- #1

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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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- Thread starter xstetsonx
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- #1

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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

PhanthomJay

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Right.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

Wrong.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

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)?

mgsinxcosx=m((v^2)/r)?

- #4

PhanthomJay

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That's getting close, but still incorrect. Please show or indicate how you are arrivig at this equation.so will be

mgsinxcosx=m((v^2)/r)?

- #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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- #7

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normal force?

- #8

PhanthomJay

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Yes. So of those 2 forces, which one has a component in the x (radial, horizontal) direction? What is that component?normal force?

- #9

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nsinx so it is the same as mgtanx?

- #10

PhanthomJay

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Yes, I'm not sure how you got that, but that is correct.nsinx so it is the same as mgtanx?

- #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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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') = ?well i think n=mg/cosx

- #13

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thanks so much otherwise i might get that wrong on the test lol....

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