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Finding the angle

  1. Jul 26, 2005 #1
    A uniform board is leaning against a smooth vertical wall. The board is at an ange above the horizontal ground the coefficient of static friction between the ground and the lower end of the board is 0.650. Find the smallest value for the angle such that the lower end of the board does not slide along the ground.

    I've tried drawing all the forces. I made the sum of the forces in the y direction equal to zero, the x direction zero, and the sum of all the torques to equal to zero. I came up with that 0.650 = the normal foce exerted by the wall divided by the force exterted by the weight of the board = the lever arm of the weight of the board in the center divided by the lever arm of the normal force of the wall. Then I used tan(theta) = 0.65 and got theta to be 33 degrees.

    However this answer isn't right. the answer is 37.6 degrees. What should I do to get this answer?

    smooth means that there is no friction exerted by the wall.
     
    Last edited: Jul 26, 2005
  2. jcsd
  3. Jul 26, 2005 #2
    Solution.

    Well..

    I don't know where you are in your physics education at present, this is the method I used to solve the problem.

    What you have to do is draw your free body diagram, you will notice since the wall is frictionless that there are no vertical forces acting on the board at the point where it comes in contact with the wall. The only force is the wall pushing the board away from it. Since this is the only force, and its horizontal, this force must be equal to the only other horizontal force, which is the force of friction on the floor.

    Now, I took the rotational equilibrium around the point where the board touched the ground (which means you can disregard forces acting on that point since they are of distance 0 from the reference point), that is, (tou) = I(alpha), since this is in equilibrium ( ie. barly staying on the wall). We can set (tou) equal to zero, therefore we must account now for all the forces that are acting on the board and they must all add to zero.

    First we take the force of gravity which is acting on the center of gravity of the board. This force is equal to (L/2)cos(alpha) * Fg (where L is the length of the board, alpha is the angle between the horizontal floor and the board, and Fg is the force of gravity, or Mass * Gravity). I followed a clockwise convention when solving this, so that this was a positive force, adding to the rotation of the board. The only other force acting on this is the force that the wall is pushing with. This force is equal to -Lsin(alpha)*Ff (or the force of friction which is Mass * gravity * mu). The calculation is now simple and follows like this:

    Mg(L/2)cos(alpha) - (mu)MgLsin(alpha)=0 -Factor MgL out and divide by it.
    1/2cos(alpha) - (mu)sin(alpha) = 0
    1/(2mu) = tan(alpha)
    therefore (alpha) = arctan(1/(2mu))

    and the answer is 37.56859202 acording to my calculator :)

    Hope you can understand this as its my first post. And hope it helps.

    Sasha.
     
  4. Jul 26, 2005 #3
    Sash, thank you so much for your help. I finally understand how you derived the answer. I used a slightely different method, but thank you.
     
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