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I got a different answer

  1. Oct 31, 2003 #1
    Response to:
    force question
    i dont understand this..the way the forces go
    if someone could show me how to do it and the steps..that would be great:
    The static friction coefficient btw a wall and a pic is 0.7 while the kinetic one is 0.3. By what force shoudl one push the 6kg pic against the wall so that it will not slide?What if he/she pushed the pic with a force making an angle of 30 degrees to the horizontal (both cases)

    1. Break the F into x-y component such as x=Fcos30 ,y = Fsin30
    2.Since friction= U times normal force .and the normal force =Fcos 30
    3.Set f=mg+Fsin30
    U(Fcos30)=6(9.8) + F sin 30
    0.7(Fcos30)=6(9.8) + F sin 30
    F=553.58N

    Am I correct ?
     
    Last edited: Oct 31, 2003
  2. jcsd
  3. Nov 1, 2003 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Well, I don't know. What was the question?

    I put it that way because there are actually 3 different questions here.

    1) What force is necessary if you push perpendicular to the wall?
    Of course, there is a downward gravitational force of -mg= -6(9.8)= -58.8 Newtons and there has to be enough friction force to overcome that: friction force= UF= .7F. Set .7F= 58.8 and solve for F.

    2) What force is necessary if you push at a 30 degree angle to the wall, angled UPWARD. the weight of the picture is still 58.8 Newtons.
    The normal force will be F cos(30) so the friction force will be
    0.7 F cos(30). Now there will be TWO forces vertically: the weight of 58.8 Newtons downward and the force F sin(30) upward of your push.
    The net force downward is 58.8- F sin(30) and that must be offset by
    0.7 F cos(30): Solve 0.7 F cos(30)= 58.8- F sin(30).

    3) What force is necessary if you push at a 30 degree angle to the wall, angles DOWNWARD (the problem said "(both ways)"). THIS is the question you answered since you added F sin(30) to mg.
     
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