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Rotational Motion maximum acceleration

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data

    7. A 130-kg refrigerator that is 2.30 m tall and 90.0 cm wide is being delivered by a truck. The center of gravity of the refrigerator is located at its geometrical center and the refrigerator is facing sideways in the truck and is prevented from sliding. What is the maximum acceleration that the truck can have before the refrigerator begins to tip over?

    a. 4.17 m/s2

    *b. 3.84 m/s2

    c. 7.68 m/s2

    d. 8.34 m/s2

    2. Relevant equations

    a=v2/r

    3. The attempt at a solution

    I have no idea what to do. I thought that the radius should be the distance between the center of mass and the center of the fridge when it starts moving, but in this case the radius will change (but it doesnt matter cause we want the start of the motion). I also think that it has to do with force and torque and breaking the equilibrium, but I dont know.
     
  2. jcsd
  3. Oct 31, 2011 #2
    Solve the problem by summing moments at the edge that would remain on the ground (center of rotation). When the moment produced by the truck accelerating equals the moment due to the weight of the refrigerator, tipping (rotation) begins. Assume the weight of the appliance is concentrated at its CG. Assume the force due to truck acceleration is also applied at the CG.
     
  4. Oct 31, 2011 #3
    How will i find the moments without the time :s
     
  5. Oct 31, 2011 #4
    Moments are force times distance. It's just like torque. The frig will tip on one of its edges. The weight assumed concentrated at its CG creates a 'moment' in one direction of rotation. The acceleration of the truck provides another force at the CG. When that moment (torque) just equals the moment (torque) due to its weight, the frig starts to tip. Maybe the term moment is no longer used. It was when I came along.
     
  6. Oct 31, 2011 #5
    Ohh i thought you ment momentum :s ups so ittl be lke
    m*a*dx=m*g*dy
    130*a*.9/2=130*9.81*2.3/2
     
  7. Oct 31, 2011 #6
    130*a*.9/2=130*9.81*2.3/2

    Take a good look at this. You have a mistake.
     
  8. Oct 31, 2011 #7
    I thought two things 1. That the acceleration should be 9.81*r and 2. That the distance is really 2.3/2 - .9/2 brcause it doesnt fall all the way
     
  9. Oct 31, 2011 #8
    130*a*.9/2=130*9.81*2.3/2

    This torque balance has the 'a' and 9.81 switched around. The height of the frig is 2.3 meters. CG therefore is located at 2.3/2 meters from truck bed. When the truck accelerates, the mass of the frig creates a torque about the rear edge (assuming truck accelerates frontward) that equals force times radius (moment arm as I call it). The force is mass times acc. The moment arm is the distance from the truck bed to the CG.

    The opposing torque is due to the weight of frig. It's radius (moment arm) is .9/2. Its force is 130 times 9.81.

    Does this help?
     
  10. Oct 31, 2011 #9
    Ohh i got it so its lke the truck is pushing the fridge through the cm but it is resisting with its weight
     
  11. Oct 31, 2011 #10
    That's it.
     
  12. Oct 31, 2011 #11
    :) thanks!
     
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