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Physic lab Torque Problem

  1. May 1, 2012 #1
    1. The problem statement, all variables and given/known data
    B. Torque Due to Perpendicular
    Component of F
    4. The tension in string 1 (on the left) is not
    known, so we will calculate the torques
    about this point (10.0 cm). (That way,
    there is no torque due to the unknown
    string 1.) Calculate the torque due to the
    weight of the meter stick about 10.0 cm.

    5. Calculate the perpendicular component
    of the tension in string two (T2sinθ) for
    the five different values of θ. Is the
    perpendicular component of the tension
    the same for each trial? Calculate the
    torques due to string 2 about x = 10.0
    cm. Is the net torque acting on the
    meterstick about 10.0 cm equal to zero
    for each trial?

    Mass of meter stick= 149g
    Clamps(these are the things that connect the string to the meter stick) = 17g
    right side weight = 200g

    left side clamp at 10.0cm
    Center of mass of ruler = 50cm
    right side clamp at 90 cm





    2. Relevant equations

    t= fl = rfsin([itex]\theta[/itex]) = Ftanr




    3. The attempt at a solution
    Calculate the torque due to the
    weight of the meter stick about 10.0 cm.

    4. The center of mass of the meter stick was 50 cm
    So the only thing that is torquing it is the gravity of the meter stick, from the 10cm point
    The weight of the meter stick(including 2 clamps = 1.79N
    t= rfsin(90)


    [itex]\ t= (.40m) (1.792 N) (sin(90)) [/itex]
    [itex]\ t= .717NM [/itex]
    Negative 717NM since the torque is clockwise
    Can someone confirm that I am doing this correctly?


    5. Im just going to do the calculations for trail 1 to see if i am doing it correct


    Calculate perpendicular component of tension in string 2
    [itex]\ t2sin = opposite [/itex]

    [itex]\ (1.98N) sin 31 = 1.02 N [/itex]


    Calculate the torques due to string 2 about x = 10.0 cm.

    t= Ftanr
    [itex]\ t= (1.02 N) (.80 m)[/itex]
    [itex]\ t= .816 NM [/itex]
    Positive since torque is counterclockwise.

    Calculate the torque due to the
    weight of the meter stick about 10.0 cm.

    For this step i just added the string 2 torque from the torque measured in 4.

    net torque = .-717NM + .816NM = .099

    Can someone check if i did this correctly?
     

    Attached Files:

    Last edited: May 1, 2012
  2. jcsd
  3. May 1, 2012 #2

    tms

    User Avatar

    What about the bit of the meter stick to the left of the string? Doesn't it have weight, too?
     
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