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Determining maximum force

  1. Apr 8, 2008 #1
    The maximum tensions that FB and FC can undergo are 5kN and 5.5kN respectively.
    Determine the maximum force of FA that can be reached without exceeding these limits.
    [​IMG]

    This is all i have so far
    0 = -FA + FBsin40
    What's the next step?
     
  2. jcsd
  3. Apr 8, 2008 #2
    Calculate the components of F_B. Use this to solve for the difference between F_B and F_C to calculate F_A.
     
  4. Apr 8, 2008 #3
    wouldnt u just calculate the x component since FC has no y component?
     
    Last edited: Apr 8, 2008
  5. Apr 8, 2008 #4
    Is this answer correct:
    5sin40=3.2139kN
    To determine tension in FC
    5cos40=3.830kN
    3.830kN does not exceed the maximum limit for FC therefore the maximum force of FA is 3.2139kN.
     
  6. Apr 8, 2008 #5
    You already have the tension in F_C. The rope, F_A, is not straight down. The sum of the forces in the x-direction must be zero, and the sum of the forces in the y-direction must be zero.
     
  7. Apr 8, 2008 #6
    5.5kN is not the tension in FC it is the maximum limit.
     
  8. Apr 8, 2008 #7
    It is the maximum limit of the tension in F_C and you want to find the maximum limit of the tension in F_A.
     
  9. Apr 8, 2008 #8
    But if FC was to have a tension of 5.5kN that would put greater tension on FB. Therefore you would have to lower FB's y component to compensate for the increase of the x component to not let the tension in FB exceed the max limit.
     
  10. Apr 9, 2008 #9
    I think in this case Fc does not have vertical component, so we need not to care about it. the diagram try to fool us i think. so FA(max)=FB(max)=5kN.
     
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