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Force q.

  1. Mar 22, 2008 #1
    1. The problem statement, all variables and given/known data
    a 5.1*10^2 kg bundle of bricks is pulled up a ramp at an incline of 14 degrees to a construction site. the force needed to move the bricks up the ramp is 4.1*10^3 N. What is the coefficient of static fiction between the bricks and the ramp?


    2. Relevant equations
    well Fsmax= 4.1*10^3 because thats how much force is needed to overcome in order to move the bricks and Fn=510*9.81*cos14 then you would have to divide to get Us and i got 0.84 but the book says its wrong. so could some1 please help me figure out what i did wrong?
    thx a ot.


    3. The attempt at a solution
     
  2. jcsd
  3. Mar 22, 2008 #2
    I think you're confusing Fsmax with force due to tension. The force of static friction = [tex]\mu[/tex]Fn

    To approach this sort of problem, ALWAYS start off with a free body diagram. I actually just started using these physics forums so I don't know how to draw one on this site but i'll try to explain to you how to do it. When you draw it, you find that the sum of the forces in the Y direction is the following:

    Sum Fy = Fn - mgcos14.[/B] You get mgcos14 by breaking mg into vector components.

    Sum of the forces in the x direction is
    Sum Fx = Tension - Ffriction - mgsin14.[/B]

    You can set Fy equal to ma since F=ma. Since the block is not being moved in the y direction, it has no acceleration in that direction so ma becomes zero.
    Therefore, Fn-mgcos14 = 0 and Fn=mgcos14

    Next, set Fx = ma. Again, you are trying to find the coefficient of STATIC friction, which means the friction that occurs while the object is not moving. So acceleration is zero again and you set Fx=0. From this you get
    T-Ffriction-mgsin14 = 0
    You know that Ffriction = [tex]\mu[/tex]Fn so the above equation becomes
    T-[tex]\mu[/tex]Fn-mgsin14 = 0

    Now you can solve for [tex]\mu[/tex]. You are given tension and mass in the problem. And you know Fn because you solved for it above. So just plug everything in and see what you get! Let me know if you get the right answer. =)
     
  4. Mar 22, 2008 #3
    ooo yup ur right. could you explain what exactly is tension? because in my book when they say like #N is needed to set something in motion the example in the book used that as Fsmax.
     
  5. Mar 23, 2008 #4
    Tension is a force that usually exists in rope when the rope is being stretched. For example, if I were to push a box up an incline, I would be exerting a force on the box. If I were to use a rope to pull a box up an incline, there would be a force pulling the box up and that force is called tension (and it is in the rope).
    Basically, don't get tension and force confused. They're pretty much the same thing, just different names. And tension is more specific, it is usually only associated with rope or string. Don't know if that makes sense or not.

    When your book says that you need #N to set something in motion, that is the tension required to OVERCOME the static friction force. In other words, the block is being kept in place by the force of friction (which opposes the tension force, if you draw a free body diagram, you can see this). So in order to move the block, you need to apply at least as much tension as there is friction force. And when you do that, the block begins to move. Remember, there's still friction acting on the block once it moves but now its kinetic friction.
    Hope this helps!
     
  6. Mar 23, 2008 #5
    but it doesnt talk about ropes in the problem? coz i thought you only use tension when there is pulleys.
     
  7. Mar 23, 2008 #6
    you have to assume that there is a rope because the problem says that bricks are PULLED up.
     
  8. Mar 23, 2008 #7
    oo i seee thanx alot for the help!
     
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