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Banked curves, friction

  1. Aug 24, 2011 #1

    Vmax = 13.1 m/s
    Vmin = 8.1 m/s

    I've spent all morning attempting part 4 of this problem :/ Can someone help me please?


    tanθ = v2 / rg
    v = sqrt(rg)tanθ = sqrt [(20)(9.8)tan30] = 10.6376996 m/s
    v = 10.6376996 m/s

    Fdue to banked curve = mv2 / r = 5.658m

    The centripetal force is the net force in the radial direction:
    Fcentripetal = mv2 / r = mgμcosθ + 5.658m

    Cancel the m's:
    v2 /r = gμcosθ + 5.658
    v2 / 20 = 9.8(0.2)cos30 + 5.658
    v = 12.1 m/s???????

    Please help :/ thank you
  2. jcsd
  3. Aug 24, 2011 #2

    Doc Al

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    That equation is only good for the no friction case.

    Hint: Apply Newton's 2nd law to derive a version that works for the friction case. (How is that equation derived? Same basic idea, only now you have an additional force.)

    Hint 2: Your free body diagram shows only one of the conditions. How does it change for minimum speed versus maximum speed cases?
  4. Aug 24, 2011 #3
    Hello Doc

    I was trying to solve for the force of the banked curve separately from the force of friction... I eventually added them together....Did you see that I did that? Or is that just not the way to go about it?
  5. Aug 24, 2011 #4
    Also, I understand that to find the other velocity, friction should be pointing the other way, right??? I just did not bother to find this other value of v because I knew that my initial approach was incorrect....
  6. Aug 24, 2011 #5

    Doc Al

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    Yes, I saw what you tried to do but it didn't really make sense. What do you mean by the 'force of the banked curve'? The normal force?

    In any case, to solve for the velocity you need to rederive the formula from scratch including friction.
  7. Aug 24, 2011 #6

    Doc Al

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  8. Aug 24, 2011 #7
    Yes sir - I meant the normal force.

    I will take a break, ponder on it, and take another stab at it tonight. I will get back to you. :)
  9. Aug 24, 2011 #8
    Ack - I can't leave this problem alone...!!!!

    This is my second attempt. I feel that I am much closer, but still off somewhere.

    Radial Direction
    FNsin 30 + Fscos30 = mv2/r
    mgcos30sin30 + mgμcos30 = mv2 / r

    Drop the m's
    gcos30sin30 + gμcos30 = v2/r

    Solve for V
    r[gcos30sin30 + gμcos30] = v2
    v = sqrt(r[gcos30sin30 + gμcos30])
    v = sqrt(20[9.8cos30sin30 + 9.8(0.2)cos30])
    v = 10.9 m/s
  10. Aug 24, 2011 #9

    Doc Al

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    Good. But express Fs in terms of FN.
    Careful! You cannot assume that FN = mgcos30. That would be true if there were no component of acceleration normal to the surface, but that doesn't apply here.

    Instead, leave FN as an unknown. Get a second equation for vertical forces.
  11. Aug 25, 2011 #10
    What is causing the acceleration normal to the surface? The bike only moves on the surface of the track. I do not see any movement in the direction of the normal force (AKA perpendicular to the surface of the track)...what I mean is, the bike isn't shooting upwards in the air in the direction of the normal force. So how can it have any type of acceleration in that way? I think not understanding this may be my problem. Is FN = mg?

    Ok so for the horizontal forces I have:

    FNsin30 + FNμcos30 = mv2 /r

    With the velocity being in the x-direction.

    For vertical forces I have:
    FNcos30 - weight - Fssin30 = mv2 /r
    FNcos30 - mg - FNμsin30 = mv2 /r

    With the velocity in the y-direction.
    Last edited: Aug 25, 2011
  12. Aug 25, 2011 #11

    Doc Al

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    The acceleration is horizontal. The normal to the surface is not vertical, so the acceleration has a component normal to the surface.

    No. (Why guess?)


    Not sure what you mean here. The important thing is that the acceleration is horizontal.

    Why are you setting vertical forces equal to mv2/r? There's no vertical acceleration.
  13. Aug 26, 2011 #12
    I just noticed that I totally neglected the force of mgsin30 down the slope this entire time.
  14. Aug 26, 2011 #13
    Well, that didn't help. Haha. I admit defeat on this problem. It's like when you say a word too many times...this problem has lost all meaning.
  15. Aug 26, 2011 #14
    HOLY MOLY...I JUST GOT IT. I had to piece some of the vague hints you gave in earlier posts together and WHAM. THANK YOU DOC AL! you never fail me. haha I've never worked on a single problem this long :))) feels so rewarding :)))
  16. Aug 26, 2011 #15
    Although I've reached the correct answer, I still do not see why FN does NOT equal mgcos30.
  17. Aug 26, 2011 #16


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    If you drive a car, at speed, through a dip in the road, the car's suspension is compressed and the passengers feel they have been pushed down into the seats. That is because at the bottom of the dip, the FN has to not only balance weight, but provide the necessary [upward] centripetal force so is much stronger than is was when driving on flat ground.

    When you drive a car around a banked curve, the FN is also larger than that needed if the car was stationary - for the same reason. [of couse if the car was stationary you would need some friction to prevent the car slipping down the banking]
  18. Aug 26, 2011 #17
    awesome thank you peterO :)
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