Calculating acceleration on level road/ when road uphill

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Homework Statement


On a level road a motorcycle slows down with an acceleration of -3.40 m/s^2. Calculate the value of its acceleration when the road goes up hill 15 degrees.

Homework Equations


Fnet = MA[/B]


The Attempt at a Solution


I tried setting up a system of equations and made friction equal.[/B]

Level road: friction force = ma
Inclined road: friction + wsin(theta) = ma2

Ma= ma2-wsin(theta)
Ma = ma2 - mgsin(theta)
A = a2 - gsin(theta)
A2 = A + gsin(theta)
A2 = -3.40 + (-9.8)(sin15)

But the answer isn't -5.94, so I don't know what I'm doing wrong. Thanks.
 

Answers and Replies

  • #2
Orodruin
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Is the friction the same in both cases? What is the limiting case when the inclination is 90 degrees?
 
  • #3
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Is the friction the same in both cases? What is the limiting case when the inclination is 90 degrees?

It does not specify. That is the whole question that I posted. I'm in the very beginning of AP physics 1, so I'm not sure what a limiting case is yet.
 
  • #4
Orodruin
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It does not specify.
I know it doesn't, it is a question for you, not for the formulation.

I'm in the very beginning of AP physics 1, so I'm not sure what a limiting case is yet.
A limiting case is a corresponding situation in which you can easily reason what the answer should be. In this case replacing the angle with a 90 degree angle.
 
  • #5
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I know it doesn't, it is a question for you, not for the formulation.


A limiting case is a corresponding situation in which you can easily reason what the answer should be. In this case replacing the angle with a 90 degree angle.

Oh okay. Thank you. I would think the friction would be the same because the Natural force and the coefficient of friction wouldn't change, it's the same surface. If the inclination was 90 degrees I suppose it would be zero acceleration, but honestly I'm not quite sure.
 
  • #6
Orodruin
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Natural force and the coefficient of friction wouldn't change
Do you mean "normal force"? In that case, why would you think it would be the same?
If the inclination was 90 degrees I suppose it would be zero acceleration, but honestly I'm not quite sure.
You are saying a bike on a vertical wall has zero acceleration?
 
  • #7
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Do you mean "normal force"? In that case, why would you think it would be the same?

You are saying a bike on a vertical wall has zero acceleration?

Yes, I mean normal force, I apologize. Since it's staying on the road and not floating/sinking, normal force would equal weight and so would be = mg. The mass would stay the same and so would gravity. Coefficient of friction is dependent on the surface, but since the surface is the same it would be equal in both scenarios.

Oh, the acceleration would be -9.8m/s^2, since gravity would pull it down.
 
  • #8
Orodruin
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normal force would equal weight
Is this true for an object which is not on a horizontal surface? What does "normal" mean and what is the role of the normal force?

Oh, the acceleration would be -9.8m/s^2, since gravity would pull it down.
Yes, but according to the same reasoning you employed for 15 degrees, you would get the result -9.8 sin(90)-3.4 = -13.2 m/s^2. What is the normal force in this case?
 
  • #9
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Is this true for an object which is not on a horizontal surface? What does "normal" mean and what is the role of the normal force?


Yes, but according to the same reasoning you employed for 15 degrees, you would get the result -9.8 sin(90)-3.4 = -13.2 m/s^2. What is the normal force in this case?

I thought it would be. The normal force is perpendicular to the surface of contact, so I thought that, though normal force and weight may have different directions, their magnitude was equal as long as the object wasn't sinking/floating.

Would the normal force be -9.8 times mass?
 
  • #10
Orodruin
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I thought it would be. The normal force is perpendicular to the surface of contact, so I thought that, though normal force and weight may have different directions, their magnitude was equal as long as the object wasn't sinking/floating.

Would the normal force be -9.8 times mass?
You have then not understood the concept of the normal force. The normal force is the reaction force from the surface that prevents an object from passing through the surface. It ensures force equilibrium in the normal direction (assuming that other forces are not accelerating the object away from the surface in which case the normal force is zero). You can find the normal force by requiring force equilibrium in the normal direction.
 
  • #11
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You have then not understood the concept of the normal force. The normal force is the reaction force from the surface that prevents an object from passing through the surface. It ensures force equilibrium in the normal direction (assuming that other forces are not accelerating the object away from the surface in which case the normal force is zero). You can find the normal force by requiring force equilibrium in the normal direction.

So normal force would be equal to the portion of weight in the opposite direction to normal force?
 
  • #12
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You have then not understood the concept of the normal force. The normal force is the reaction force from the surface that prevents an object from passing through the surface. It ensures force equilibrium in the normal direction (assuming that other forces are not accelerating the object away from the surface in which case the normal force is zero). You can find the normal force by requiring force equilibrium in the normal direction.

The normal force would be equiviliant to weight in the first scenario, correct? In the second scenario, normal force would equal weight in the y direction? I tried drawing a picture and a free body diagram...
 
  • #13
Orodruin
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So normal force would be equal to the portion of weight in the opposite direction to normal force?
If by "portion" you mean "component", yes.

The normal force would be equiviliant to weight in the first scenario, correct? In the second scenario, normal force would equal weight in the y direction? I tried drawing a picture and a free body diagram...
It is unclear to me what cases you refer to and which direction you have designated as y. Please keep in mind that I cannot see your drawing.
 
  • #14
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It is unclear to me what cases you refer to and which direction you have designated as y. Please keep in mind that I cannot see your drawing.[/QUOTE]

By first scenario I mean when the motorcycle is traveling on a flat road. And second scenario is up the hill. The normal force is along the Y axis and friction is along the X.

I think I could link the two equations together by coefficient of friction.
 
  • #15
Orodruin
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The normal force is along the Y axis and friction is along the X.
You still have not defined your axes. Again, I cannot see your drawing.
 
  • #16
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You still have not defined your axes. Again, I cannot see your drawing.

How do I define my axes?
 
  • #17
Orodruin
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How do I define my axes?
You specify in which direction the x and y axes are pointing. Again, I can guess, but if I guess wrong and you have understood something wrong then I might end up just confusing you. It is therefore very important that you give all of the information and do not just assume I know everything that you can see.
 
  • #18
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You specify in which direction the x and y axes are pointing. Again, I can guess, but if I guess wrong and you have understood something wrong then I might end up just confusing you. It is therefore very important that you give all of the information and do not just assume I know everything that you can see.

The positive x axis is pointing up the hill, and the negative x axis is pointing down the hill. The motorcycle is at the origin. The normal force lies on the positive y axis.
 
  • #19
Orodruin
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Then yes as long as by "portion" you mean "component" in post #12.
 
  • #20
Orodruin
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I suggest you try to redo the problem with this knowledge.
 
  • #21
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I suggest you try to redo the problem with this knowledge.

Flat road:

Ff=ma
N=W=mg

uN = ma
umg = ma
u = a/g

Hill:

N = Wy
N= wcos(theta)

Ff= ma - wsin(theta)
uN = ma - wsin(theta)
uwcos(theta) = ma - wsin(theta)
Umgcos(theta) = ma - mgsin(theta)
u= (ma - mgsin(theta)) / mgcos(theta)
U= (a - gsin(theta)) / gcos(theta)

(a1/g) = (a2- gsin(theta)) / gcos(theta)


a1gcos(theta) = ga2 - g2sin(theta)

A2= a1cos(theta) + gsin(theta)
A2 = -3.40cos15 - 9.8sin15
A2 = -5.82

Does that seem accurate? It's so close to my prior answer...
 
  • #22
Orodruin
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It is very difficult to follow what you are doing because you do not describe what you are doing or explain your notation.

The end result looks reasonable. Try checking it against the limiting case of a vertical wall.
 
  • #23
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It is very difficult to follow what you are doing because you do not describe what you are doing or explain your notation.

The end result looks reasonable. Try checking it against the limiting case of a vertical wall.

It was right! Thank you SO much!
 
  • #24
haruspex
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It was right! Thank you SO much!
Yes, the right answer according to the book, but what has been called friction throughout this thread might not be that.
We are not told the reason for the slowing down. There are three possible causes: rolling resistance (the motorcycle is freewheeling), normal braking (the tyres are still in rolling contact) or kinetic friction, i.e. skidding (brakes on hard, wheels locked).
The first and last behave the same way in that the resulting force is a constant coefficient multiplied by the normal force on the tyres. So you can assume either of those and arrive at that answer. However, 3.4m/s2 is much too much for rolling resistance and not enough for skidding, which should be closer to 8m/s2, unless the surface is poor.
Normal braking is another matter. The retarding force is not affected by the slope, making your original answer correct.
 
  • #25
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Yes, the right answer according to the book, but what has been called friction throughout this thread might not be that.
We are not told the reason for the slowing down. There are three possible causes: rolling resistance (the motorcycle is freewheeling), normal braking (the tyres are still in rolling contact) or kinetic friction, i.e. skidding (brakes on hard, wheels locked).
The first and last behave the same way in that the resulting force is a constant coefficient multiplied by the normal force on the tyres. So you can assume either of those and arrive at that answer. However, 3.4m/s2 is much too much for rolling resistance and not enough for skidding, which should be closer to 8m/s2, unless the surface is poor.
Normal braking is another matter. The retarding force is not affected by the slope, making your original answer correct.

Interesting, thank you. I didn't realize that they would lead to different results.
 

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