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Torque/statics problems

  1. Feb 2, 2012 #1
    It's my first time posting here; I'm usually able to get help from my teacher and friends at school, but I'm in a bind: I was absent for most of the torque/statics unit this week and I have a test tomorrow that can't be re-scheduled. Here's a few problems similar to the ones I'm to be tested on; if I could be guided through these I'd be eternally grateful.

    1a. The problem statement, all variables and given/known data

    A brick is placed on a bike pedal to provide torque and cause the bike to roll. If the brick weighs 8kg and the pedal is 16cm away from the crankshaft's axis of rotation, find the torque applied to the crankshaft when the pedal is in each of the following positions: straight forward, 45 degrees towards the front and down, and straight down.

    1b. Relevant equations

    t = m g x sin( theta) (is this right?)

    1c. The attempt at a solution

    Straight down-zero (no distance between the center of rotation and the point at which the weight acts)
    at 45 degrees t = 8 * 9.8 * 0.16 * sin(45deg) = 8.87 Nm
    and at 90 t = 8 * 9.8 * 0.16 * 1 = 12.5 Nm

    Really not sure if these are right...but seems somewhat logical...

    2a. The problem statement, all variables and given/known data

    A basic draw bridge over a river is operated by two chains, one attached to each upper corner of the bridge. The bridge's mass is 300kg and is stationary at a 55 degree angle upwards from the horizontal. At this angle the chains are perfectly horizontal. Find the tension in each chain, the force exerted by each of the hinges on the bridge (assuming there are two identical hinges, one at each bottom corner), and the tension in each chain required to begin lifting the bridge from a far/open position.

    2b. Relevant equations

    See below...it's a start...

    2c. The attempt at a solution

    I don't have the length of the chain, or the door? Does this matter?

    Assuming the length of the door = 1, the height of the chain, when horizontal (with a 55degree right traingle) = cos 35 = .819

    The new triangle, when the door is horizontal is A (door) = 1, B (height) = .819

    the angle between the chain and the door = invtan(.819/1) = 39.3175

    I know the weight on the door = 300kg, acting at .5 so the vertical component of the force on the two chains combined = 150kg.

    So the total tension on the two chains = the vertical force (150kg) / sin 39.3175 = 150kg / .6336 = 236.736kg

    Each chain then has a total tension of 236.736/2 = 118.368

    Not really sure to go from here?
     
  2. jcsd
  3. Feb 3, 2012 #2

    Simon Bridge

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    This is not correct - the brick is always a fixed distance from the center of rotation. You mean the perpendicular distance between the applied force and the center of rotation is zero.

    The other way to think about this is to resolve the applied force into radial and tangential components. This way helps for complicated systems.


    By "door" you mean "bridge"? Put the length of the bridge as L and work it out.

    Weight of bridge acts at center of mass - you spotted that.
    Analyse by balancing the torques then resolve into radial and tangential components to get the tension in the the chains and the forces the hinges have to balance.
     
  4. Feb 3, 2012 #3
    That makes sense, I mis-spoke. Thanks! Other than that does the work for #1 seem correct?

    Radial and tangential components? You lost me there. I tried using torque but I'm not too confident in these numbers:

    Tnet=0
    Tnet=3000N(L/2)-Lcos55(x)
    1500L=Lcos55(x)
    1500=cos55(x)
    x=2615.17N
    So 1307N per chain?

    For the hinges:

    3000cos55 = 1720.73
    so 860N each?

    And I have one more question (simpler), can you let me know if you agree with the answer? Just want to be sure I have the basics down.

    Question

    A defective teeter-totter, 8.4m long weighing 8kg, is located in a playground. The fulcrum was accidentally placed 22cm to the left of the real center of mass of the bar. If a 43kg boy sits at the far end of the short side, where should an identically weighted child sit to balance the teeter totter (measured from the far end of the long side)?

    Work

    Tnet=0
    Tnet= 8.4m(80N) + 3.98m(430N) - x(430N)
    430x=2383.4
    x=5.54m

    8.4-5.54=2.86m from the long side

    Thanks again for all the help.
     
  5. Feb 3, 2012 #4

    Simon Bridge

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    "radial" = pointing along the radius.
    We are talking about circular motion.
    Draw a line from the center of the rotation to the place the force is applied. This is the radial direction. Perpendicular to that is the tangential direction. Resolve the applied force against these directions.

    I don't do numbers - just physics :)
    The bridge is the arm of the rotation.
    forces perpendicular to the bridge provide torques.
    There are three torques to balance.

    If the fulcrum is 22cm left of the balance point then the com is 22cm to the right of the fulcrum. So there are two torques to the right and one to the left. The length of the moment arm should be measured from the center of the rotation - i.e. the position of the fulcrum.

    Remember to explain your reasoning. It is hard to tell what you intend from just the numbers.
     
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