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Force Friction at an angle

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data

    What force would you have to exert on a 326-N trunk up a 21.0° inclined plane so that it would slide down the plane with a constant velocity? What would be the direction of the force? (The coefficient of friction between the plane and the trunk is 0.322.)

    2. Relevant equations

    not sure what to use by
    Ff=coefficient friction *Fn

    3. The attempt at a solution
    Ihave tried drawing a picture of this
    but what i done is
    figure that the Fw=326N
    from there I am not sure where to use it or what equations to
  2. jcsd
  3. Oct 21, 2008 #2


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    Hi halo9909,

    From your picture, what force components are parallel to the incline? what force components are perpendicular to the incline? Once you have those, you can write two force equations. What do you get?
  4. Oct 21, 2008 #3
    well since 326N is the Fw
    so i would use
    so Fnormal=326+115.82=442.8
    Ff=.322*442.8=142.59 since Ff=mu*Fnormal

    Fnet= Fa-Ff
    From here I am unsure what to do,
  5. Oct 21, 2008 #4


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    Didn't you get 116.828N for this?

    These are fine; these are the components of the weight force parallel (x) and perpendicular (y) to the incline.

    Now you can use this in the force equations. I would start with the perpendicular (y) direction. What forces have components in the y direction? And what is the acceleration in the y direction? Once you have those you use them in:

    (sum of all y-components of all forces ) = (mass) (acceleration in y direction)

    which is what Fnet,y=may means.

    So you should be able to find the normal force from that equation. What do you get for that? (Remember that the components can be positive or negative.)

    After you have the normal force, repeat the process for the parallel (x) direction. Using the normal force and the equation for kinetic friction in it, you should be able to find F. What do you get?
  6. Oct 21, 2008 #5
    so the Fnormal for the vertical force would be 442.82=326+116.82
    and the horizontal would still be 161.74=394.34-142.6
  7. Oct 21, 2008 #6


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    No; the equation you need to start with is essentially:

    (y component of normal force) + (y component of weight) + (y component of friction)+(y component of applied force) = (mass) (acceleration in y direction)

    Now they tell you the acceleration is zero, and the normal force is always perpendicular to the plane, so this becomes:

    (normal force) + (y component of weight) + (y component of friction)+(y component of F)=0

    You have already found the weight's y component (magnitude); what is the y component of the friction and the force F? (It will turn out that finding the normal force is simpler than your expression.)
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