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Theory for acceleration of a cart: lab experiment

  1. Oct 19, 2011 #1
    1. The problem statement, all variables and given/known data
    In a lab experiment in my introductory physics class, we are asked to verify Newton's Second Law by taking data from an experiment and then comparing that data to a theory. We are given a cart of unknown mass m[itex]_{}1[/itex] is put on a horizontal track with a string attached over a pulley to a hanging mass m[itex]_{}2[/itex]. It is given that both masses are moving as one system and therefore have the same acceleration. My problem lies with the theory part of this experiment: how am I supposed to solve for the acceleration with known values? Let the acceleration of the system be a, the rolling frictional force f[itex]_{}r[/itex], the coefficient of rolling friction μ[itex]_{}r[/itex], the normal force F[itex]_{}N[/itex], the tension in the string T, and the acceleration due to gravity g.
    This picture is similar to what our experiment looks like: http://www.physicssource.ca/images/cart_forcesensor_track.gif [Broken]

    2. Relevant equations
    Newton's Second Law, F=ma


    3. The attempt at a solution
    Perhaps there is something intuitive about the mass of the cart and the coefficient of rolling friction that I am not seeing but I just cant figure it out. Here is what I have so far:

    First I set up a free body diagram of the mass of the cart m[itex]_{}1[/itex] to show that:
    ƩF[itex]_{}x[/itex]=f[itex]_{}r[/itex]-T=m[itex]_{}1[/itex]*a (Equation 1)
    ƩF[itex]_{}y[/itex]=F[itex]_{}N[/itex]-m[itex]_{}1[/itex]*g=0, So F[itex]_{}N[/itex]=m[itex]_{}1[/itex]*g
    Also, we know that f[itex]_{}r[/itex]=μ[itex]_{}r[/itex]*F[itex]_{}N[/itex], So f[itex]_{}r[/itex]=μ[itex]_{}r[/itex]*m[itex]_{}1[/itex]*g

    Now I set up a free body diagram of the hanging mass to show:
    ƩF[itex]_{}x[/itex]=0
    ƩF[itex]_{}y[/itex]=T-m[itex]_{}2[/itex]*g, so T=m[itex]_{}2[/itex]*(g+a)

    Substituting all back into Equation 1:
    μ[itex]_{}r[/itex]*m[itex]_{}1[/itex]*g-m[itex]_{}2[/itex]*(g+a)=m[itex]_{}1[/itex]*a
    μ[itex]_{}r[/itex]*m[itex]_{}1[/itex]*g-m[itex]_{}2[/itex]*g-m[itex]_{}2[/itex]*a=m[itex]_{}1[/itex]*a
    g*(μ[itex]_{}r[/itex]*m[itex]_{}1[/itex]-m[itex]_{}2[/itex])=a*(m[itex]_{}1[/itex]+m[itex]_{}2[/itex])
    (g*(μ[itex]_{}r[/itex]*m[itex]_{}1[/itex]+m[itex]_{}2[/itex]))/(m[itex]_{}1[/itex]+m[itex]_{}2[/itex])=a

    This is where I am stuck: How am I to get rid of or solve for these two unknown values μ[itex]_{}r[/itex] and m[itex]_{}1[/itex]?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 20, 2011 #2
    Except for a minus sign I got what you have. I'm guessing you measure acceleration and are supposed to figure out m1? Without knowing μ_r you have done as much as you can.
     
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