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Homework Help: Equilibrium/friction problem

  1. Sep 28, 2007 #1
    1. There is a block parallel and touching a vertical wall, a force F is applied at a 40 deg. angle from -x,-y. the weight of the block is 88.9N. The coefficient of static friction between the block and the wall is 0.560.
    a) what is the minimum force F required to prevent the block from sliding down the wall?
    b) what is the minimum force F required to start the block moving up the wall?

    2. Relevant equations

    3.a) Fs(max)=(0.560)(88.9N)=49.8N

    b) 88.9N +49.8 N=138.7N=Fy
  2. jcsd
  3. Sep 28, 2007 #2
    You should be more specific...

    I saw an error on your equations. The friction equation is defined as F[tex]_{sf}[/tex]=[tex]\mu[/tex][tex]_{s}[/tex]*N.

    N is the normal force of the surface. In your case is the Normal force from the wall and you used the weight of the box. You should use as N the x component of the force F you should apply.

    Try again knowing this fact.

    Note: Superscript should be underscript.
  4. Sep 28, 2007 #3
  5. Sep 28, 2007 #4
    Are you guessing?

    Here, the least force applied to prevent motion is when friction vector is pointing upward.
    Write both of your equations, [tex]\Sigma[/tex][tex]\vec{F}[/tex]=0 x and y.

    Solve equations simultaneously. You unknowns are N and F and you have two equations.
  6. Sep 28, 2007 #5
    I don't understand.
  7. Sep 28, 2007 #6
    Due to my english? It can't be that bad.
  8. Sep 28, 2007 #7
    No, I think I am having trouble because I am not sure where to place the applied force vector on my free body diagram. I have fs pointing +y, mg -y, Fn=-x, F 40 deg. in the (-x,-y)quadrant.
  9. Sep 28, 2007 #8
    Ahh. That's they way it's suppose to be. Now you make use the following equations

    [tex]\sum Fx=0[/tex]


    Note: I assumed the angle given is the angle between the force and the x axis, otherwise you should use sin

    [tex]\sum Fy=0[/tex]

    [tex]Fs-m*g+F*sin(40)=0 [/tex] [tex]Fs=\mu s*Fn[/tex] so first equation for Fn you should have:

    [tex] Fn=F*cos(40)[/tex] and you substitute this equation on the [tex]\sum Fy=0[/tex] equation:

    [tex]F*cos(40)-m*g+F*sin(40)=0[/tex] and then [tex]F=\frac{m*g}{cos(40)+sin(40)}[/tex]

    Last edited: Sep 28, 2007
  10. Oct 20, 2009 #9


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    I recently came across this question while looking for solutions to mine (which was essentially the same thing). After solving it, I thought that I'd come back and add on to the solution, in case there were still some questions or somebody else is in need of assistance.

    Fcos(x)-w+usFn=0 and Fsin(x)-Fn=0

    After simplifying the two equations that, together, find the solution to the problem, one ends up with:


    where Theta is the angle measurement, Mu is the coefficient, and W is the weight (or force).
  11. Oct 20, 2009 #10


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    Gold Member

    Likewise, for the second part, simply subtract the second term in the denominator rather than adding, because the frictional force is in the opposite direction.
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