Khan Question on Tension - is this a mistake?

[SSerenity]

I copied two questions. For the first question, how can there be two different tensions for elements in series?

And the second question only take into account one mass (the one on TB)? iF we use the reasoning from the previous question, wouldn't the TB be experiencing a force at a given acceleration proportional to both masses?


Whats wrong with my reasoning here?

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[BerkReviewTeach]

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[bobeanie95]

This problem is pretty much a box and string physics problem in disguise. Essentially you have two boxes (or two sections) being pulled by string (or actin). Since the two sections are attached, they will have equal acceleration. Then you need to identify in which direction the tensions are pulling.

For M2: You have a tension pulling to the right TB and TA pulling to the left, thus the equation is: TB-TA= m2*a

For M1: You only have a tension pulling to the left so: TA=m1*a

Note that acceleration is equal so you set your equations equal to each other and solve: TB=3/2TA. There is your answer for the first question.

For the second part of the problem, they want you to find the maximum acceleration, so to get the maximum acceleration you want the section that has the tension applied only in one direction, so you use equation #2 (TA=m1*a).

You plug in: 9E-12N=(60kDa*1.6E-24kg)*a

 

[SSerenity]

Thank you.

I still don't understand why we use TA for the second problem. Didn't we just establish that TB will experience 3/2 * TA of tension at any given acceleration? I don't understand why the TB string wouldn't break first

 

[bobeanie95]

It's because they're asking you about maximum acceleration.

Let's simplify the numbers and the problem a bit, say you had a piece of paper (.1kg) that you are trying to pull with a rope and you want to achieve maximum acceleration. The rope has a maximum tension of 10 N before tearing. Sounds familiar?

In instance #1 you have a one rope pulling the paper to the right (arbitrary). If you're pulling at the maximum force of 10N, the paper will have a maximum acceleration of 100m/s^2

In instance #2 you have a rope pulling the paper to left and to the right. If you're pulling at the maximum force of 10N in both sides, the net force will be 0 and the acceleration is 0 m/s^2. Even if the opposing forces aren't equal (and non-zero), the maximum acceleration will never reach 100m/s^ because they're opposing.

Now back to our original problem. You are correct that TB will experience a tension greater than TA, but the key point is that they are acting on M2 in opposite directions (similar to instance #2), due to this you could still experience an acceleration but not the maximum (instance #1). Therefore, you use the equation relating to M2 "TA=m1*a", making "D" correct. Hope this helps!