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Frictional force for bank curves

  1. Jan 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A 1453kg car rounds a curve of 122-m radius at a speed of 48 km/h. How large must the force of friction between tires and pavement to prevent the car from skidding?

    2. Relevant equations
    F=(coeficient)mg
    Net force=Ff+Fcen*cos (-) +mgsin(-)
    tan(-)=(v)^2 / gr


    3. The attempt at a solution
    All solutions failed!

    Please help me because i need to pass it tomorrow
     
  2. jcsd
  3. Jan 21, 2015 #2

    The problem statement doesn't mention banking, in which case the solution would be pretty straight forward. Are you sure banking is to be considered?
     
  4. Jan 21, 2015 #3
    What would be the best solution then?
     
  5. Jan 21, 2015 #4
    If there is no banking, then the centrifugal force will be balanced by friction. You can get the velocity by equating the centrifugal and frictional forces.
     
  6. Jan 21, 2015 #5
    My teacher said there is banking
     
  7. Jan 21, 2015 #6

    DEvens

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    Did he?

    When I was in undergrad I was taking a class in group theory. And everybody in the class was having a very tough time doing one of the homework assignments. We were all struggling to follow what our notes from class said. We were pestering each other for hints and all going crazy.

    Except for this one guy. He was sitting in the study hall drinking coffee and looking out the window. What's your story? Why are you not worried. Oh, I got a textbook on group theory out of the library and did the homework. But the prof said to do it this way! Oh, he said. I never pay too much attention to what the prof says.

    That guy now has a senior position at a prestigious research institution.
     
  8. Jan 21, 2015 #7
    Alright. So if there is banking, the frictional force will be inclined. It will have two components, a vertically downward one and a horizontal one.

    Similar is the case with the normal reaction from the ground. It will have an inclination as the road has banking. It can be resolved into a vertically upward and a horizontal component.

    Are you able to follow me till here theofficialack?
     
  9. Jan 21, 2015 #8

    BvU

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    Does not count as an attempt at solution. How do you know anyway? -- what have you tried so far ? What do you mean with (-) ?

    Good luck tomorrow :). Or do you have to hand in this particular exercise :( ?
     
  10. Jan 21, 2015 #9
    Well the velocity is stated.. What does it mean when it says force of friction?
     
  11. Jan 21, 2015 #10

    BvU

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    Precisely what it says: How many Newtons does the force of friction have to be in order that this 1453 kg car can follow a circular trajectory with radius 122 m while having a speed of 48 km/h ?

    And what siddharth says, I would formulate differently: If ... then the centripetal force will have to be provided by friction.

    You sneakily avoided showing any of your attempts at solution. Let's have a deal:
    Answer the questions in post #8: How do you know anyway? -- what have you tried so far ? What do you mean with (-) ?

    And just to be clear: " i need to pass it tomorrow "now means: "I have to hand in my answer for this exercise today" ?

    There's something ingrained in PF culture that we all want to help like crazy, but we try to stear clear of doing your exercises for you.
    Simple reason: If I let you run to train for my marathon, it doesn't help me at all to make it to the finish.
     
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