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Rotational dynamic equilibrium problem on an angle

  1. Oct 28, 2012 #1
    A man drags a 67.0-kg crate across the floor at a constant velocity by pulling on a strap attached to the bottom of the crate. The crate is tilted 24.0° above the horizontal, and the strap is inclined 64.0° above the horizontal. The center of gravity of the crate coincides with its geometrical center, as indicated in the drawing. Find the magnitude of the tension in the strap.



    Translational and rotational equilibrum (net torque is zero and net force is 0) for both x and y axes
    center of gravity equation
    torque-lever arm multiplied by force
    torque=moment of inertia multiplied by angular accelaration
    force=mass x acceleration

    I do not have an attempt at the solution. I tried different things and nothing seemed to work
     
  2. jcsd
  3. Oct 28, 2012 #2

    TSny

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    Hi mx2ko. Welcome to PF.

    You gotta give us something. :devil:

    So, tell us more specifically what you tried. Or, tell us what basic physics principles you think apply to this problem and why you're having difficulty with applying those principles.
     
  4. Oct 28, 2012 #3
    I know that the net force in the x and y direction both equal zero and the net torque equals zero because the object is in equilibrium since he is pulling the crate at a constant velocity, which means there is no acceleration.
    I tried to solve for the tension force by making the net torque and forces equal to zero and then finding T based off of everything else that I could find. The double angle I think is what is messing me up. I know the weight of the crate divided both the length and the width/height of the crate in half I think and the normal force on the crate is acting on the corner that is touching the ground I think.
     
  5. Oct 28, 2012 #4

    TSny

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    Great! You're on the right track. :smile:

    Let's see. You have 3 unknown forces: the tension in the strap acting at one of the lower ends of the crate and the normal force and friction force acting at the other lower end of the crate that touches the ground.

    Can you see a good place to take the origin (axis) for calculating the torques?
     
  6. Oct 28, 2012 #5
    is the good place placing it on the spot where the crate touches the ground so the normal force and the friction are both zero? and its a typical axis not one that is tilted on an angle
     
  7. Oct 28, 2012 #6

    TSny

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    Yes, that's a good place to pick the axis. Then the torques due to the normal force and friction will be zero. The axis will not be tilted. It will run along the edge of the crate that touches the floor.
     
  8. Oct 29, 2012 #7
    so then there would only be the weight and the tension that have torques. the weights lever arm is half the length x cosine24degrees i think. but what would the lever arm for the tension be since that would be needed to find the tension
     
  9. Oct 29, 2012 #8
    actually I don't even think the lever arm of the weight is half the length x cos24degrees, but i could be wrong. so in that case I don't know how to find either of the lever arms
     
  10. Oct 29, 2012 #9

    TSny

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    Yes, the only nonzero torques about the axis will be the torques due to the tension and the weight.

    The lever arm of the tension can be expressed in terms of the length of the crate and the given angles. If the length of the crate is not given, then you will have to leave it in the expression as an unknown and hope it cancels out later .

    To find an expression for the lever arm of the weight, you will need to know something about the height of the crate. Is anything given about the height? For example, is the height the same as the length?
     
  11. Oct 29, 2012 #10
    the length of the crate is .9 m and the height is .4 m. I don't understand what you mean by the lever arm of the tension is expressed in terms of the length of the crate and the given angles. Also, since the height is given, what would the lever arm for the weight be.
     
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