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Horizontal force outward inclined plan.

  1. Jan 5, 2014 #1
    1. The problem statement, all variables and given/known data
    A block of 5 kg is on plane that is inclined of 37°, an outward (It is not directed toward the plane but it drags the the block horizontally away) horizontal force of 50 N is applied on the . The friction coefficient of that plane is 0,30.

    What is the total force that is applied in the perpendicular direction to the plane? (The book gives 9,0 N answer)

    2. Relevant equations
    F=m*a


    3. The attempt at a solution
    So...there is something i can't understand. I know that on anything that is on a plane, there is a normal force N that is the reaction. But i don't in this case...Shouldn't be the total force in the (y) direction 0? Shouldn't N be Py (Weight force in y direction) - Fy (Component in y direction of the horizontal force)?

    If the total force in y direction isn't 0, then doesn't the body goes away from the plane itself?
     
  2. jcsd
  3. Jan 5, 2014 #2

    ehild

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    There are forces acting on the block, and forces, acting on the plane. The problem asks about the normal component of the total force acting on the plane.

    You speak about total force but do not say what is the object the forces are exerted.

    So what is you speak about? The block or the plane?
     
  4. Jan 5, 2014 #3
    I need to find the total force applied along y (The direction perpendicular to the table).

    The horizontal force is applied on the block.

    Ps (The problem gives as answer 9,0) It's not clear whether it wants to total force on the plane or on the block but i assume the total force on y is 0? Or not because of Fy (y component of the horizontal force)?
     
  5. Jan 5, 2014 #4

    tiny-tim

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    Hi Bedeirnur! Welcome to PF! :smile:
    Yes … that's the starting-point for all static or dynamic problems on a flat surface. :wink:
    Yes … isn't that 9 ? :confused:

    (cos = 0.8, sin = 0.6)

    Show us your calculations. :smile:
     
  6. Jan 5, 2014 #5
    Yeah actually. The problem answer says that the total force on y is 9. But in my calculations it is 0.

    Fy (Component on y of the horizontal force) + N (reaction) - Fp (Weight force on y) = 0

    50sin37+N-mgcos37=0

    That's what i've arrived to. And i can't actually understand how the total force on y is 9 while i get 0.
     
  7. Jan 5, 2014 #6

    tiny-tim

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    oh i see!

    yes i agree the total force has to be 0, since as you say the block stays on the plane!

    but i guess they're asking for N :confused:
     
  8. Jan 5, 2014 #7
    :confused:Well, the problem doesn't say the block stays on the plane. I've just assumed so.

    How can i know if the block detaches from the plane? Maybe that's the problem, the block detaches from the plane, or not?

    I've thought as the component in y of the weight force is greater than the component in y of the horizontal force that it doesn't detach (?)

    Yep you're right...The only thing that comes up to be 9Newton is N, can't understand how that is the "total force" in the y direction.

    Ps. Thanks for the forum welcoming :)
     
    Last edited: Jan 5, 2014
  9. Jan 5, 2014 #8

    ehild

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    You need to read the question carefully.
    So the y component of force that acts on the plane....

    ehild
     
  10. Jan 5, 2014 #9

    CWatters

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    The question is badly worded or you transcribed it badly. I believe they are asking you to calculate N the normal force on the block. You will need that later to calculate the friction force (eg to check the block isn't sliding).

    Rearrange your equation and you get..

    N = mgcos37 - 50sin37
    = 5*9.81cos37 - 50sin37
    = 9 Newtons

    Your choice of polarity suggests that is towards the plane so the block isn't pulled off the surface.

    However you also need to check that the friction force is enough to stop it sliding. Write an equation for the forces acting parallel to the plane.
     
  11. Jan 5, 2014 #10
    Thank you for the answer.
    First of all i'd like to show you an image so it is clearer.
    jTqcy4C.png
    Did you assume they are asking the N normal force by the answer or by what? Because "Total force" doesn't actually mean N or not? Is the problem in how they asked the question?

    For the rest, i've understood what happens in the parallel direction to the plane, just had that problem in the y direction.


    On the plan only acts Fy (y component of weight force), and surely that is not 9 N, so they're not asking this.
     
    Last edited: Jan 5, 2014
  12. Jan 5, 2014 #11

    ehild

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    The block exerts force on the plane. That is equal in magnitude with the force the plane exerts of the block, that is the normal force.

    The block stays on the plane, so it does not move in the direction normal to the plane. As you pointed out, the y component of the net force on the block is zero N+Fy-Gy=0.


    ehild
     
    Last edited: Jan 5, 2014
  13. Jan 5, 2014 #12

    ehild

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    No, the weight of the block is the force of gravity exerted on the block. (It would be the normal force in case of a horizontal surface).

    The plane pushes the block, the block pushes the plane, according to Newton's Third Law. The force of interaction between them is called the normal force.

    ehild
     
  14. Jan 5, 2014 #13

    CWatters

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    The wording..

    ...is unclear.

    Why have they used the word "applied"? That suggests they might want the component of the 50N force that acts perpendicular to the plane.... but they also use the word "total" which suggests some kind of sum. The other components of that sum might nor be considered "applied". Is gravity considered "applied"?

    If the block is stationary the only way I could get an answer of 9N is if I assume they mean the total normal force acting on the block resulting from the applied 50N and gravity.

    However I don't think the block is stationary. I only had a quick look earlier but it appears it will be sliding/accelerating down the slope. In which case there is an issue with using g in mgcos37.
     
  15. Jan 5, 2014 #14
    In facts it'll slide/accelerate. All clear thanks :)

    What do you mean by saying there is an issue with using g in mgcos37? Isn't that the component in y of the Weight force in the y direction?
     
  16. Jan 5, 2014 #15

    ehild

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    The forces acting on the block are: gravity; the horizontal applied force F; the normal force from the plane.

    The y (normal to the plane) component of F is less than the normal component of gravity, so the plane has to exert an outward normal force N , and the block stays on the plane (sliding down).

    The block exerts also force on the plane: It is -N. The problem asks that force.

    ehild
     
  17. Jan 5, 2014 #16

    ehild

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    There are two interacting objects: the plane and the block. If you speak about total force you need to name the object. The problem asks the force applied on the plane.
    The only force applied on the plane is the force exerted by the block and it is of the same magnitude as the normal force.


    ehild
     
    Last edited: Jan 5, 2014
  18. Jan 6, 2014 #17
    In this case ehild i just can't understand you, sorry but for my little knowledge i can't arrive to your point.

    Let me explain what i've understood and someone tell me if that's what was meant.
    For newton's third law : "For every action, there is an equal and opposite reaction."

    Let's call the y component of gravity 'Fp'
    So what is pushing towards the plane? Fp-Fy.
    (-39+30=-9 N)
    That is -9 Newtons towards the plane.

    Then to compensate the plane exerts a force that is equal but contrary to -9 so it exerts a force of 9 Newtons.

    Is that correct?

    Thank you for all your answers, you've all been super helpful!
     
  19. Jan 6, 2014 #18

    CWatters

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    If it is accelerating down the slope it's partially in free fall so it's weight is not mg. It's weight would be m(g-a) where "a" is the vertical component of the acceleration.
     
  20. Jan 6, 2014 #19

    ehild

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    Almost. The y components of the forces acting on the block are the y component of gravity, the y component of the horizontal force and N, the normal force of the plane. They add up to zero. You get N=9 N, the force exerted by the plane. According to Newton's Third Law, the same but opposite force is exerted by the block on the plane and that was the question.

    When you speak about forces always define the object the forces are applied on. Newton's Third Law is about interacting bodies. If one exerts a force F on an other one, that other body exerts -F force on the first one.

    ehild
     
  21. Jan 6, 2014 #20

    ehild

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    Fp-Fy-N=0 is the sum of y components of the forces exerted on the block.

    And, according to Newton's Third Law, the block exerts -9 N force on the plane. And that is the answer: the block pushes the plane with 9 N force. .

    ehild
     
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