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Homework Help: More A.P. French Mechanics

  1. Sep 16, 2005 #1
    A. P. French has a knack for creating simple problems I can't solve. I would appreciate some suggestions. (This is 6-8, for those who have his book on mechanics.)

    In part 1, a car, m=10^3 kg, is traveling at 28 m/s on a horizontal straight road. The driver sees a tree 100 m away, and with a reaction time of .75 seconds, applies the brakes and stops 9 m short of the tree. What is the decelerating force?

    This was not hard, I got -5600 N for the force. F/mg = 4/7, which is important later.

    Part b is the part I can't seem to solve. I think I have it right, but it doesn't agree with the answer in the book.

    B) Same situation as before, same decelerating force as before, but this time it is on a grade of arcsin (1/10). What speed does the car hit the tree?

    For the sum of the X forces, I get f_grav - f_fric = f_car, or since F_fric/mg = 4/7,

    mg sin A - (4/7)mg cos A = ma

    980 - 5600 cos A = 10^3 a, or a = 4.59

    I use v^2 = 28^2 - 2(4.59))(79), and get v^2 = 58.78.

    (79 m is the distance to the tree, after subtracting the reaction time distance of D = 28 m/s (.75) = 21 from the total distance of 100 m.)

    This all seems right to me, but Mr. French gives the value of v^2 as 42.

    Could someone point out where I have gone wrong? (This is not homework, I am using the French text for extra reading and problems.)

    Thank you,
  2. jcsd
  3. Sep 16, 2005 #2
    Since the car is supposed to hit the tree I'll assume it's a 'descending' slope. If you're viewing the horizontal force balance taking the direction tangential to the slope as horizontal direction, the equation you used (which I quoted here) has an error in it. It should be:

    mg sin A - (4/7)mg = ma,

    since the car is applying the braking force in the opposite direction of its motion, which is the direction you took to be horizontal.
  4. Sep 16, 2005 #3
    Thanks. I don't understand this though. Wouldn't the normal force be less on the incline? This is the same value for the frictional force as one would have on a perfectly horizontal surface. And it also doesn't deliver the answer in the book, which I am beginning to think might be wrong.

  5. Sep 16, 2005 #4
    Well, in the second question you stated that the car exerts the same braking force as in the horizontal situation. No mention of any normal force or friction coefficient. I suppose that, if you assume the car applied maximum braking force in the horizontal case, you could calculate a new maximum braking force which is valid for the inclined plane - proportional to the normal force the plane exerts on the car.

    But since the braking force was stated as being the same as in the first situation, I don't see why that'd need to be done.
  6. Sep 16, 2005 #5
    Yes, this makes sense, and is how I started out, but when I didn't get the correct answer, I tried other ways, this seemed the most sensible. However, nothing I have done has worked out the way French thinks it should.

    It's frustrating to feel that I am missing some crucial insight in what really seems like a trivial problem.

  7. Sep 17, 2005 #6
    Could you possibly post the exact question as stated in the book? Maybe there's an extra subtle difference between the two situations.
  8. Sep 17, 2005 #7
    Sure. Thanks for your time and help with this.

    A car of mass 10^3 kg is traveling at 28 m/sec (a little over 60 mph) along a horizontal straight road when the driver suddenly sees a fallen tree blocking the road 100 m ahead. The driver applies the brakes as soon as his reaction time (0.75 sec) allows and comes to rest 9 m short of the tree.

    (a) Assuming constant deceleration caused by the brakes, what is the decelerating force? What fract is it of the weight of the car (take g = 9.8 m/sec^2)?

    (b) If the car had been on a downward grade of sin^{-1} (1/10) with the brakes supplying the same decelerating force as before, with what speed would the car have hit the tree.
    The answer to (b), which is what is causing me so much grief, is given in the book as the square root of 42, approximately 15 mph.

    Thanks again,
  9. Sep 17, 2005 #8
    Right. So, the equation valid for the forces parallel to the slope is:

    [tex]F_{brake} + Mg\cdot \sin(\alpha) = M \cdot a[/tex]

    -5600 + 1000 * 9.8 * sin(arcsin(1/10)) = 1000 * a

    -4620 = 1000 * a

    a = -4.62 m/s^2


    [tex]s = v_{0}t + \frac{1}{2}at^{2}[/tex]


    [tex]v_{crash} = v_{0} + at[/tex]

    rewriting the above equation for s into a polynomial of the form at^2 + bt + c = 0 and solving for t using the abc-formula yields:

    [tex]t = \frac{v_{0} +/- \sqrt{v_{0}^2 + 2sa}}{-a}[/tex]

    filling in all parameters yields: t = 4.47 s or t = 7.65 s. The value for t valid here must be smaller than what we got in the first question, because the effective braking force is smaller - so t = 4.47 s is the one we need. Now, v(final) = v(0) + at = 28 - 4.62*4.47 = 7.35 m/s, which comes quite close to sqrt(54) m/s. The precise value, if no rounding is performed, seems to be sqrt(54.04) m/s.

    Funny, I get another answer still. I can't find a mistake in my answer, though...
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