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Torque required to tip a safe

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data

    A safe is in the back of a truck resting against a ledge that is .0381 m high. The truck is traveling at a velocity of approx. 29.1 m/s (about 65 mph). The driver then is forced to make an emergency stop, causing a negative acceleration of -6.86 m/s2. (solved for the acceleration with friction and velocity).

    What will the safe do?

    The safe's mass: 1063kg
    Friction Coefficient (safe on truck bed): .8

    2. Relevant equations

    Force(net)x = 0

    Force(net)y = 0

    Torque(net) = r χ F

    3. The attempt at a solution

    As it stands, I have tried to construct an extended Force & Torque diagram. The pivot point of the system is located at the top of the ledge. The object is not accelerating in either the x or y direction so the net force in those directions is 0.

    My main issue, is why is the normal force shifted (I have looked for explanations, they show the normal force has shifted towards the pivot, but they don't say why). Additionally, I do not know how to solve the equation for the minimum force required to tip the safe.
     
  2. jcsd
  3. Mar 24, 2013 #2

    Simon Bridge

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    Welcome to PF;
    ... I cannot tell what you have tried from just this.
    You sound confused - start again from the beginning and go through it slowly.

    The truck slams on the breaks ... the truck slows sharply... but the law of inertia means...
     
  4. Mar 25, 2013 #3

    CWatters

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    Is that all the info given? Is there a missing diagram?

    Which side of the safe is the ledge? In front or behind? (It's not clear why you are given the coefficient of friction and a ledge).

    Is the width and height of the safe given?

    Why do you say that? If the safe isn't decelerating then there is no force causing it to tip over.

    Because if it's just about to tip up "all the weight" is at the edge near the pivot and none onder the back edge. See youtube for video of motor bike doing a nose stand. As the back wheel is just about to lift off ground all the weight is on the front wheel.

    At the point where it's just about to tip over the torque will sum to zero. Any more torque in the forward direction and it will tip over.
     
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