Forces (possibly resolving)

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

    [​IMG]

    a) The tension in the cable
    b) The magnitude of the components of foce exerted by the wall on the beam

    3. The attempt at a solution
    I really don't know where to start:confused:. Would somebody be able to guide me through step by step?
     
  2. jcsd
  3. cepheid

    cepheid 5,190
    Staff Emeritus
    Science Advisor
    Gold Member

    I can't emphasize this enough...all statics problems should be approached using the same basic method: DRAW A FREE BODY DIAGRAM. Don't do anything else until you have done this.

    In this case you want to draw a free body diagram for the beam. That way, you will know all of the forces acting on the beam, in both the x and the y directions. Why is this useful? Think about it this way: why is this a statics problem? Because things are stable. The bar is presumably in equilibrium, meaning that it is not translating (moving) up and down or left to right. This implies that the sum of the forces acting on it in each coordinate direction is equal to zero: [itex] \sum F_x = 0[/itex], [itex] \sum F_y = 0[/itex]. (The bar is not rotating either, which means that the sum of torques on it is also zero. You will probably need to use this fact in order to obtain enough equations to solve for all of your unknowns). So all you have to do is draw a picture that will help you keep a tally of all the x and y forces, sum the appropriate ones to zero, and use the equations you obtain to solve for your unknowns. *It's really that simple.* I'll be nice and list the forces that I can see here:

    vertical forces
    --------------

    1. The weight of the beam acting downward (ie in the negative y direction). It can be considered to act entirely at the beam's centre of mass (ie it should be placed at x = 4.0 m).

    2. The vertical component of the force due to the tension in the cable (probably acting upward ie in the positive y direction).

    3. The vertical "reaction force" at the pin support (ie the force the pin exerts on the beam). You can assume either the + or - y direction for this force, it doesn't matter. If your assumption was wrong, you'll simply get a negative answer when solving for this force.

    4. The weight of the dude, acting downward.

    horizontal forces
    ----------------

    1. The horizontal force on the beam due to the tension in the cable (probably acting to the left or in the negative x direction).

    2. The horizontal "reaction force" at the pin support (assume either left or right, once again it doesn't matter).

    Check this list against the forces in your free body diagram to make sure I didn't miss anything!!!

    I'll let you do the number crunching...
     
  4. I let this question sit for a bit, but now I'm coming back to it.
    In order to find tension, we solve the sum of either the verticle/horizontal components equal to zero.

    If for instance I take verticle, then I get the following:
    Tsin(53deg)+N-600g-2000g=0

    If I take the horizontal, then I get the following:
    -R-Tcos(53deg)=0

    Now two equations would be ringing bells in my head and telling me simultaneous, but here there are two unknowns. What can I do?

    Additional question, does the guy standing 2m from the pin have any significance to the question?
     
  5. Doc Al

    Staff: Mentor

    If there were only two unknowns, you'd be golden. Alas there are three unknowns: T, N, R. (I assume N and R are the vertical and horizontal components of the force exerted by the wall on the beam.)

    So you need a third equation. Hint: Consider rotational equilibrium.

    Absolutely!
     
  6. Question: Why isn't torques CCW = 8Tsin(53)?
     
  7. Doc Al

    Staff: Mentor

    Are you calculating torques about the pin? If so, then that would be correct.
     
  8. Yep got it, thanks to all! :tongue2:
     
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