1. Limited time only! Sign up for a free 30min personal 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!

Tension Question

  1. Apr 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A hungry bear weighing 720 N walks out on a beam in an attempt to retrieve a basket of food hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 5.40 m long; the basket weighs 80.0 N.

    a) When the bear is at x = 1.00 m, find the tension in the wire and the components of the force exerted by the wall on the left end of the beam.
    N (tension)
    N ( Fx)
    N ( Fy)

    3. The attempt at a solution

    Okay well I sort of understood this question and created a balanced torque equation. After doing this I got that:

    Tension = 345.66
    Fx = 172.83

    But I need help with finiding Fy Please.
    I only have like one more chance to answer the question on my homework so I want to make sure I can get the right answer.
    I know I have to write out some equation like maybe:

    Rsin(theta) = 720 + 200 + 80 - 345.66sin(60)

    Is this correct? And if its not, am I on the right track?
    Last edited: Apr 27, 2009
  2. jcsd
  3. Apr 27, 2009 #2
    is the beam supported by a horizontal string on each end?
  4. Apr 27, 2009 #3
    Need more data on how the beam is supported.

    Sounds like a cantilever beam with a tension wire at the other end, but is this tension wire vertical or anchored to the wall.
  5. Apr 27, 2009 #4
  6. Apr 27, 2009 #5


    User Avatar
    Science Advisor
    Homework Helper

    Hi yb1013! :smile:

    I think you've calculated the y-component of the tension, instead of the y-component of the force at the wall. :redface:
  7. Apr 27, 2009 #6
  8. Apr 27, 2009 #7
    would this be closer to the y-component for the force?

    Rsin(theta) = 720 + 200 + 80 - 345.66sin(60)
  9. Apr 27, 2009 #8
    what a dumb bear...

    Anyways, I broke the componets into x and y directions. The y dir. is Tcos(theta) -mg =0 due to thier being no motion.

    also, thier are 3 masses. bear, basket, and beam.
  10. Apr 27, 2009 #9


    User Avatar
    Science Advisor
    Homework Helper

    He's smarter than the average bear!! :biggrin:
    Yup … if the tension is right, then that looks right too! :smile:
  11. Apr 27, 2009 #10

    So would it go like 345.66cos(60) - 1020 = 0 ??
  12. Apr 27, 2009 #11
    Ok well the numbers in that question aren't actually my exact numbers, but with my numbers I come out with:

    Rsin(theta) = 720.65

    would that mean that its:

    Rsin(60) = 720.65
    R = 832.13

    So 832.13 would be the force in the y-component??
    I only have one more guess left on my homework, so I wanna make sure lol
  13. Apr 27, 2009 #12


    User Avatar
    Science Advisor
    Homework Helper

    uhh? :confused:

    perhaps I misunderstood your Rsin(theta) …

    I assumed you meant that to be the y-component of the reaction force at the wall:

    Ry = 720 + 200 + 80 - 345.66sin(60)

    (ooh, and it's 1000 not 1020 :wink:)
  14. Apr 27, 2009 #13
    Yea forget what I just said, I had a brain fart for a second there lol

    Its just 720, sorry about the confusion.

    Thanks again!
  15. Apr 27, 2009 #14


    User Avatar
    Science Advisor
    Homework Helper

    better check that …
  16. Apr 27, 2009 #15
    well its right, I stated in one of my other posts that the numbers were exactly what my actual problem contained. the 720 N of the bear is actually 740, which is why when I added all the numbers up I got 1020 instead of 1000.. So in the end the force is equal to 720

    Sorry for all the confusion
  17. Apr 28, 2009 #16


    User Avatar
    Science Advisor
    Homework Helper

    stuffing down more goodies than the average bear …

    ah :rolleyes:heavier than the average bear!! :biggrin:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Tension Question
  1. Tension Question (Replies: 1)

  2. Tension question (Replies: 2)

  3. Cable Tension question (Replies: 1)

  4. Tension question (Replies: 2)

  5. Tension question. (Replies: 2)