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Find the force on the cable, beam, and pole

  1. Mar 7, 2009 #1
    1. The problem statement, all variables and given/known data

    a mass of 18kg is supported at the end of a horizontal beam hinged to a vertical pole, a cable at 32 degrees with the beam is attached to the pole to support the mass.

    Find the force on the cable, beam, and pole.

    2. Relevant equations


    f=ma


    3. The attempt at a solution

    I dont know how to calculate force with an angle do you use cos or sin and what would you multiple it by ???
     
  2. jcsd
  3. Mar 7, 2009 #2
    Did you draw out a picture of the situation?
     
  4. Mar 7, 2009 #3
    yes im not sure how to post a picture though
     
  5. Mar 7, 2009 #4

    tiny-tim

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    HI Joe91090! :smile:

    It's always cos (of the angle between the force and the particular direction) …

    you only write sin if for some reason you're using 90º minus that angle

    (for example, if you want the vertical component, and if θ is the angle from the horizontal, then the relevant angle is 90º - θ, so you use cos(90º - θ), which is sinθ :wink:)
     
  6. Mar 7, 2009 #5
    so is the equation F=m(cosθ)x a ??
     
  7. Mar 7, 2009 #6

    tiny-tim

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    Is that supposed to be Newton's second law? :confused:

    If so, forget it … everything is stationary, so a = 0.

    Hint: do a force diagram, and then take components or moments :smile:
     
  8. Mar 7, 2009 #7
    in the problem is says the acceleration is 9.8m/s
     
  9. Mar 7, 2009 #8
    Well, yes that is what is feeling due to gravity but the particle isn't moving so that can't be the only acceleration it's experience, or more precisely, the only force.
     
  10. Mar 7, 2009 #9

    tiny-tim

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    Yes, that is g, the acceleration due to gravity.

    Look, you need an equation relating all three forces …

    what do you think that equation is going to be?
     
  11. Mar 7, 2009 #10
    would 15.56N be too small for the force on the cable. I did 18cos(32).
     
  12. Mar 7, 2009 #11
    I thought F=mg x cos(θ) but now im thinking thats wrong
     
  13. Mar 7, 2009 #12

    tiny-tim

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    Joe91090, I've no idea what you're doing! :cry:

    Please show us your full calculations if you want us to check them :smile:
     
  14. Mar 7, 2009 #13
    ok so im trying to solve for the force on the cable first. I really have no idea what formula to use all i know is that it is at 32 degrees from the horizontal with a 18 kg mass hanging from the end of it.
     
  15. Mar 7, 2009 #14
    You should probably end up with two equations 1 for the y direction and 1 for the x direction, the net force is zero, so identify the forces in each direction and set it equal to zero.
     
  16. Mar 7, 2009 #15
    the question seems so simple im just looking for what formula or formulas i would use
     
  17. Mar 7, 2009 #16
    Most of these problems are all the same thing with small differences. Haven't you done a problem like this before or atleast seen the teacher go over a similar one?
     
  18. Mar 7, 2009 #17
    No i miss last class and my professor assigns homework online and it is due tomorrow
     
  19. Mar 7, 2009 #18
    http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed06.htm [Broken]
     
    Last edited by a moderator: May 4, 2017
  20. Mar 7, 2009 #19
    First draw a force diagram, mark on all the forces, after that since its only three forces equilibrium so to use an equilibrium triangle is the easiest way to do this.
    With the triangle, just mark on everything then solve it using basic trig.

    I seen you made the mistake of thinking that it is only 18N going downward, but its not, you know the formula use it, :) F= ma

    Since the object is not moving so the cable must provide an upward force of the same as the downward force.

    With the force on the beam, just think about what i said above. :P

    If you want to know the formula, its still F = ma, but since it's 2D, you must use vector.
    So the formula is [tex]\sum{F}[/tex] = ma
     
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