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Critical angle for a box to start moving on an incline

  1. Mar 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A ramp with an angle given by ø (from the horizon) and a box of mass represented by "m" with a given µ=0.55 between the ramp and the box, find the minimum (critical) ø for which the box will begin to move down the ramp. Wether µ is kinetic or static is not given in the question.

    2. Relevant equations
    These equation are all in respect to the x and y axes being rotated such that they are lined up with the ramps surface
    Fg=mg
    Fgx=(sinø•mg)
    Fgy=-(cosø•mg)
    Fn=-(Fgy) which equals cosø•mg
    Ff=µFn which equals 0.55(cosø•mg) and since we label down the ramp as +, Ff=-0.55(cosø•mg)

    I think that is all we will need + some trig identities

    3. The attempt at a solution

    I'm really not to sure how to set this question up. Any and all help is much appreciated. If you could label all forces acting upon the box with the same terms that I have, that would be very helpful.
     
  2. jcsd
  3. Mar 15, 2015 #2
    You've listed a bunch of forces, but you need to structure them into single equations for the x and y components of the forces acting on the block.
    [itex]\sum F_x[/itex]
    [itex]\sum F_y[/itex]
    Draw a force diagram and work out how they all sum together.
     
  4. Mar 15, 2015 #3
    Doing that I get:

    ∑Fx= Fgx+Ff
    ∑Fy=Fn+Fgy=0

    Where do I go from here?
     
  5. Mar 15, 2015 #4
    Think about your coordinate system you're using. Is the force of friction going in the same direction as the x-component of the gravitational force. And is the normal force in the same direction as the y-component of the gravitational force? Also it would help to put the forces in terms of values you know., such as mass and gravity.
     
  6. Mar 15, 2015 #5
    I am not actually given any mass, this is supposed to be a purely algebraic solution for the answer. With that being said, putting in values for g will be redundant. In terms of the coordinate system, if I plug in vector quantities I get
    ∑Fx=(sinø•mg)-0.55(cosø•mg)
    ∑Fy=cosø•mg-(cosø•mg)=0--- although the ∑Fy is useless at this point as we already have the Fn and thus the Ff as given above.

    Would it also be important to make ∑Fx=ma, even though the question is not specifically asking for acceleration?
     
  7. Mar 15, 2015 #6
    At [itex]\theta_{max}[/itex] Before the block slides what would the acceleration be in any direction?
     
  8. Mar 15, 2015 #7
    Oh okay, I see what you're getting at. It would have to be 0 in the x and y directions as all forces will equal each other. From this we get:

    ∑Fx=0*m=0
    0=(sinø•mg)-0.55(cosø•mg)
    -(sinø•mg)=-0.55(cosø•mg)
    now dividing both sides by mg we get
    -sinø=-0.55cosø

    Should I keep going and get a zero on either side? My problem with that is that if I bring either terms to the same side as the other term, I will have something along the lines of -sinø/-0.55cosø= 0, If the quotient of two trig ratios equals 0, what does that tell me? Also, at what point (if any) will we have to take the inverse sine/cosine. How will I know which inverse to take given that I have both cosine and sine.
     
  9. Mar 15, 2015 #8
    I would get rid of that negative sign too. So since we're thinking about the ratio of these forces, and their relationship with theta, how would you go about working with the y-component and combining the two equations?
     
  10. Mar 15, 2015 #9
    Im not sure if this is what you mean, but say you get to the point where you have sinø/0.55cosø=0, you could now break up sine and cosine into their ratios so:

    [(opp/hyp)ø]/[(0.55adj)/(hyp)ø] in which case you could eliminate the hyp from both terms to get opp/0.55adj. Since the tan ratio is = opp/adj we could say that what we have there is = tan/0.55

    Not sure if thats correct, but thats what I would do.
     
  11. Mar 15, 2015 #10
    You have:
    [itex]sin(\theta)=0.55cos(\theta)[/itex]
    How would you get theta by itself from this? Think of getting theta into one term.
     
  12. Mar 15, 2015 #11
    I'm not too sure, everything I've done eliminates the theta completely.
     
  13. Mar 15, 2015 #12
    If we get the coefficient of friction by itself we get:
    [itex]\frac{sin(\theta)} {cos(\theta)}=0.55[/itex]
    What can you substitute for the ratio of sine and cosine of an angle?
     
  14. Mar 15, 2015 #13
    Oh okay, that makes sense, so then we could say:
    sin(θ)/cos(θ)=0.55
    [(opp/hyp)ø]/[(0.55adj)/(hyp)ø]
    Which will cancel out the hyp's and we are left with
    oppø/adjø=0.55

    I am now inclined (no pun intend) to say that the tanø=0.55
    and from there say that øcritical= tan-1(0.55)= aprox 29º which also happens to be the correct answer
    but since there is a ø/ø, wouldn't we just have tan=0.55 which really doesn't tell us anything?
    Essentially, why don't the ø's cancel out?
     
  15. Mar 15, 2015 #14
    The theta's don't cancel like a variable would. The sine and cosine aren't constants in front of theta, they're functions in themselves. Instead simply think of [itex]tan(\theta)[/itex] as the ratio of [itex] sin(\theta)[/itex] and [itex]cos(\theta)[/itex].
     
  16. Mar 15, 2015 #15
    Very interesting, this makes much more sense now. I think it was θmax and the fact that there won't be any acceleration at that point that really messed me up. Thanks for the help @Cake :smile:
     
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