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circular banked track (1 Viewer)

Hikari Clover

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the circular bend on a bike track has a constant radius of 20 m and is banked at a constant angle of 30 degree to the horizontal. a bicycle rider can safely negotiate the bend if the maximum sideways thrust F, up or down the slope, is at most one-tenth of the normal reaction N . by resolving the forces vertically and horizontally, show that the range of speeds V, correct to 2 decimal places and in m/s, at which the bend can be safely negotiated is
9.50 <= V <= 11.99
take g=10 m/s^2


thx:wave:
 

blink376

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if u let x=30 degrees, N=reaction , F=friction and resolve forces horizontally and vertically, you end up with

Equation 1: Nsinx + Fcosx= (mv^2)/r
Equation 2: Ncosx - Fsinx = mg

(Equation 1 times sinx) + (Equation 2 times cosx)= N
(Equation 1 times cosx) - (Equation 2 times sinx) = F

With the expressions you have for N and F, let 10F=N to find the maximum speed and let 10F= -N to find the minimum speed (since the friction will direct the opposite way if it is below optimum speed), which gives you the inequality 9.50<V<11.99
 

Tmer

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alright that does give you the answer, but i don't understand why simply putting N=10F into eqn 1 and 2 and solving them together doesn't work? can some 1 explain this?
 

JasonNg1025

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Question asks for an inequality and it says the max sideways thrust F up or down the slope at 1/10 of N (which is directed from the friction as blink376 said)

So N/10 = F => N = 10F would give max speed

The question also says up OR down slope so that N = -10F would be minimum speed.

Hope that cleared it up a bit :)


 

Tmer

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no that was not my question... some 1 else please explian?
 

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