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Incline Plane

  1. Oct 31, 2008 #1
    1. The problem statement, all variables and given/known data

    The block shown in Fig. 4-48 lies on a smooth plane tilted at an angle = 25.0 degrees to the horizontal. Ignore Friction.

    Determine the acceleration of the block as it slides down the plane.

    Picture: http://img504.imageshack.us/my.php?image=webassignplanesa0.jpg

    2. Relevant equations



    3. The attempt at a solution

    Our teacher taught us to solve the problem by using right triangle trig, SOHCAHTOA etc, but I can't do that since I don't have enough information...

    Does anyone know how I can begin this?
     
  2. jcsd
  3. Oct 31, 2008 #2
    You do need to use right angled trig, and you do have enough information.

    I would start by defining all of the forces acting on the block (a free body diagram can help).

    Then sum up the forces in each direction, divide by mass and you have your answer.
     
  4. Oct 31, 2008 #3
    The problem doesn't give you any mass...

    The problem only gives you two angles!
     
  5. Oct 31, 2008 #4
    Ahh, but you don't need to know what the mass is to solve it! Take the advice from my first post; sum up the forces in each direction and tell me what you get.
     
  6. Oct 31, 2008 #5
    I'm not given any forces....

    How do I calculate a force?
     
  7. Oct 31, 2008 #6
    Okay, first we'll write this problem out algebraically.
    You need to use your basic knowledge to find out where the forces are in this question.

    We can assume that there is going to be a force on the mass due to the acceleration due to gravity. Fgravity = mg.

    There is going to also be a normal reaction force, the nature of this normal force is that it will be perpendicular to the plane it's sitting on. Can you work out the normal reaction in terms of Fgravity?
     
  8. Oct 31, 2008 #7
    Fn = Fgravity?

    I have no clue
     
  9. Oct 31, 2008 #8
    No, Fn is going to be the component of gravity that is perpendicular to the plane. We have the hypotenuse (Fgravity) and we have an angle (25 degrees). Can you use trigonometry to find Fn in terms of Fgravity? Let me know if you aren't understanding this too.
     
  10. Oct 31, 2008 #9
    I'm confused how I can use Fg to find Fn, when I don't know what Fg is.
     
  11. Oct 31, 2008 #10
    I tried 9.8Sin25 and 9.8Cos25, and both didn't work...
     
  12. Oct 31, 2008 #11
    Okay, we have to understand that we're not solving for numbers yet. We are just solving algebraically, using letters to represent the forces.

    I can see where you're a little confused, Fg = mg, but we don't know m. (g of course = 9.8ms-2)

    Try to solve for Fn in terms of Fg using trigonometry, there are more details in my previous post.

    Hint: The answer will look like either: [tex] F_gsin(\theta) [/tex], [tex] F_gcos(\theta) [/tex] or [tex] F_gtan(\theta) [/tex]. Can you work out which one and why?
     
  13. Oct 31, 2008 #12
    FgTan?

    Because I just tried FgSin and FgCos and both didn't work.
     
  14. Oct 31, 2008 #13
  15. Oct 31, 2008 #14
    We're not at the answer yet. This is just the normal reaction force. It is infact [tex] F_gcos(\theta) [/tex]. This is because the if Fg is the hypotenuse, Fn is the adjacent angle to theta. Using SOHCAHTOA we can derive [tex] F_gcos(\theta) [/tex].

    Now, we have the normal reaction force and the gravitational force. We can see though, that these two forces do not balance out (they are not equal and opposite), which hints that there must be another component of force in the system. This is true.
    It is going to be the component of gravity that is parallel to the plane.
    We have the component perpendicular to the plane ([tex] F_gcos(\theta) [/tex]), how can we use trigonometry now to solve for the component parallel?

    Once you have worked out this force in terms of Fg (mg) you can divide by mass to achieve your answer.
     
  16. Oct 31, 2008 #15
    Can you draw a picture?

    My picture is wrong then?
     
  17. Oct 31, 2008 #16
    Not at all, your picture is correct. Here is a modified version of it:

    [​IMG]

    pardon the poor drawing skills.. An important concept (but sometimes hard to get a grip of initially) to realise is that just as the three sides of the triangle are proportional to the angles of the triangle, so are the forces due to gravity and normal force proportional to the same angle.

    I have drawn another triangle with the same angle in it, and you can see that Fg is the hypotenuse, Fn is the adjacent vertex, and the ? vertex is the one that we are solving for.

    What would this vertex be in terms of Fg (mg) and a trigonometric function?

    I hope you are understanding this, let me know if you aren't.
     
  18. Oct 31, 2008 #17
    The ? would be equal to FgSin
     
  19. Oct 31, 2008 #18
    Bingo!
    We also know that Fg = mg (right?)

    Therefore, we can say that [tex] F_{parallel} = mgsin(\theta) [/tex]

    We also know that F = ma. Therefore, [tex] mgsin(\theta) = ma [/tex] and we are solving for a (acceleration). Can you solve this?
     
  20. Oct 31, 2008 #19
    Yes

    Devide by M on both sides and the m's cancel each other

    Therefore

    gsin(theta) = a

    But I already tried that..

    9.8sin(25) = 4.14
     
  21. Oct 31, 2008 #20
    Hmm you will want to get a second opinion on this then, I interpreted the question as solving for the acceleration in the x plane (possibly because of the axes they provided).
     
    Last edited: Oct 31, 2008
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