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Confusion about type and quantity of force being exerted

  1. Dec 28, 2015 #1
    1. The problem statement, all variables and given/known data
    A 500-N person stands 2.5 m from a wall against which a horizontal beam is attached. The beam is 6 m long and weighs 200 N (see diagram below). A cable attached to the free end of the beam makes an angle of 45 degrees to the horizontal and is attached to the wall.

    a) draw a free-body diagram of the beam
    b) Determine the magnitude of the tension in the cable.
    c) Determine the reaction force that the wall exerts on the beam.

    2. Relevant equations
    F = ma
    Torque = (distance)(F)(sin (angle))
    3. The attempt at a solution
    I attempted this problem just like any torque/ force problem, but I misunderstood the behavior of the reaction force of the wall on the beam. I was able to correctly identify the weight of the person, the weight of the beam, and the tension of the cable as forces applied to the beam; however, I incorrectly identified the force of the wall on the beam as a contact force perpendicular to the wall. Because of this incorrect identification, my answer part c was incorrect. As I reviewed my answer in the "answers explained" section of the textbook, I saw that instead of a single normal force exerted by the wall on the beam, there is a single "reaction force," with magnitude denoted R, at some angle. After this, I pondered the situation a bit and realized that there could indeed also be a friction force exerted on the beam since the wall isn't smooth or frictionless. Still, a friction force and a normal force would be two separate forces rather than one "reaction force."



    Can someone please explain to me why the force of the wall on the beam is labeled as one single force by my textbook? Also, what type of force is this? If the "reaction force" is indeed just a resultant force of a friction and a normal force, would it be wrong to label the free body diagram with two separate forces and call them friction and normal?
     

    Attached Files:

  2. jcsd
  3. Dec 28, 2015 #2

    haruspex

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    A reaction force is whatever is necessary to maintain, or attempt to maintain, physical integrity. The beam is not merely resting against the wall, it is attached to the wall by a joint. The joint prevents the beam end moving in any direction, up, down, left, right, but allows rotation. So the reaction force can be in any direction, but there is no reaction torque.
    If the beam end were attached rigidly to the wall there could also be a reaction torque, but in this case there would not be enough information to deduce the tension. You would need data on moduli of elsasticity.
     
  4. Dec 28, 2015 #3

    BvU

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    Hello Chozen, :welcome:

    You posted two clear and concise posts as an entry in PF; well done !

    It's a bit more than that: the problem statement says the beam is attached to the wall ! So it's not really friction, but more like yet another normal force.
    Whether you consider the vertical and normal component separately or not, the wall exerts only one force with one magnitude and one direction on the point where the beam is attached. If you replaced the attachment (hinge, bolt or whatever) by an inclined and 100% smooth surface perpendicular to the direction of the combined force, then the beam should still be in equilibrium (albeit not stable). And that is all your exercise wants calculated.
     
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