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Car going up a hill. Find the Angle of the hill.

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data
    A 1550-kg car experiences a combined force of air resistance and friction that has the same magnitude whether the car goes up or down a hill at 21 m/s. Going up a hill, the car's engine produces 44 hp more power to sustain the constant velocity than it does going down the same hill. At what angle is the hill inclined above the horizontal?


    2. Relevant equations
    P=W/t=FvCosθ
    W=fΔx


    3. The attempt at a solution
    I drew out my free body diagram and I have the Normal force, friction, and the components of the weight force (mgCosθ in the y-direction and mgSinθ in the x-direction)
    I figured I need to find out the F in P=FvCosθ. I'm just not sure what F it is I have to find.
     
  2. jcsd
  3. Oct 12, 2012 #2

    TSny

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    There's another force you need to include. What force "propels" the car up the hill?
     
  4. Oct 12, 2012 #3
    i thought wheels generated friction which caused a recipriocal force on the car by the road, which caused a car to move.
     
  5. Oct 12, 2012 #4

    TSny

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    Exactly. I'm not sure that the "friction" mentioned in the problem is the same as the friction generated by the wheels to drive the car up the hill. I think they might have been talking about internal friction that acts along with the air resistance. But, anyway, I think we're clear on the forces.

    Think of the force generated by the wheels as ultimately coming from the engine. So, if you are going to set up an expression for the power that the engine produces, you are going to need the force generated by the engine (i.e., the friction force generated by the wheels.) Can you think of a way to get an expression for that force?
     
  6. Oct 12, 2012 #5
    Well I'm given power and I'm velocity. Would it be right to use a rearrangement of P=Fv, F=P/v?
     
  7. Oct 12, 2012 #6

    TSny

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    OK, but note they only give you the change in power when switching between going up and coming back down.

    Can you bring in something else that would let you relate F to the angle of the slope?
     
  8. Oct 12, 2012 #7
    The components of the Weight force?
     
  9. Oct 12, 2012 #8

    TSny

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    I was thinking more along the lines of a principle or law that you could apply. Note that they tell you the velocity is constant.
     
  10. Oct 12, 2012 #9
    Well if velocity is constant it cant be newton's 2nd Law. I'm stumped.
     
  11. Oct 12, 2012 #10

    TSny

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    Oh yes it can :)
     
  12. Oct 12, 2012 #11
    Well if it's at a constant velocity wouldn't that make F=0 due to the lack of acceleration?
     
  13. Oct 12, 2012 #12

    TSny

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    But what does F represent in the 2nd law?
     
  14. Oct 12, 2012 #13
    Net Force?
     
  15. Oct 12, 2012 #14

    TSny

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    Sure. That's important. F is always the sum of all of the forces acting on the object when setting up the 2nd law. Or, if you're working with components, it says that the sum of all the components of the forces in a certain direction equals the mass times the component of the acceleration in that direction.
     
  16. Oct 12, 2012 #15
    So do I use the component(s) of the Weight force to relate F and theta?
     
  17. Oct 12, 2012 #16

    TSny

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    Let's pick a specific symbol for each force: f for the air resistance, mg for the weight, Fw for the force from the wheels, and Fn for the normal force. How would you write an expression for the net force acting upward along the slope?
     
  18. Oct 12, 2012 #17
    ƩF=Fw-f-mgcosθ
     
  19. Oct 12, 2012 #18

    TSny

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    Good. Can you use the 2nd law to express Fw in terms of the other forces? [Oops: You didn't quite get that right. Think again about the component of the weight.]
     
    Last edited: Oct 12, 2012
  20. Oct 12, 2012 #19
    Yes, because ƩF=0, (because I know that the car is moving at a constant v), Fw = f + mgCosθ
     
  21. Oct 12, 2012 #20

    TSny

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    Yes, but you'll need to correct the mgCos(theta) term.
     
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