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Cart in equilibrium on an incline.

  1. Nov 19, 2009 #1
    Right i'm not stuck on the whole question it's just one bit i need advice on rather than answers.


    I have a cart with 2 wheels on a 30degree slope and it is in equilibrium, i know how to work out the forces directly down onto the slope mg*cos30 and the force pulling it down the slope mg*sin30. as its it equilibrium the opposing forces are the same.

    however where i am stuck is that a force is applied to the handle facing up the slope, i know this would be the same as the force pulling down the slope...

    BUT the handle is 1 metre lower than the line parallel to the incline which runs through the carts centre of gravity which gives the forces up and down the slope mg*sin30.

    how to i account for this lower handle to work out the force needed to be applied to it to hold the cart in equilibrium?

    any advice would be great as i feel i'm stuck on something so simple.

    Michael.


    The line through the centre of gravity is 2 metres above the slope if that makes a difference.
     
    Last edited: Nov 19, 2009
  2. jcsd
  3. Nov 19, 2009 #2
    Your problem is stated very poorly. You assert that the cart has only two wheels but where are the two wheels located? One wheel in front of the other? Both side by side at the rear? Both side by side in the front? Both side by side below the center of mass?

    regardless, remember that because the cart is in equilibrium:

    [tex]\Sigma F = 0[/tex]

    &

    [tex]\Sigma T = 0[/tex]

    where T = total torque.
    (Are we also ignoring friction?)
     
    Last edited: Nov 19, 2009
  4. Nov 19, 2009 #3
    hi sorry, i didnt think the wheels where necessary as they seem to be mainly for the next part of the question finding out the load at each one, one is 1 metre up the slope from the carts centre of gravity and the rear wheel is 2metres before the centre of gravity.

    it may be easier to think of it as just a solid block on an incline being held by a handle and friction is not stated atall in the question so i'm guessing that as its not moving, in static equilibrium it's not playing any part.

    The force pulling it down the slope is 9.81kN therefore the force required to hold it static along the line passing through its centre of gravity parallel to the slope is 9.81kN in the opposite as you stated there.

    As the handle is 1metre below this centre of gravity line which is 2 metres above the slope i'm not sure how this affects the required opposing force on the handle to keeps x-axis forces totalling to 0.
     
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