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Help with set up

  1. Feb 10, 2005 #1
    http://www.quantumninja.com/hw/random/2.JPG

    I was hoping someone could help me set up this problem. I am not quite sure how to start it.
     
  2. jcsd
  3. Feb 10, 2005 #2

    learningphysics

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    Generally for these types of problems without, you can get only 3 equations.

    sum of forces in x direction =0
    sum of forces in y direction =0
    sum of moments about any point=0

    You may not need all the above equations. For moments... if you do need that equation, choose the right point to take the moment about (gives you the information you need). You can technically take the moment about any point, but it'll make the math harder.
     
  4. Feb 10, 2005 #3
    So is it

    [tex]Fy= N[/tex]

    [tex]Fx=\mu N [/tex]
    [tex]Fx=\mu Fy [/tex]
    [tex] 0 = L * N * sin 53 - 0.4 L 200 * sin(37) [/tex]
     
    Last edited: Feb 10, 2005
  5. Feb 10, 2005 #4
    then you couls say

    [tex] \frac{80 sin 37}{sin 53} = N[/tex]

    sorry I am getting confused
     
    Last edited: Feb 10, 2005
  6. Feb 10, 2005 #5
    Wait isn't it this instead
    So is it

    [tex]Fy= 200[/tex]

    [tex]Fx=N [/tex]

    [tex] 0 = L * N * sin 57 - 0.4 L 200 * sin(37) [/tex][/QUOTE]

    [tex] \frac{80 sin 37}{sin 53} = N = Fx[/tex]
     
    Last edited: Feb 10, 2005
  7. Feb 10, 2005 #6
    Can someone see if my work is correct for
    a.) = N = [tex] \frac{80 sin 37}{sin 53} = N = Fx[/tex] = [tex] 80 tan(37) [/tex]

    b.) [tex] N=\mu W [/tex] so

    [tex]\frac{80 tan(37)}{200} [/tex]
     
  8. Feb 11, 2005 #7

    VietDao29

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    Hi,
    Your picture is WRONG, recheck it.
    How come an angle is both 37 degrees and 53 degrees?
    Remember that the 4 forces must go through 1 point, so that the sum of them is 0.
    Find out where the P and the N cut each other. The sum of Fx, And Fy must go through that point.
    You have Fy (200 N), you will find Fx (Friction force), and will find out the smallest coefficient of friction needed to support the ladder.
    Hope it help,
    Viet Dao,
     
  9. Feb 11, 2005 #8

    VietDao29

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    Hi,
    Or you can use moment to solve it. A lot faster.
    Hope it help,
    Viet Dao,
     
  10. Feb 11, 2005 #9

    minger

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    Yes, I would definately use summation of moments. Your perpendicular (to the axis) force from the wall will just be (.4)*(200cos37). To get the force in the x direction, which will be the load from the wall, divide that by cos53. Now you will sum forces in the x direction. Since the only other force in that direction is the friction force, it will be equal and opposite to the force from the wall.
     
  11. Feb 11, 2005 #10

    learningphysics

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    A couple of things... as Viet says, you've labeled the same angle as both 53 and 37, so I'm not able to tell if your moment equation is correct.
    If the angle between the horizontal and the ladder is 53 then your equation is correct. If the angle is 37, then your equation is wrong and you have to switch around the 53 and 37.

    So from the moment equation you can solve for N.

    sum of force in x direction=0:

    Fx-N=0

    Since you have N, now you can get Fx. That's part a of the question. Don't confuse the normal force at the top N with Fy normal force at the bottom. They are different.

    For part b, you need Fy.

    Sum of forces in y direction = 0:

    Fy-200=0
    Fy=200.

    you have friction force Fx. you have the normal force at the bottom Fy. So the coefficient of friction is Fx/Fy.

    I apologize if you already understood all this. I was confused by the variables you used, so I thought I just go through it.
     
    Last edited: Feb 11, 2005
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