Static friction and car and unbanked curve

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  • #1
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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
 

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  • #2
PhanthomJay
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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
Right.
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
Wrong.

The centripetal acceleration is directed radially (horizontally) inward toward the center of the curve.
 
  • #3
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so will be
mgsinxcosx=m((v^2)/r)?
 
  • #4
PhanthomJay
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so will be
mgsinxcosx=m((v^2)/r)?
That's getting close, but still incorrect. Please show or indicate how you are arrivig at this equation.
 
  • #5
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That's getting close, but still incorrect. Please show or indicate how you are arrivig at this equation.

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
 
  • #6
PhanthomJay
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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??
 
  • #7
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normal force?
 
  • #8
PhanthomJay
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normal force?
Yes. So of those 2 forces, which one has a component in the x (radial, horizontal) direction? What is that component?
 
  • #9
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nsinx so it is the same as mgtanx?
 
  • #10
PhanthomJay
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nsinx so it is the same as mgtanx?
Yes, I'm not sure how you got that, but that is correct.
 
  • #11
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well i think n=mg/cosx so plug it back into the equation (mg/cosx)(sinx)=mgtanx right?
 
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  • #12
PhanthomJay
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well i think n=mg/cosx
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') = ?
 
  • #13
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thanks so much otherwise i might get that wrong on the test lol....
 

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