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Two stacked blocks, push using one block, static friction between blocks

  1. Sep 30, 2008 #1
    Newtonian mechanics problem with blocks

    Problem:

    [tex]m_{1}[/tex] and [tex]m_{2}[/tex] each interact with [tex]m_{3}[/tex] via static friction, with the same [tex]\mu_{s}[/tex]. The horizontal surface below [tex]m_{3}[/tex] is frictionless. An external force [tex]F_{ext}[/tex] acts on [tex]m_{1}[/tex] from the left and the entire system of three connected masses moves to the right (and obviously accelerates). The idea is to provide a suitable magnitude of [tex]F_{ext}[/tex] as to prevent both [tex]m_{1}[/tex] and [tex]m_{2}[/tex] from moving with respect to [tex]m_{3}[/tex] during the acceleration, i.e. we don’t want [tex]m_{1}[/tex] to slide down along, nor [tex]m_{2}[/tex] to slide “back” along [tex]m_{3}[/tex]. To make life easy, we let [tex]m_{1}[/tex], [tex]m_{2}[/tex] and [tex]m_{3}[/tex] all have the same mass [tex]m[/tex].

    a) Find in terms of relevant parameters, the possible range of [tex]F_{ext}[/tex] which will allow the desired motion to take place.
    b) It might be that, if [tex]\mu_{s}[/tex] is less than some critical value [tex]\mu_{s(critical)}[/tex], no value of [tex]F_{ext}[/tex] will allow the desired motion. Give a simple argument why this might be true, and if so, determine [tex]\mu_{s(critical)}[/tex] in terms of relevant parameters.

    Relevant equations:

    [tex]\sum F = ma[/tex]

    Here is my attempt at the problem:

    Part A:
    For [tex]m_{2}[/tex]:
    [tex]N_{2} = m_{2}g[/tex]
    [tex]F_{fr2} = m_{2}a[/tex]
    [tex]\mu_{s}m_{2}g = m_{2}a[/tex]
    [tex]\mu_{s}g = a[/tex]
    [tex]\mu_{s} = \frac{a}{g}[/tex]

    For [tex]m_{1}[/tex]:
    [tex]F_{ext} - N_{1} = m_{1}a[/tex]
    [tex]N_{1} = F_{ext} - m_{1}a[/tex]
    [tex]F_{fr1} = m_{1}g[/tex]
    [tex]\mu_{s} (F_{ext} - m_{1}a) = m_{1}g[/tex]

    For [tex]m_{3}[/tex]:
    [tex]N_{3} - F_{fr1} - N_{2} = m_{3}g[/tex]
    [tex]N_{1} - F_{fr2} = m_{2}a[/tex]

    Substitute in for [tex]N_{1}[/tex] and [tex]F_{fr2}[/tex]:
    [tex]N_{1} - \mu_{s}m_{2}g = m_{2}a[/tex]
    [tex]F_{ext} - m_{1}a - \mu_{s}m_{2}g = m_{2}a[/tex]
    [tex]F_{ext} = m_{3}a + m_{1}a + \mu_{s}m_{2}g[/tex]

    Since [tex]\mu_{s}m_{2}g = m_{2}a[/tex]:
    [tex]F_{ext} = m_{3}a + m_{1}a + m_{2}a[/tex]

    Since [tex]m_{1} = m_{2} = m_{3} = m[/tex]:
    [tex]F_{ext} = 3ma[/tex]

    Part B:
    My guess is that if [tex]\mu_{s}[/tex] is infinitely small so that friction is negligible, any magnitude of force applied on the blocks would cause block 1 to slide down and block 2 to slide backwards relative to block 3.

    I don't know where to start with the parameters, but here is what I have so far:
    [tex]\mu_{s(critical)} < \mu_{s}[/tex]
    [tex]F_{fr(critical)} < F_{ext} < F_{fr}[/tex]

    Any help would be highly appreciated!
     

    Attached Files:

  2. jcsd
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