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Expression for the magnitude

  1. Oct 12, 2004 #1
    Find an expression for the magnitude of the horizontal force in the figure for which does not slip either up or down along the wedge. All surfaces are frictionless.




    I am a little confused, what is this question really asking? i have attempted sevral trig equations but i cant get the answer correct...anyone point me in the right way?
     

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  2. jcsd
  3. Oct 12, 2004 #2

    Doc Al

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    Staff: Mentor

    First identify all the forces acting on the block. Then apply Newton's 2nd law. (Hint: vertical forces must add to zero, otherwise the block will accelerate in the vertical direction.)
     
  4. Oct 12, 2004 #3
    im still so confused on this problem....ok i have the forces for the block and the wedge the same, they both have a line pointing up and down, and pointing right for the force...but im completly lost on how to begin solving this problem.
     
  5. Oct 12, 2004 #4
    If [tex]m_1[/tex] is not to slip in any direction, it will move forward together with [tex]m_2[/tex] with accelerelation a. Upward and right direction is choosen to be y and x axis.
    Consider [tex]m_1[/tex]:
    y component :
    [tex]N_{1}sin(90-\theta)+(-m_{1}*g)=0[/tex]
    [tex]N_{1}cos\theta=m_{1}*g[/tex]…(1)
    x component :
    [tex]N_{1}cos(90-\theta)-m_{1}*a=0[/tex]
    [tex]N_{1}sin\theta=m_{1}*a[/tex]…(2)
    Consider [tex]m_2[/tex]:
    x component:
    [tex]F+(-N_{1}sin\theta)=m_{2}*a[/tex]
    [tex]F-N_{1}sin\theta=m_{2}*a[/tex]…(3)
    (2) into (3):
    [tex]F-m_{1}*a=m_{2}*a[/tex]
    [tex]F =a*(m_{1}+ m_{2})[/tex]…(4)
    (2) divided by (1) :
    [tex]tan\theta=\frac{a}{g}[/tex]…(5)
    (5) into (4) :
    [tex]F=gtan\theta(m_{1}+ m_{2})[/tex]
     

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    Last edited: Oct 12, 2004
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