1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Particle equilibrium

  1. Sep 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Hibbler.ch3.f3.jpg

    If the 4.4-kg block is suspended from the pulley B and the sag of the cord is d = 0.15 m, determine the force in cord ABC. Neglect the size of the pulley. (Figure 1)
    Express your answer to three significant figures and include the appropriate units.


    2. Relevant equations


    3. The attempt at a solution
    upload_2016-9-30_16-18-6.png

    r(BC) = (x^2 + .15^2)^.5
    r(AB) = [(.4-x)^2 + (.15^2)]

    T(x) = r(BC)cos(theta) - r(AB)cos(theta)
    T(y) = r(BC)sin(theta) + r(AB)sin(theta) - 43.164

    I tried adding r(BC) + r(AB) and then grphing to find the zeros but that was wrong.

    I also have; x/cos(theta) + .4-x/cos(theta) = .4cos(theta)

    not sure how to proceed.
     

    Attached Files:

  2. jcsd
  3. Sep 30, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi Robb,

    Is there a reason to assume ##x\ne0.2## m ?
     
  4. Sep 30, 2016 #3
    Can I assume it's centered?
     
  5. Oct 1, 2016 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You are given that A and C are on the same horizontal level. You have correctly presumed that the angle is the same each side (if anything, that was the step that needs some physics to justify). If the vertical through B meets AC at E, what can you say about triangles AEB, CEB?
     
  6. Oct 1, 2016 #5
    I suppose they are equal. I guess I don't like to make that assumption without proof. I suppose they would have to be though, given that the pulley would be assumed to be frictionless. Is that a correct assumption?
     
  7. Oct 1, 2016 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes. If there were static friction in the pulley then there would be a range of answers.
     
  8. Oct 1, 2016 #7
    So, shouldn't T(ABC) = -43.164N?
     
  9. Oct 1, 2016 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Not what I get, and it certainly cannot be negative.
    Please post your working.
     
  10. Oct 1, 2016 #9
    F(x) = 0; F(BC)cos37 - F(BA)cos37
    F(y) =0; F(BC)sin37 + F(BA)sin37 - 43.164
    F(BC) = F(BA)
    -.6435F(BC) - .6435F(BC) = 43.164
    F(BC) = 33.5359N
     
  11. Oct 1, 2016 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Ok.
    Two issues.
    1. Where did those minus signs come from in the line before last (which you then ignored to get the last line)?
    2. There is no need to find the angle. Doing so has introduced some rounding error. The sine and cosine are exactly 0.6 and 0.8.
     
  12. Oct 2, 2016 #11
    1. good question

    I see what you're saying about not needing the angle.

    F(x)=0; F(BC)(0.8) - F(BA)0.8)= 0
    F(y)=0; F(BC)(0.6) + F(BA)(0.6) -43.164=0

    F(BC)=F(BA)
    1.2F(AC) = 43.164
    F(AC) = 35.97

    I assume this is the force in cord ABC?
     
  13. Oct 2, 2016 #12

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes. But I would not have written F(AC); that's a bit confusing since there is nothing acting along the straight line AC. You could write F(ABC), or just leave it as F(AB).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Particle equilibrium
Loading...