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Applying Newton's Laws question.

  1. Sep 23, 2012 #1
    I will try my best to explain this question:

    Block S (the sliding block) with mass M = 3.3 kg, is free to move along a horizontal friction-less surface and connected, by a cord that wraps over a friction-less pulley, to a second block H (the hanging block), with mass m = 2.1 kg. The cord and pulley have negligible masses. The hanging block H falls as the sliding block S accelerates to the right. Find A) the acceleration of block S, B) the acceleration of block H, and C) the tension in the cord.

    Okay so T = Ma, got it. I will call it equation #1.
    Now T-mg = -ma, got it as well and I will call it equation #2.

    My book goes on: "Now note that Eqs. #1 and #2 are simultaneous equations with the same two unknowns, T and a. Subtracting these equations eliminates T. Then solving for a yields:
    a = (m/M+m)(g)
    Substituting this result into Eq. #1 yields:
    T=((Mm)/(M+m))(g)
    Putting in the numbers gives, for these two quantities: a = 9.8m/s^2 and T = 13 N."

    Here is is my questions, why do I subtract the two equations? My book says nothing about simultaneous equations and therefore gives no explanations for why I should subtract them when I find them. My second question is could someone explain how exactly they are substituting the equations which has been solved for a into equation #1 and why I would want to do that anyway? Also this I am assuming the acceleration of 9.8m/s^2 is for both block S and black H right? Which makes sense because the table is friction-less and so is the pulley.

    Its kind of ironic my book wastes a full page explaining how to set up the stupid diagrams and then when it comes to making the actual calculations there's no explanation what so ever. This makes me think that there is some simple logic behind it that I have yet to grasp.
     
  2. jcsd
  3. Sep 23, 2012 #2

    BruceW

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    Homework Helper

    Once you see that you have two equations with two unknowns, you should recognise that you might be able to eliminate one to solve for the other. That is what the problem was testing. This is one of those things that once you have practised a lot of them, it immediately comes to mind. Also, you might not have noticed, but they have used the same value of acceleration for both objects! This is an assumption that must be made at the start of the problem. Can you think of why they assume this? (Hint: It is really an assumption about the cord between the two objects).

    They have got this equation (let's call it equation z):
    [tex]a= \frac{m}{M+m} g [/tex]
    and they also have equation 1: [itex]T=Ma[/itex] So what they have done is replaced the 'a' in equation 1 with the entire right-hand-side of equation z, since this is all equal to 'a'. The reason they have done this is to get an equation which gives the tension in terms of known parameters.

    Also, 9.8m/s^2 is the value of g (which is weight divided by mass). So this is not the value of the acceleration of either of the objects. Their actual acceleration will be less than this. But your intuition is correct that the two objects both have the same acceleration.

    And about these types of question generally, the more you practise them, the more you get a feel for them and they will become easier to do. So don't give up!
     
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