kkps's numbers were correct, just the interpretation that was wrong. I've edited it to avoid making that assumption.kuruman said:For ##m_1 = 1## kg and ##m_2 = 2## kg, the acceleration is $$a=\frac{m_2-m_1}{m_2+m_1}g=\frac{1}{3}g$$and the tension is 13##\frac{1}{3}## N as indicated in post #85.
Thanks for pointing out the mistakekuruman said:Let's examine and interpret the equations in your drawing. You have
View attachment 316230
Equation 1: Ft - 10 = 1*a
Here Ft is the tension, 10 (N) is the weight of the mass on the left and 1*a is mass*acceleration
Equation 2: Ft - 20 = 1*(-a)
Here Ft is the tension, 20 (N) is the weight of the mass on the right and 1*(-a) ##\dots~## is what ? If the mass is 2 kg, the term should be 2*(-a) to be correct. Then one has
Ft - 10 (N) = a
Ft - 20 (N) = -2a
If the first equation is doubled and added it to the second, one gets
3 Ft - 40 (N) = 0 ⇒ Ft = 40/3 (N) = 13.3 (N) and that is the correct tension in the string connecting the masses. More generally, $$F_T=\frac{2m_1m_2}{m_1+m_2}g.