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Homework Help: Equilibrium and Torque and minimum mass

  1. Jul 17, 2008 #1
    1. The problem statement, all variables and given/known data

    To make larger pictures, an enlarger can be moved to the edge of a table so that the image is projected onto the floor, as shown.


    mass of bass = 2.0 kg
    mass of support = 2.0 kg
    mass of enlarger = 6.0 kg

    a) What is the minimum mass that needs to be added on the base at point A to keep the enlarger from tipping off the table?

    b) With this mass in place, what will be the reaction (normal) force at B?

    2. Relevant equations

    T is Torque, F is Force

    T = 0
    F = 0

    3. The attempt at a solution

    base (b)
    support (s)
    enlarger (e)

    if A is the pivot and counterclockwise is positive:

    -Tb - Ts - Te = 0

    did i set the question up correctly? i don't know if A is the pivot point, or if i've got the negatives right.

    this torque business is difficult.
  2. jcsd
  3. Jul 17, 2008 #2


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    You can choose any point about which to sum torques. You chose point A, which is fine. But your plusses and minuses are off. And you don't show the distances from the forces to the chosen pivot point. And you don't show all the forces. There's the weight of the base, the weight of the support, the weight of the enlarger, and the Normal force that acts up at point B. Each of these forces produce a torque about point A. By summing these torques and setting them equal to 0, you can solve for N, the Normal force, then solve for the force required at A by noting that the sum of all forces in the vertical direction must be 0. Alternatively, you could solve for A by summing torques about B. Note also that the weight forces act thru their center of mass. Please show some work so that we can be of further assistance.
  4. Jul 17, 2008 #3
    thanks jay.

    here's what i've got:

    i changed the pivot to the bottom of the support (right above B)--(easier to calculate angles and r)

    Fs = mg = 2(9.81) = 19.62
    Fe = 6(9.81) = 58.86
    Fb = 2(9.81) = 19.62

    normal is n

    T = rFsinθ where r is the distance from the pivot to where the force acts and θ is the angle between r and F

    Tb - Ts - Te + Tn = 0

    0.15(19.62) - 0.4(19.62) - 0.8(58.86)sin20 + ___?

    if the bottom of the support is the pivot, what would the distance to the normal force be?
  5. Jul 18, 2008 #4


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    See my comments above in red. You're getting there...
  6. Jul 18, 2008 #5
    Tb - Ts - Te + Ta = 0

    0.15(19.62) - 0.4(19.62)sin20 - 0.8(58.86)sin20 + 0.3Fa = 0

    before i go any further, are these the correct positive/negative signs?
  7. Jul 18, 2008 #6


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    Yes, looks real good, good job. Now just solve for the force at A, then you need to find the normal force at B.
  8. Jul 18, 2008 #7
    solving A:

    -15.846 + 0.3Fa = 0
    0.3mg = 15.846
    9.81m = 52.82
    m = 5.4

    for the normal force:

    F = 0
    Fn = Fa + Fb + Fs + Fe
    Fn = 52.82 + 19.62 + 19.62 + 58.86
    Fn = 150.9

    the normal force only has a vertical component, right?
  9. Jul 18, 2008 #8


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    Yes, that looks perfect. Normal forces act perpendicular to the surfaces on which they act, and as such, it only has a vertical component in this problem. Looks like you're ready now to get into the torque business!
  10. Jul 18, 2008 #9
    great! thanks

    i will probably be back though. :shy:
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