1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Gold Member

    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


    User Avatar
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Equilibrium/friction problem