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Max acceleration of car up ramp

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A car is driving up a slope at angle 16 deg. to the horizontal, trying to accelerate as much as possible. The static and kinetic friction coefficients are .90 and .80, respectively. Find the maximum possible acceleration (assuming a sufficiently powerful engine).


    2. Relevant equations
    Fs,max = .90 x normal force
    Max acceleration = -mg(sin16) + fs,max / mass

    3. The attempt at a solution
    Ok, I've seen a problem similar to this one and (plugging in the values from this question) it was solved using this first step:

    Max acceleration = -mg(sin16) + fs,max / mass.

    My first question is, where did the (sin16) come from? I drew a free body force diagram to look like this:

    car.jpg

    EDIT: Sorry, in the image, f,static should actually be f,kinetic, since the car is accelerating.
     
  2. jcsd
  3. Mar 21, 2009 #2
    Doesn't that depend on how much power the engine can generate?
     
  4. Mar 21, 2009 #3
    The question says "Assuming a sufficiently powerful engine", so I'm guessing... no? :confused:
     
  5. Mar 21, 2009 #4
    Anyone? Still have no clue where the sin16 came from. Opposite over hypotenuse, I don't see it.
     
  6. Mar 21, 2009 #5

    Doc Al

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    If you break the weight (mg, down) into components, the component parallel to the incline will be mg sinθ (down the incline). (If you think in terms of a right triangle, mg will be the hypotenuse.)
     
  7. Mar 21, 2009 #6
    I feel like I'm never going to grasp this stuff. I just cannot see where you're getting this from.

    Could you perhaps draw a diagram of what you mean? If not, thanks anyway.
     
  8. Mar 21, 2009 #7

    Doc Al

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    Study this: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm" [Broken]
     
    Last edited by a moderator: May 4, 2017
  9. Mar 21, 2009 #8
    That was extremely helpful, thanks so much. I have 3 physics textbooks and none of them seem to explain that very well at all.

    I'll give the question another shot now on my own and see what I get.
     
    Last edited by a moderator: May 4, 2017
  10. Mar 21, 2009 #9
    Ok, I understand what I'm doing now (I think), but I can only seem to solve for max acceleration in terms of n or m. Can't seem to get a direct answer without knowing the mass of the car. Here's my work:

    1. Separate the components of mg.

    2. To find the maximum acceleration in the presence of friction, use the equation a[tex]_{max}[/tex] = mg(sin[tex]\theta[/tex]) + f[tex]_{s,max}[/tex] / m.

    3. We're only given the coefficient of static friction, so to find f[tex]_{s,max}[/tex], we use f[tex]_{s,max}[/tex] = [tex]\mu[/tex][tex]_{s}[/tex] x n.

    4. Plugging in the coefficient of static friction we get f[tex]_{s,max}[/tex] = (.90)n

    5. Plug into original equation: a[tex]_{max}[/tex] = mg(sin16) + (.90)n / m.

    6. Mass cancels, plug in g: a[tex]_{max}[/tex] = 9.8(sin16) + (.90)n.

    7. Simplify: a[tex]_{max}[/tex] = 2.70 + (.90)n.

    That's as far as I can go, don't know how to get rid of n without knowing the mass of the car. Or am I doing something wrong?
     
  11. Mar 21, 2009 #10

    Doc Al

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    You don't need the mass to find the acceleration. It cancels out. Hint: Express the normal force in terms of mg.

    You need both components of mg, parallel and perpendicular to the ramp. The perpendicular component will tell you the normal force.
     
  12. Mar 21, 2009 #11
    Right. The perpendicular component equals the normal force because they have to balance each other out. The equation for the perpendicular component is mg(cos16), still leaving me with an extra m. Should there be two m's in the denominator, say, m[tex]_{1}[/tex] + m[tex]_{2}[/tex], one for each component?
     
  13. Mar 21, 2009 #12

    Doc Al

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    It's not "extra". It's just what you need so that the m's cancel nicely.
    Nope. I think you're making the same algebra error that you made in this thread: https://www.physicsforums.com/showthread.php?p=2124470#post2124470
     
  14. Mar 21, 2009 #13
    Ok, so we have:

    a[tex]_{max}[/tex] = mg(sin16) + (.90)mg(cos16) / m

    Which can be factored into:

    a[tex]_{max}[/tex] = m(g[sin16] + .90g[cos16]) / m

    m's cancel, and plug in 9.8 for g:

    a[tex]_{max}[/tex] = (9.8[sin16] + .90(9.8)[cos16])

    a[tex]_{max}[/tex] = 11.17m/s[tex]^{2}[/tex]
     
  15. Mar 21, 2009 #14

    Doc Al

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    You left out a minus sign in from of the mg(sin16). (Otherwise: Good!)
     
  16. Mar 21, 2009 #15
    Oh, so rather, it would look like:

    (-9.8[sin16] + .90(9.8)[cos16])

    = 5.78 m/s[tex]_{2}[/tex]
     
  17. Mar 21, 2009 #16

    Doc Al

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    Good!
     
  18. Mar 22, 2009 #17
    Hey Doc Al, shouldn't the force diagram have static friction going UP the hill. Because the static friction is what drives cars?
     
  19. Mar 22, 2009 #18

    Doc Al

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    Absolutely. The diagram is wrong. I meant to point that out, but forgot. Thanks!

    (But the equation ended up OK.)
     
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