Recent content by kkps

Thanks for pointing out the mistake What I wanted to show by this example is that the following statement made by PeroK is wrong "B cannot be pulling on the string with 20N and the string pulling back with 15N. That violates Newton's third law." In my FBD of B, there is 20N down and 13.3N up...

please see post #80 IMO, post #1 is asking the same question as shown in the diagram of post #80 (as I mentioned in post #81, some simplification of original question can be applied to give a meaningful answer without further questioning - force from ground etc. are irrelevant to the original...

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. Most textbooks have similar questions where, say, A pulls something to...

to make it simple, translate the question to this: A(1Kg)hanging on a rope on left of pulley---massless ROPE---B(2Kg)hanging on a rope on right of pulley in the above (pulley) case, the entire system (A-rope-B) will fall to right This is an Atwood machine problem The original question is...

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? FBD of A 10N<-A->15N A is moving to right with a=5, if m of A=1kg...

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...

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 problem I have still not answered the "rope with...

10N<-A---ROPE---B->20N is not different from <-A(say 1Kg)hanging on a rope on left of pulley---ROPE---B(say 2Kg)hanging on a rope on right of pulley this is same as Atwood machine problem of course, we can add force on ground etc. to problem and complicate, but it is not needed to solve the...

restating the question. A is pulling a massless rope with 10N, B is pulling in opposite direction of A with 20N. What is tension on string? It is 15N. 10N<---A--->------<---B--->20N Let us assume mass of A and B = 1kg. Let us denote tension on rope = Ft. Let us denote acceleration of A&B=a...