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Force exerted on bar by pin support

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data
    The 8kg slender bar is released from rest in the horizontal position. When it has fallen 45 degrees, what are the x and y components of force exerted on the bar by the pin support A?

    2. Relevant equations
    F=ma where a= a t_hatt + v^2/p n_hat

    3. The attempt at a solution
    The force exerted on bar by pin support is the inverted force acting on the bar downward. I see it as pretty much the same as a box on a floor, in order for the box not to go through the floor an equal force must be pushing it upward, so this should also be the case for the bar and pin support.

    Thus I assume: F+mg.

    In attached picture I show each force I know and from there I can see that these forces acting in each direction;
    a_hat:mv^2/p - mg/sqrt(2)

    Now, the way I've put out my t-axis and n-axis I'll should get my inverted F as; F=-ma+mg/sqrt(2) t_hat + mv^2/p - mg/sqrt(2)
    This however is wrong. There is nothing else in this question I don't understand but this.

    My professors answer on this is: F=ma-mg/sqrt(2) t_hat + mv^2/p + mg/sqrt(2), and for some reason I just can't understand what he has done.

    Can someone please explain why gravity acts negative in t_hat and positive in n_hat while the rest of F stays the same? I think I've wasted like 3 hours on this problem :P

    Attached Files:

  2. jcsd
  3. Aug 16, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I don't understand this statement. Are you assuming equilibrium?

    This is fine. (Except for the 'thus', as it doesn't seem to follow from the previous statement.) All you are saying here is that the total force on the bar is the weight plus the support force from the pivot.

    I don't understand what you're doing here. I assume you intend these to be the forces from the pivot? (Your diagram doesn't really show forces, except for mg.)

    Do this. What's ΣF in each direction? Apply Newton's 2nd law.

    For example, calling the tangential force from the pivot Ft, the total tangential force is:
    ΣF(tangential) = Ft + mg/sqrt(2)

    Applying Newton's 2nd law:
    Ft + mg/sqrt(2) = ma

    Thus: Ft = ma - mg/sqrt(2)

    Do the same for the other component.
  4. Aug 16, 2011 #3
    Indeed you are freggin correct! I'm gonna try this method on more example exercises 'till I fully understand it!
    Thank you, sir/mam!
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