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A torque question help please

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Questions are included in the photo


    2. Relevant equations



    3. The attempt at a solution

    tumblr_mdf3r5MoR51r2g5epo1_500.jpg
     
    Last edited: Nov 13, 2012
  2. jcsd
  3. Nov 13, 2012 #2
    11. There are 3 torques acting on the beam. The sum of these need to be zero.
     
  4. Nov 13, 2012 #3
    You should be showing an attempt to a solution before getting help but I'll get you started.

    Both questions rely on the same thing: information has been provided to locate the center of mass of the beam (11) and the broom (12).
     
  5. Nov 13, 2012 #4
    For question 11.

    Total clockwise t = total anti clock wise t
    (10 x 3) = (28 x 2) (Fb x ??)

    Total tension will be: 28N + 10N + Beam weight force (?)
    Problem here is that I don't know the weight force of the beam...

    That was my attempt before I post this online...


    For question 12.

    Total support force for a finger must be: (1.5 x 9.8) + (0.4 x 9.8) = 18.62N (?)
    I can solve this question if I knew where any two forces were being applied...
     
    Last edited: Nov 13, 2012
  6. Nov 13, 2012 #5
    Total clockwise t = total anti clock wise t
    (10 x 3) = (28 x 2) (Fb x ??)
    I assume Fb is the weight of the beam. Its torque will act in the middle of the beam.
     
  7. Nov 13, 2012 #6
    Using point X as the pivot point then:

    Total clock wise t = total anti clock wise t
    (Fb x 1.5) + (10 x 3) = (28 x 2)
    1.5 Fb + 30 = 56
    1.5 Fb = 56 -30
    Fb = (56-30)/1.5 = 17.3333

    17.33N

    The answer says 52N...
     
  8. Nov 13, 2012 #7
    The distance to the middle is 0.5 not 1.5
     
  9. Nov 13, 2012 #8
    Oh man... stupid mistake... please help me do the question 12!
     
  10. Nov 13, 2012 #9
    total mass 1.5kg
    center of mass at 1.4m
    imagine the broom is a 2.8m rod instead.

    if you add 0.4kg to one side, what is the distance that you need to move from the center, in order to balance the rod.

    a logical answer will be <1.4m
     
  11. Nov 13, 2012 #10

    The answer says its 1.1m...
     
  12. Nov 13, 2012 #11
    The centre of gravity or centre of mass is located at 1.4 meter. If it was not the broom would have rotated off her finger. She would need to balance it a bit to the right of this point with the additional 400 gram. Call this distance x and recalculate about this new point .
     
  13. Nov 13, 2012 #12
    this answer is correct,

    next hint,
    half of 1.5kg is 0.75 kg
    if center is 1.4m, you will need another 1.4m to make this statement true, 2.8m
    add 0.4kg to 1.5kg, which make this broom now 1.9kg

    with these 3 values, you can easily work the answer as 1.1m
     
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