Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basic Misconception tension in rope and vectors

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data
    I am trying to learn enough physics to understand forces in climbing anchors and am stuck with an exercise problem that I am unable to understand my misconception.
    Mass-less rope, no friction.


    2. Relevant equations

    The problem question is diagram D
    SB is a spring balance
    T1 = 98n, correct
    T3 = 98n, correct
    SB = 196n, wrong, ( 98n is the correct answer )

    3. The attempt at a solution

    Isolating each side as in diagram B
    T1 = 98n = TB1 = TB2
    TB1 force is to the left, TB2 force is to the left so they will not be added together
    ( from a experimental view I could hold the force exerted 98n )

    Putting them back together again as in diagram C
    T1 = 98n = T3
    TC1 = T1 = 98n
    TC2 = T3 = 98n
    TC1 + TC2 = 196n
    TC1 force is to the left, TC2 force is to the right, so they add together
    ( from a experimental view I would not attempt the experiment, could be painful )

    The tension force in the rope has to be equal in Diagram A
    T1 = T2 = T3
    but T1 and T3 are downward forces that have to be transmitted somewhere and the only place is T2 so what is T2 ?
    T2 = T1 + T3 = 2x98n = 196n
    as T1 = T2 = T3 , ( T1 = 98n, T2 = 98n, T3 = 196n, therefore the tension will balance to 392n / 3 = 130n )

    Hopefully you can see I have tried to understand but have come to the conclusion that I have a fundamental misconception as regards tension force, could it be something similar to the common mass and weight misconception. ?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 22, 2012 #2
    what is the question asking for
  4. Feb 22, 2012 #3
    Why is the spring balance 98n and not 2 x 98n
  5. Feb 22, 2012 #4
    ok rockclimber
    I have a question for you
    suppose a a spring balance is hung on a wall with the help of a string
    and mass m hanging down with the help of a string
    what will spring balance show ? reading
  6. Feb 23, 2012 #5
    Hi Kushan

    or .98 x mass

    as isolated in diagram B

    I am sure that right :smile:
  7. Feb 23, 2012 #6
    So you got the answer ?
  8. Feb 23, 2012 #7
    Sorry no

    the question is two masses are connected as diagram D

    just because the two masses are connected that does not make the force of one of the masses disappear or does it ?

    We have two downward forces

    Ok, I will try again

    One mass hanging has to have a force of 98n, the rope always transmits its tension evenly, therefore it is of no consequence how it is terminated, either by a wall or ( in diagram D ) another mass hanging in opposition .

    So one of the forces generated by the mass does not disappear, it is opposed by the other force.
  9. Feb 23, 2012 #8
    Do you find any similarity in this figure and you rquestion ?

    in terms of tension aboce the spring and below the spring?

    Attached Files:

  10. Feb 23, 2012 #9
    Yes, the similarity is that the spring is just in a different position,
    so the spring will read the same no matter where it is in the system.
    Ta = Tb

    Is therefore tension force at any point in the rope a pair ( action and reaction ) ( I looked this up ! )
    To find the tension in the rope it is either the action force or reaction force of the entire system.

    Is the answer to the original question what is the tension, its the reaction force 98n
    The spring balance will read 98n because it part of the rope system

    Should I be looking more closely at the direction of the arrows in your diagram ?

  11. Feb 23, 2012 #10
    Yea and your welcome :)
  12. Feb 23, 2012 #11
    Hi Kushan
    Many thanks for the help
    Wow this is a really friendly place
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook