Find Banking Angle and Normal/Friction Force on Curved Road

[logan3]

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?

$r = 65m$
$v = 22 m/s$
$g = 9.8 m/s^2$
$m = 750 kg$

Homework Equations

$F_{Ny} = F_N cos\theta = mg \Rightarrow F_N = \frac {mg}{cos\theta}$
$\vec F_c = \frac {mv^2} {r} = F_N sin\theta \Rightarrow (\frac {mg}{cos\theta}) sin\theta = \frac {mv^2} {r} \Rightarrow gtan\theta = \frac {v^2} {r} \Rightarrow \theta = tan^{-1}(\frac {v^2} {gr})$
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) $\theta = tan^{-1}(\frac {(22 m/s)^2} {(9.8 m/s^2)(65m)}) = 37.227^o \sim 37^o$
(b) $F_N = \frac {mg}{cos\theta} = \frac {(750 kg)(9.8 m/s^2)}{(cos 37.227^o)} = 9,230.949 N \sim 9,200N$
$\vec F_f = \vec F_c = F_N sin\theta \Rightarrow (9,230.949 N)(sin 37.227^o) = 5,584.48N \sim 5,600N$

Thank-you

 

[haruspex]

$\theta = tan^{-1}(\frac {v^2} {gr})$

Good.

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

That's only true on a horizontal surface. Here, those two forces act in different directions.

The Attempt at a Solution

I went ahead and treated "radius of curvature of 65 m" as simply r = 65m.

Good move.

$\vec F_f = \vec F_c$

Think about the logic/FBD behind your equation $\vec F_c = F_N sin\theta$. What does that imply for $\vec F_f$?

 

[logan3]

I guess I'm a bit confused. Isn't there a different speed in order to prevent the car from sliding down the bank versus sliding up the bank? Where does the 22m/s fit in? Do I set the sliding down and up equations equal to each other and solve for the coefficient of friction, then use that to find the friction force?

 

[haruspex]

I guess I'm a bit confused. Isn't there a different speed in order to prevent the car from sliding down the bank versus sliding up the bank?

For a given coefficient of friction, yes. But what is meant by 'proper angle' here? You solved that part ok, yet friction never appeared in the analysis. Why do you think that is?

 

[HalfLostAndSearching]

for your question. I would like to provide a response to your content. Based on the information provided, your calculations for the banking angle and normal/friction force appear to be correct. The radius of curvature refers to the distance from the center of the curved road to the center of the circle that forms the curve. In this case, it is given as 65m. It is important to note that this is a simplified calculation and in real-life scenarios, there may be other factors that need to be considered, such as the weight distribution of the car and the banking angle of the road itself. I would recommend conducting further research and considering these factors to ensure the most accurate results. Overall, your approach and calculations seem to be sound and provide a good starting point for understanding the concept of banking angles and normal/friction forces on curved roads.