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Applications of Newton's Laws

  1. Dec 22, 2009 #1
    1. The problem statement, all variables and given/known data

    8. (II) A car can decelrate at -4.80 m/s^2 without skidding when coming to rest on a level road. What would its decelration be if the road were inclined at 13 degrees uphill? Assume the same static friction force.

    2. Relevant equations

    Ffr = MuFn
    F = ma

    3. The attempt at a solution

    Ok I don't see what I'm doing wrong here.

    The first think I did was find Mu on the horizontal surface

    F = ma = Ffr = Mu m g
    mass canceled out
    a = Mu g

    Mu therefore = a/g
    Mu = 4.8/9.8 = .4898

    Then for the up hill

    F = ma = Fgx + Ffr
    = mg sin (theta) + Mu mg cos (theta)

    mass canceled out

    a = g sin (theta) + Mu g cos (theta)
    a = 9.8 sin (13) + .4898 (9.8) cos (13) = 6.9 m/s^2 then it would be negative 6.9 m/s^2
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 22, 2009 #2

    kuruman

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    Gold Member

    It looks fine. What makes you think it is incorrect?
     
  4. Dec 22, 2009 #3
    The book said negative 7.00 m/s so I just wanted to make sure that I was not going crzy

    Thanks!

    See here's what some other person told me and got 7.00 m/s^2 and I'm not sure how they got it or even follow it at all

    The deceleration rate is the ratio of applied force (backwards along the direction of motion) to the mass.

    The mass remains the same. Therefore the acceleration is higher by the ratio
    (Fstatic + M g sin 13)/Fstatic

    You also know that the decelerating force applied by the tires is
    Fstatic = M*4.80,
    (from Newton's law), and that it stays the same

    Therefore
    (deceleration-uphill)/(deceleration-level)
    = 1 + M g sin 13/M*4.80
    = 1 + g sin 13/4.80 = 1.4597
    (deceleration-uphill) = 1.4597*(-4.80)
    = -7.007 m/s^2

    If I had used 9.80 instead of 9.81 m/s^2 for g, I would have gotten -7.004 as an answer

    --------------------------------------------------
    I did not follow this solution at all and for some reason this person got the answer in the book...
     
  5. Dec 22, 2009 #4
    I do not see what is wrong with my soultion though... aparently something is and there is some other way to solve it that I can not make sense of
     
  6. Dec 22, 2009 #5
    Both ways are fine;

    for the second one, your friend is using this reasoning;

    [tex] \frac{F_{uphill}}{F_{horizontal}} = \frac{ma_{uphill}}{ma_{horizontal}}[/tex]

    the m's cancel therefore the new acceleration you wish to find out, [tex] a_{uphill}[/tex]

    is equal to [tex] a_{uphill} = \frac{F_{uphill}}{F_{horizontal}} a_{horizontal} = -\frac{F_{uphill}}{F_{horizontal}}4.8 [/tex]

    Also on a side note, while using the coefficient of friction did give you the right answer, I question the logic in using it here;

    think about it physically, when you apply the brakes, what happens? The wheels are forced to slow down by the brakes, in the question it says that the wheels do not skid, so the coefficient of friction (the maximum resistance) cannot be applied here i'm afraid.

    Incidentally, there was no need at all to consider the coefficent of friction remember;

    [tex] F_{NET} = ma [/tex] so therefore, I simply consider the net force, which is the force of the brakes, -4.8m, and add the force of gravity, -mgsin(13);

    so -4.8m - mg sin(13) = ma

    -4.8 -9.8sin(13) = a

    a much nicer way, no?
     
    Last edited: Dec 22, 2009
  7. Dec 22, 2009 #6
    I think the sign change is caused by how you defined your y and x axes
     
  8. Dec 22, 2009 #7

    kuruman

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    You are correct and the other method is incorrect. Ratios don't always work when you solve physics problems, and this is one case where they don't. If you do it (correctly) as you have, the ratio of uphill to flat surface accelerations is

    [tex]\frac{a_{uphill}}{a_{flat}}=\frac{\mu g cos\theta + g sin\theta}{\mu g}=cos\theta+\frac{sin\theta}{\mu}[/tex]

    If you do it the other (incorrect way), and use aflat = μg, you get

    [tex]\frac{a_{uphil}}{a_{flat}}=1+\frac{sin \theta}{\mu}[/tex]

    Do you see the difference? When the angle is zero, both expressions correctly reduce to 1. However, as the angle increases, so does the discrepancy between answers. At 13o , the cosine is 0.974 - not very far from 1 - that is why your answer is close (but smaller) than the other answer.

    My advice to you is to forget that you ever saw the other solution. You did it right in the first place. :wink:

    *** Addendum on edit ***
    There are two ways to look at the problem. The statement "Assume the same static friction force" may require interpretation. If it is to be taken literally and the force of friction is independent of the angle of incline, then method of ratios is correct as chewy0087 pointed out. This assumption may be unphysical. It is, perhaps, more appropriate to assume that the coefficient of kinetic friction is the same in which case GreenPrint's original solution is correct.
     
    Last edited: Dec 22, 2009
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