Why Can't I Solve for a2 Instead of a1 in Connected Objects Problem?

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CVRIV
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I attached an image of the problem and solution from the book and my attempt at solving it. I don't need help solving the problem the way the book explains it; i understand that.

What I don't understand is why I can't solve for (a2) instead of (a1). The book says a2 = -a1, which makes sense to me, because the magnitudes are the same except the polarity for a2 is negative.

What about a1 = -a2? I tried solving the problem the same way except I substituted -a2 for a1 instead of the other way around and it doesn't work. I just don't understand why it wouldn't work.

Please help me understand this.
 
Attachments
  • Textbook page showing Example 4.14, Connected Objects: a block and ball connected by a string over a pulley, with a diagram of forces and a solution worked out in equations for acceleration and tension.
    Exp 414.jpg
    42.6 KB · Views: 400
  • Handwritten derivation of tension and acceleration for two connected masses with kinetic friction, solving for a2 = g(m2 - μK m1)/(m2 - m1) = 18.9 using values 7 and 4 kg and μK 0.3.
    Capture.JPG
    37.3 KB · Views: 409
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You need to keep the signs consistent. In your first equation a positive a1 means the mass accelerates towards the edge. In your second equation a positive a2 means the mass accelerates downwards. That is possible - but then you have a1=a2 without the minus sign.
 
Why would positive a2 accelerate downwards? I thought positive a2 accelerates upwards?

In equation 2 I divided both sides by -1 so that I could cancel out the tension T. Was that not right?
 
Oh wow. I think I know what I did wrong. How stupid of me. :confused:
 
CVRIV said:
Why would positive a2 accelerate downwards? I thought positive a2 accelerates upwards?
It is arbitrary which direction you choose, but you have to be consistent.
 
Finally! The problem was that from the very start I was inverting the polarity of m1a2. I kept writing down T - m2g = -m2a2. I did that because I had it in my head that a2 was negative, which it is, but only after solving for a2. By assigning it as negative from the start, I was in fact just screw it all up. Also... I had it stuck in my head that I had to add the two equations together. I went back to the previous problem in the book, which as a Atwood's Machine problem, and it was subtracting the equations. I I thought I was doing it wrong so I told myself I have to subtract the problems. That's when I realized, fully realized, that it totally depended on the cancellation of T. I get it now. Thanks for your help.
 
Attachments
  • Handwritten derivation on lined paper: equations 1 and 2 give T - μK M1 g = M1 a1 and T - M2 g = M2 a2, then a1 = -a2 and M1(-a2) = T - μK M1 g, ending with a2 = g(μK M1 - M2)/(M2 + M1) = -5.17 m/s² boxed.
    0310190012_2.jpg
    46.4 KB · Views: 387