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    Rotation: Tipping or sliding.

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    I have a 72" tall air-compressor that weights 120 pounds with a 20" base. I have a cart that is 36"wx72"l the surface of the cart where the compressor will be mounted is 10" off the ground. The cart wheels are in the corners, it is rated to carry 3600 pounds and weights 114 pounds. If the compressor is placed centered on the cart will it have a tip over problem because of a high center of gravity.

    © BrainMass Inc. brainmass.com December 24, 2021, 5:13 pm ad1c9bdddf
    https://brainmass.com/physics/rotation/rotation-tipping-sliding-36258

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    I have a 72" tall air compressor that weights 120 pounds with a 20" base. I have a cart that is 36"wx72"l the surface of the cart where the compressor will be mounted is 10" off the ground. The cart wheels are in the corners, it is rated to carry 3600 pounds and weights 114 pounds. If the compressor is placed centered on the cart will it have a tip over problem because of a high center of gravity.

    As the compressor is mounted in the center and if the cart is moving with constant velocity there is no chance of tipping over of the compressor. The problem will arise only when the cart is accelerating. At the time of accelerating of the cart the compressor is to be accelerated with it, this is done by the force of friction between the surface of the cart and the compressor. This force is also create a torque which is trying to rotate the compressor anticlockwise about its center of mass and due to which the line of action of the normal reaction of the surface on the compressor (N = mg) is shifting to left, providing a torque to balance the torque due to the friction. As the acceleration increases the normal reaction shifts and finally having a shift = d/2. A further increase in the acceleration will cause the tipping about point O.
    If the acceleration a of the cart is such that the compressor is just not tipping i.e. at the verge of tipping the torques due to the normal reaction and due to friction about the center of mass of the compressor are just balancing. Gives us
    ma (h/2) = mg(d/2)
    or a = (d/h)g = (20/72)9.8 = 2.72 m/s2

    Therefore to avoid the toppling of compressor the acceleration of the cart should be less then 2.72 m/s2.

    Note:
    [For this analysis the friction between the surface of the cart and the compressor is considered sufficient to avoid any sliding. Other before toppling it will slid backward. For this also we can say that coefficient of static friction must be greater then (a/g).]

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 5:13 pm ad1c9bdddf>
    https://brainmass.com/physics/rotation/rotation-tipping-sliding-36258

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