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Frictional force

  1. Jun 10, 2007 #1
    1. The problem statement, all variables and given/known data
    A block is at rest on the incline (32 degrees from the horizontal). The coefficients of static and kinetic friction are Us=0.73 and Uk=0.62 respectively. The acceleration of gravity is 9.8 m/s^2. What is the frictional force acting on the 39 kg mass?

    PART 2: What is the largest angle which the incline can have so that the mass does not slide down the incline?

    2. Relevant equations

    force of friction = (Uk) Fn
    force of friction = (Us) Fn

    3. The attempt at a solution

    My initial thought was that the frictional force would be 0 because the block is at rest?

    I have no idea how to start part 2 either. Please help if you can!
  2. jcsd
  3. Jun 10, 2007 #2
    First, start out by drawing a diagram with all the forces that are acting on the block. Figure where the normal force, force of gravity, and frictional forces fit in.

    As far as your thought about it being zero goes, think about simply standing on the floor. Sure you are not moving, but since gravity is perpetually acting on you, then what prevents gravity from not pushing you through? A counter-active force, which you probably know as the normal force. How do you think this extends to the block?
  4. Jun 10, 2007 #3
    So if the normal force is 39 X 9.8 m/s^2 = 382.2 N
    how can the frictional force be found? Sorry if this is such a simple question, I'm really lost though.
  5. Jun 11, 2007 #4
    Ah, so there is one problem already. In this case, the normal force is not actually mg because it is on an incline. The normal force is actually mgcosØ, which can be found through geometry because the gravitational force will point straight downwards, but the normal force is perpendicular to the plane (the incline) and will form an angle relative to the gravitation force.

    You should verify this geometry and work it out yourself as you will need it for actually determining the force of friction acting on the block.
  6. Jun 11, 2007 #5
    [tex]F_{fs} = (F_N)(\mu_{s})[/tex]

    Because of this relation the friction force cannot be zero because the force of gravity is acting on the system, which inturn affects the normal force, which affects the friction force.
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