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Rotational Torque, Forces, Tipping

  • Thread starter johnnnnyyy
  • Start date
  • #1
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Homework Statement


A refrigerator has a vertical center of mass .5m above the floor, and a horizontal center of mass of .5m which is in the horizontal center, the mass of the fridge is 200kg. What is the maximum initial horizontal force a machine can apply to the fridge which is at rest for the fridge to slide and not tip? The machaine is applying the force 1.5m above the floor. Coefficient of static friction is 0.2.


Homework Equations



F=umg
F=mgL/2r


The Attempt at a Solution


Minimum force required for refrigerator to tip-
F=mgL/2r: F=(200kg)(9.8)(1)/2(1)=980N
Minimum force required for refrigerator to tip
F=umg: F=(.2)(200)(9.8)=392N
I have no clue how to solve what will happen when a force more than the minimum force is applied by the machine.
 

Answers and Replies

  • #2
haruspex
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I'm not making much sense of your equations. Pls define L, r and F. Are you using F to mean two different things?
You have two calculations for "Minimum force required for refrigerator to tip" leading to two different answers. Is one the descriptions wrong?
 

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