Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Static friction

  1. May 5, 2010 #1
    A car, travelling at speed v, can safely negotiate an unbanked curve with a 50.0-m radius when the coefficient of friction between the tires and the road is 0.8. How much bank would a curve with the same radius require if the car is to safely go around it at the same speed, v, without relying on friction?

    my work:
    Msmg=Fr=m((v^2)/r)
    (0.8)(9.8m/s^2)=(v^2)/(50m)=19.79m/s

    Mgsinx=Fr=m((v^2)/r)
    (9.8)sinx=(19.79m/s)/(50m)=53.13 degree


    please tell me if i am wrong or right
     
  2. jcsd
  3. May 5, 2010 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Right.
    Wrong.

    The centripetal acceleration is directed radially (horizontally) inward toward the center of the curve.
     
  4. May 5, 2010 #3
    so will be
    mgsinxcosx=m((v^2)/r)?
     
  5. May 5, 2010 #4

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That's getting close, but still incorrect. Please show or indicate how you are arrivig at this equation.
     
  6. May 5, 2010 #5
    well the only Fr is the gsinx but since you said Fr is directly horizontal to the center so i did cos
    but i might be wrong now i am kinda confuse too
     
  7. May 5, 2010 #6

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    the radial acceleration is not gsin theta. Draw a free body diagram of the car. There are 2 forces acting on the car. One is its weight, acting vertically down. What's the other??
     
  8. May 5, 2010 #7
    normal force?
     
  9. May 5, 2010 #8

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes. So of those 2 forces, which one has a component in the x (radial, horizontal) direction? What is that component?
     
  10. May 5, 2010 #9
    nsinx so it is the same as mgtanx?
     
  11. May 5, 2010 #10

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, I'm not sure how you got that, but that is correct.
     
  12. May 5, 2010 #11
    well i think n=mg/cosx so plug it back into the equation (mg/cosx)(sinx)=mgtanx right?
     
    Last edited: May 5, 2010
  13. May 5, 2010 #12

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, from Newton 1 in the y direction [/quote]so plug it back into the equation (mg/cosx)(sinx)=mgtanx right?[/QUOTE]yes, solve for theta (or the angle you are calling 'x') = ?
     
  14. May 5, 2010 #13
    thanks so much otherwise i might get that wrong on the test lol....
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook