Finding the horizontal force that the road applies to the car

[rugerts]

Homework Statement

Given weight car = 2500lb, acceleration = constant = 2 ft/s^2, radius of curvature at C = 1000 ft, distance between A and B = 1300 ft, distance between B and C = 900 ft

Find horizontal force of road on car at B and C.

Homework Equations

F = ma in normal-tangential coordinate system

The Attempt at a Solution

I tried using kinematics, since I've got constant acceleration, to find the velocity, in order to use that to find the centripetal force. From there, I'm trying to find the Normal force. I'm using normal tangential coordinate system. I'm having trouble recognizing the geometry of the situation, since I seem to need to know angles. I tried using the arc sector formula. The answer I'm getting is way off of the one shown below in the image.

(Ignore the cos(0.9), that's in radians by accident. I forgot to change that on the image)

 

[Merlin3189]

Given weight car = 2500lb,

Is that the weight or the mass? They don't actually say, just, "a 2500lb car"

acceleration = constant = 2 ft/s^2
, radius of curvature at C = 1000 ft,
distance between A and B = 1300 ft,
distance between B and C = 900 ft

The diagram gives one more important piece of info about B.

Find horizontal force of road on car at B and C.[

F = ma in normal-tangential coordinate system

I'm not sure what you mean by "normal-tangential coordinate system" but that's probably my ignorance. F=ma seems a reasonable relation to use.

I tried using kinematics, since I've got constant acceleration, to find the velocity,

So let's see that for starters.

in order to use that to find the centripetal force.

So when you have the speeds, let's see that.

From there, I'm trying to find the Normal force. I'm using normal tangential coordinate system. I'm having trouble recognizing the geometry of the situation, since I seem to need to know angles.

I don't know why you need angles.

I think maybe it would be good to get as far as the speed. Then say what you are thinking about centripetal forces.

 

[rugerts]

Is that the weight or the mass? They don't actually say, just, "a 2500lb car"

The diagram gives one more important piece of info about B.
I'm not sure what you mean by "normal-tangential coordinate system" but that's probably my ignorance. F=ma seems a reasonable relation to use.
So let's see that for starters.
So when you have the speeds, let's see that.
I don't know why you need angles.

I think maybe it would be good to get as far as the speed. Then say what you are thinking about centripetal forces.

2500 lb is a weight yes. We'd divide by 32.2 ft/s^2 to get the mass. Is the extra piece of info pertaining to the inflection point?
Here's my work showing the velocities and also what I mean by normal tangential coordinate system.

 

[Merlin3189]

Is the extra piece of info pertaining to the inflection point?

Yes.

2500 lb is a weight yes. We'd divide by 32.2 ft/s^2 to get the mass.

So the weight is 2500 lb ft/sec² and we divide by g to get a mass of 2500 / 32.2 = 77.6 lb
That seems very light for a car. So I'd have thought the mass would be 2500 lb (just over a ton) and the weight would be mg.

You don't say what you are doing here (IMO v.bad!)
but from the u=0 and s=1300 I guess it's A to B
but from the r=1000 I guess it is C

That could be part of your difficulty: you are using the speed at B to calculate the force at C.

 

[rugerts]

Yes.
So the weight is 2500 lb ft/sec² and we divide by g to get a mass of 2500 / 32.2 = 77.6 lb
That seems very light for a car. So I'd have thought the mass would be 2500 lb (just over a ton) and the weight would be mg.
View attachment 232795
You don't say what you are doing here (IMO v.bad!)
but from the u=0 and s=1300 I guess it's A to B
but from the r=1000 I guess it is C

That could be part of your difficulty: you are using the speed at B to calculate the force at C.

I need the force at B first. Which is why I calculate the speed at B. It's stated in the problem. I think you're confusing lb force and slugs. Slugs is a unit of mass I'm using.

 

[Merlin3189]

I need the force at B first. Which is why I calculate the speed at B. It's stated in the problem.

But if you want the force at B and calculate the speed at B, why use the radius at C ?

I think you're confusing lb force and slugs. Slugs is a unit of mass I'm using.

Wow! That might be the cause of my confusion. I've never heard of slugs (and I'm old, British and used imperial measures when I was younger.)
So the car is 77.64 slug and you are using lb to stand for pound force rather than pound mass. That'll help me follow your work.

Anyhow, look very carefully at the question again to work out what the radius of curvature is at B.

 

[rugerts]

But if you want the force at B and calculate the speed at B, why use the radius at C ?

Wow! That might be the cause of my confusion. I've never heard of slugs (and I'm old, British and used imperial measures when I was younger.)
So the car is 77.64 slug and you are using lb to stand for pound force rather than pound mass. That'll help me follow your work.

Anyhow, look very carefully at the question again to work out what the radius of curvature is at B.

Yeah, I think that's what my problem was. I wasn't using the correct radius of curvature. Would an arc-sector relationship formula be useful here to find that radius of curvature? I made an assumption that the radius of curvature would somehow be the same, even though it didn't feel right.

 

[Merlin3189]

Do you know what a point of inflexion is?

Otherwise, which way is the centripetal acceleration between A and B, then which way is it from B to C?
What does that tell you about B?

 

[rugerts]

Do you know what a point of inflexion is?

Otherwise, which way is the centripetal acceleration between A and B, then which way is it from B to C?
What does that tell you about B?

Doesn't a point of inflection say something about symmetry?

 

[CWatters]

Yes it does.

Sorry to dive in here but perhaps you are missing something obvious...

To calculate the force on the car (at any point) you need to know how it's accelerating rather than its velocity. The acceleration could be linear, centripetal or some combination of both. So what's the acceleration at B?

 

[haruspex]

Doesn't a point of inflection say something about symmetry?

Not really. It just says that the curvature is changing sign.

 

[rugerts]

Do you know what a point of inflexion is?

Otherwise, which way is the centripetal acceleration between A and B, then which way is it from B to C?
What does that tell you about B?

it'll just say the acceleration (centripetal) is pointing inward, which will be different for both sides, since the curvature of the road changes. I knew this before. I'm having trouble calculating actual values because I'm unsure about angles and radius of curvature where it's not given (so for C it's given as 1000ft)

 

[haruspex]

will be different for both sides, since the curvature of the road changes

if the curvature is changing smoothly, has one sign for A to B, and the opposite sign for B to C, what is the curvature at B?

 

[CWatters]

it'll just say the acceleration (centripetal) is pointing inward, which will be different for both sides, since the curvature of the road changes. I knew this before. I'm having trouble calculating actual values because I'm unsure about angles and radius of curvature where it's not given (so for C it's given as 1000ft)

You don't actually need to know the radius before or after B to work out the acceleration at B. Think more about what an inflection point is. If the we were to call the radius before B positive and after B it's negative, what would the radius be at B?