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What in the name of inclined planes?

  1. Sep 20, 2014 #1
    Physics3.03.PNG (The degrees is 35.8)

    So far I thought that i'd solve it like this:
    Weight of block W = mg
    Component of W parallel to slope = Wsinθ
    Component of W perpendicular to slope = Wcosθ

    Call R the normal reaction force of the slope on the block.

    In the direction perpendicular to the slope, the force are balanced:
    R = F + Wcosθ

    The limiting frictional force, Fr is:
    Fr = μR = μ(F + Wcosθ)

    In the direction parallel to the slope, when on the point of slipping the limiting frictional force is just balanced by the component of W parallel to slope:
    Fr = Wsinθ
    μ(F + Wcosθ) = Wsinθ
    μF + μWcosθ) = Wsinθ
    μF = Wsinθ - μWcosθ)
    . . = W(sinθ - μcosθ)
    F = W(sinθ - μcosθ)/μ
    . = mg(sinθ – μcosθ)/μ

    However, when I plug in all of my numbers, I get -215N. And when I put that into my homework, it is wrong. Is this a computational error, was my formulas flawed?
     
  2. jcsd
  3. Sep 20, 2014 #2

    Bandersnatch

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    Which direction does each of the forces act?
     
  4. Sep 20, 2014 #3
    For Wsinθ it is the Y component of the block. The F force is acting INTO the plane (perpendicular)
     
  5. Sep 20, 2014 #4

    Bandersnatch

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    Wsinθ is the force that tries to pull the block down the incline. Fr is the frictional force. It is not the F force.
    What direction is the frictional force usually acting?
     
  6. Sep 20, 2014 #5
    Friction always opposes direction. So it's parallel to the incline (going upwards).
     
  7. Sep 20, 2014 #6

    Bandersnatch

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    O.k.
    Does the equation I quoted earlier show that the friction opposes direction? How do we usually indicate that?
     
  8. Sep 20, 2014 #7
    Friction should be negative? But you also said that sin is going into the plane...
    That opposing F should be Normal force. Should cos be negative then?
     
  9. Sep 20, 2014 #8

    Bandersnatch

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    Of course it should. Otherwise you've got an equation that says the frictional force helps pull the block down the incline.
    Remember that forces are vectors. That's why it's always good to draw a free body diagram, so that you can easily identify which force needs what sign.

    All your reasoning looks correct, apart from that one minus sign.
    The Wsinθ force is parallel to the incline. Did I ever say otherwise?
     
  10. Sep 20, 2014 #9
    Oh, I read it wrong. So my final function to solve for F would be:

    = mg(-sinθ – μcosθ)/μ

    ???
     
  11. Sep 20, 2014 #10

    Bandersnatch

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    Looks that way. Why won't you plug the numbers in and see if the answer is correct?
     
  12. Sep 20, 2014 #11

    PhanthomJay

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    GrimesA : your original equation looks good to me. But check your math for errors! You should be getting a positive number for F.
     
  13. Sep 20, 2014 #12
    This turned out to still give me a wrong number when I put it into my spot.
     
  14. Sep 20, 2014 #13
    The sine(35.8) is -.9 something and cosine(35.8) is almost -.2 so that gives me negative in the parenthesis which makes everything else negative.
     
  15. Sep 20, 2014 #14

    PhanthomJay

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    you are having math problems. Here is your original equation in your first post:

    F = mg(sinθ – μcosθ)/μ

    It is correct. Now plug in the numbers and solve. Set calculator trig functions to degrees.
     
  16. Sep 21, 2014 #15

    Bandersnatch

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    Whoa. PhantomJay is absolutely right. My advice was completely misleading.

    Just so you know where I made a mistake and not get forever confused by what I said earlier:
    When adding forces, you need to make sure the directions are indicated. In your example, the two forces must add to 0, so Fr-Wsinθ=0 (the left-hand side has got all the forces present listed, the right-hand side has got the resultant net force)
    They have opposite directions, hence opposite signs. But once you take on of the forces to the other side of the equation, the sign changes, and you do get your Fr=Wsinθ correctly.

    Apologies for causing more confusion than helping.

    And the problem seems to be indeed you using angles in radians rather than degrees.
     
  17. Sep 21, 2014 #16
    Wow... Just wow... That's what I get for borrowing my girlfriend's calculator...My calculator never switches to radians. Thanks a ton to the both of you!
     
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