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Incline Equilibrium with Unique Applied Force

  1. Oct 4, 2005 #1
    Wondering if someone could help me with my last set problem? I've been looking at it for quite some time now. The incoming force on the block is confusing me...as a result, I'm not sure how to set it up. I've worked it with the force parallel to the incline as well as perpendicular - but not as shown in this diagram (attached) - the force on block is horizontal to ground.

    A 2.00-kg block is held in equilibrium on an incline of angle 60.0 degrees by a horizontal force applied in the direction shown. If the coefficient of static friction between block and incline is 0.300, determine (a) the minimum value of the applied Force and (b) the normal force exerted by the incline on the block.

    Would you still set this problem up as F = mg X sin (angle) - (static friction coeff.) X mg X cos (angle)?

    And how do you solve part (b)?

    Also, I assume I would need to convert the 2 kg into Newtons - by multiplying 2.0 kg X 9.807 = 19.614N?

    Any help would be appreciated.
    Thanks in advance
     

    Attached Files:

  2. jcsd
  3. Oct 4, 2005 #2
    I'm a little rusty on this stuff, but basically, you're missing a term in the equation F = mg X sin (angle) - (static friction coeff.) X mg X cos (angle) that includes the given horizontal force. What you have right now is just F_total-parallel = F_gravity-parallel - F_friction. What you need is F_total = F_gravity-parallel - F_friction + F_horizontal-parallel. (The signs here might be off...)

    For part b, you need to factor in that the force is pressing the box into the incline, which would intuitively increase the normal:

    F_total-perpendicular = N - F_gravity-perpen - F_horiz-perpen

    I might still be missing something though. I believe the frictional force is proprotional to the normal, which is not just mgcos(o), so that really complicates things, unless of course I'm wrong.

    On a minor note about semantics, you don't "convert" kg into N since one is units of mass and the other is units of force.
     
    Last edited: Oct 4, 2005
  4. Oct 4, 2005 #3
    Thanks for the info - I'm not sure I follow completely, but I'll be sure to look at your response in detail. So, am I to assume that the kg mass can be multiplied by gravity and that should suffice? I gues it's just 2kg and not N because it's in equilibrium on the incline? The normal force explanation you gave definitely makes sense...

    Anyway, thanks for the help and I'll see if I can make this work...
     
  5. Oct 4, 2005 #4
    Shed More Light?

    Well, I'm still not having any luck! Can anyone else shed some light?
     
  6. Oct 4, 2005 #5

    mukundpa

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    First of all the normal reaction is not [tex] mg cos \theta [/tex] but it has to balance the normal component of the applied force as well. Hence the normal reaction of the incline is

    [tex] N = mg cos \theta + F sin \theta [/tex]

    We required the minimum force for that friction should be up the incline and this with the component of F in upward direction will balencing the component of weight. so

    [tex] mgsin \theta - \mu N - F cos \theta = 0 [/tex]

    Solve the two equation for F
     
  7. Oct 4, 2005 #6
    ?

    Thanks. OK, but if I am to solve for F in the first equation, what is the variable N? And kind of a stupid question, but for the mg part of the equation, is it the 2.00 kg mass X 9.8?

    Can someone please help a little further, I'm almost there - my calcs are not correct!

    Thanks
     
  8. Oct 4, 2005 #7

    mukundpa

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    No until you know N how you can find F using that equation.

    N is the force exerted by the incline on the block for not allowing it inside it, called normal reaction. Friction force depend on it, and the nature of the surface given by [tex] \mu [/tex]


    yes it is 2 x 9.8 = 19.6 newton.
     
  9. Oct 4, 2005 #8
    Thank you! Unfortunately, I didn't get it worked out in time for the grade, but I'll be ready for the next one that's similar. Thank you for the help!
     
  10. Oct 5, 2005 #9

    mukundpa

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    Sorry, as I am from India the time difference...........
     
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