1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Net reaction force

  1. May 23, 2008 #1


    User Avatar
    Science Advisor
    Homework Helper

    I'm trying to write a Library entry on Reaction Force, and I've discovered I can't answer this very simple question. :redface:
    Two equal blocks are on a horizontal table, with a gap between them.

    A straight rod is placed above the gap, but not symmetrically, resting on both blocks, with an extensive area of contact with each block.

    At what points do the two net reaction forces act? :confused:
  2. jcsd
  3. May 23, 2008 #2
    Wouldn't the rod actually have an equally distributed pressure acting against it due to each rod? Assuming a normal gravitational environment, the gravitational force on the rod would be [tex]mg[/tex]. The normal forces on the rod would be as follows:

    [tex]l \leftarrow[/tex]The length of rod on the two rods (not the length of the rod)
    [tex]d_{1} \leftarrow[/tex] The length of the rod on cube 1
    [tex]d_{2} \leftarrow[/tex] The length of the rod on cube 2
    [tex]d_{t} \leftarrow[/tex] The width of the rod
    [tex]P_{r} \leftarrow[/tex] The pressure exerted by the cubes on the rod

    [tex]\Sigma F_{y}= 0[/tex]
    [tex]mg = P_{r}*d_{t}*(d_{1}+d_{2})[/tex]

    Note that

    [tex]d_{1}+d_{2} = l[/tex]

    In a FBD, you could then simulate the two equally distributed pressures as point forces acting at the midpoint of their applied length, i.e.

    [tex]\frac{d_{1}}{2}[/tex] and [tex]\frac{d_{2}}{2}[/tex]

  4. May 23, 2008 #3

    Shooting Star

    User Avatar
    Homework Helper

    Hi tiny-tim,

    Let's just think of a one-level simpler case first. Suppose a rod is resting on a flat surface but a bit of it is protruding out. Let the weights of the portions on and off the table be W1 and W2, and the distances of the CMs of these two parts from the edge of the table be x1 and x2 respectively.

    Suppose the net reaction N, acting upward, is at a distance x from the edge. W1>W2, so that the rod does not topple, and x1 and x2 are known.

    Then N = W1 +W2 and W1*x1 = W2*x2 + N*x, taking the moment about the edge. (Of course, moment can be taken about any point.)

    This gives you a unique x, and the location of N.

    Of course you know this, but I’m sure this can be generalised to your case. I’m sorry that I can’t verify that right now, but I’ve to go, but surely I'll do it later and see what comes out. Let us know how it goes. Best of luck.
  5. May 23, 2008 #4
    I agree that it's an equilibrium problem and that you use both Newton's 2nd Law in translational and rotational form... but I don't think one should associate the net reaction force with a position, because it's not a contact force. The two reaction forces are both contact forces should be considered separately.

    The problem is that there are only two equations and four unknowns (both forces and their positions make four). I think that it's actually fairly common to have statics problems that are under determined.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook