- #1

logan3

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## Homework Statement

A curved portion of highway has a radius of curvature of 65 m. As a highway engineer,

you want to bank this curve at the proper angle for a steady speed of 22 m/s.

(a) What banking angle should you specify for this curve?

(b) At the proper banking angle, what normal force and what friction force does the

highway exert on a 750-kg car going around the curve at the proper speed?

[itex]r = 65m[/itex]

[itex]v = 22 m/s[/itex]

[itex]g = 9.8 m/s^2[/itex]

[itex]m = 750 kg[/itex]

## Homework Equations

[itex]F_{Ny} = F_N cos\theta = mg \Rightarrow F_N = \frac {mg}{cos\theta}[/itex]

[itex]\vec F_c = \frac {mv^2} {r} = F_N sin\theta \Rightarrow (\frac {mg}{cos\theta}) sin\theta = \frac {mv^2} {r} \Rightarrow gtan\theta = \frac {v^2} {r} \Rightarrow \theta = tan^{-1}(\frac {v^2} {gr})[/itex]

When the object is moving at constant velocity, the centripetal force will be the same as the force of friction.

## The Attempt at a Solution

The wording "radius of curvature" is throwing me off, because we've never had it. I looked it up and pictures showed it as part of the circumference of the curved road, not the distance the to center. Nonetheless, I went ahead and treated "radius of curvature of 65 m" as simply r = 65m.

(a) [itex]\theta = tan^{-1}(\frac {(22 m/s)^2} {(9.8 m/s^2)(65m)}) = 37.227^o \sim 37^o[/itex]

(b) [itex]F_N = \frac {mg}{cos\theta} = \frac {(750 kg)(9.8 m/s^2)}{(cos 37.227^o)} = 9,230.949 N \sim 9,200N[/itex]

[itex]\vec F_f = \vec F_c = F_N sin\theta \Rightarrow (9,230.949 N)(sin 37.227^o) = 5,584.48N \sim 5,600N[/itex]

Thank-you