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Homework Help: Newton's Second Law Lab with Pulley's and Blocks

  1. Sep 18, 2011 #1
    I did a lab earlier this week on Newton's second law with a pulley and carts. M2 was the block that traveled horizontally and M1 was the hanging mass. The mass of the cart that was moving horizontally was constant all experiment with about 0.2 kg. The mass of the hanging gradually increased. From this data, we found time, velocity, and acceleration using kinematic equations. We also plotted a graph of Acceleration vs Mass of the hanging block. Here I included a picture of what this experiment looked like: http://i54.tinypic.com/20kw1w.gif

    1. The problem statement, all variables and given/known data

    Draw a diagram showing the forces acting on each block and use Newton's 2nd law to derive the equation for acceleration of the blocks and the tension in the string.

    Compare the acceleration of the equation above and the general equation for a straight line to prove that the graph you made should be a straight line. Use this to show what the slope and y -int. of the graph should be in terms of 'g' and 'M_total.' M_total is the mass of the whole system (we assume the thread that traveled throughout the pulley was negligble and this system used a frictionless surface).

    2. Relevant equations

    For deriving acceleration and tension, I used the equations T = m2a and -T + m1g = m1a. Then, m1g = m1a + m2a and then I solved for 'a'.

    I believe these equations are correct but I am not sure how I am supposed to prove this is a straight line and what the slope and y int of the graph should be in terms of g and M_total. The equation I got from my data using the line of best fit is Y = 21.4x + 0.0139 and my mass (kg) was on the x and the acceleration (m/s^2) on the y.

    3. The attempt at a solution

    The above ^^.
  2. jcsd
  3. Sep 18, 2011 #2
    Any "line" is defined by [itex] y = mx+n [/itex], meaning that you have a coefficient, the slope, multiplying your dependant variable, x, and possibly, and auxiliary term, n.
    Try extracting "a" from your equation and see if that matches this description, and note what should the variable be in such an equation.
  4. Sep 18, 2011 #3
    a = (m1g)/(m1+m2)

    Would a be the slope and m1 + m2 be the y intercept?
  5. Sep 18, 2011 #4
    Lets look at your statements:
    So that means, that the variable in your case, was M_total, because, as you said, it was subject to change during the experiment.
    Do you think there should be an intersection with y(other than y = 0)? think about what would happen were m1 to be zero...
    In order to find the slope, you have to inquire what coefficients of the dependant variable are(see above).
    Also, remember that you were asked to plot "a", so in this case, a is actually a function of your variable(M_total or whichever).
  6. Sep 18, 2011 #5
    Well the y-int according to the line of best fit for the graph of Acceleration vs the mass of hanging is 0.0139. I believe we have this number because of experimental error? So the "x" is m1 + m2?
  7. Sep 18, 2011 #6
    You're right... and of course, there shouldn't be any intersection since if m1 goes to zero so does the acceleration.
    If by your definition(and you have to elaborate on that) M_total = m1+m2, then yes.. but that makes the slope not so easily calculable, you'd get an inverse of x, in fact.
  8. Sep 18, 2011 #7
    So it has an inverse relationship for the slope? So it's just 1/(m1+ m2)?
  9. Sep 18, 2011 #8
    In this case, yes, it's best to find the inverse, or 1/a.
    However please make sure you refer exactly to what your assignment asks you, and what does it mean by M total...
  10. Sep 18, 2011 #9
    M_total refers to the mass of m2 and m1
  11. Sep 19, 2011 #10
    Then you've done everything properly.
  12. Sep 19, 2011 #11
    I'm going to ask one more thing, it is asking to find acceleration due to gravity (g) from the slope of my graph and % error. I know how to use percent error but I don't know exactly how to find acceleration due to gravity. My equation for my line is as follows Y = 21.4x + 0.0139
  13. Sep 19, 2011 #12
    If you look at your equation for a, you have a = (m1/(m1+m2))*g, by getting two values for m1, with m2 and the respective "a"s known, you can calculate g. An error is estimated by: value/range_of_measurement.
    Feel free to ask as much, and as frequently as you like :), so long as I am of any use :D?
  14. Sep 19, 2011 #13
    So basically plug in one of my datas for the m's and then solve for g? So my "a" is 21.4?

    If so, I am getting 447.91 for g :\?
  15. Sep 20, 2011 #14
    You've said :
    That means, that (m1*g)/(m1+m2) is your dependent variable times the slope.
    In other words, if we were to assume, that this function obeys some law like:
    [itex]\Large \frac{m_1g}{m_1+m_2} = k\cdot{m_1} [/itex], Then k, the slope of your best fit, is in fact that illusive "21.4".
    Plug that in, with the neceesary M_total(m1+m2), and find g, in other words:
    [itex] g = k(m_1+m_2) [/itex]
  16. Sep 20, 2011 #15
    Yay, I think I got now, thanks! Why do you think the gravity value that I am getting is less than gravity? I would say the string, pulley, and air resistance play roles in affecting the acceleration due to gravity.

    You've been really helpful, and I promise this will be my last question on this matter :].
  17. Sep 21, 2011 #16
    Now that's a promise I'd rather you didn't keep! :D...
    If you take a look, one important factor, is the pulley. It's not ideal, and as a result there's varying tension on the string, causing what's known as torque. Since the pulley, not being a massless object, therefore has a moment of inertia, once it's already rotating, actually facilitates movement, and along with the torque acting on it, accelerates the mass' downward motion. Air resistance also negates gravity by acting upward, along with the string; The table also has resistance, in the form of friction, and so on, and so on...
    By-the-by, do your values really diverge that greatly?
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