# Fuzzy Dice Angle on a Hill

1. Mar 26, 2012

### mhz

1. The problem statement, all variables and given/known data
A sports car is accelerating at a rate of 4.23m/s^2 up a 14 degree hill. If there are dice hanging from its rear-view mirror, what angle will they swing on?

2. Relevant equations
ƩF = ma
Kinematic Equations

3. The attempt at a solution
Well, I'm thinking that I need to take into account the influence these two factors will have on the dice's angle (and the tension on the string that holds the dice):

1. Acceleration of the car
2. Force of gravity

I'm still working on a solid equation that encompasses the forces acting on the dice, so I don't want to post any unfinished math yet.

Any help or comments are very much welcome! :)

2. Mar 26, 2012

### tiny-tim

welcome to pf!

hi mhz! welcome to pf!

(try using the X2 button just above the Reply box )
have you done fictitious forces?

if so, you can regard the fictitious force as an extra gravity, and add it to the actual gravity to make a "total gravity" in which the dice will hang straight down

if not, remember that the acceleration of the dice (in equilibrium) is the same as the acceleration of the car, and use it in F=ma

3. Mar 26, 2012

### mhz

Alright, so I made this image to better show the problem:

http://img803.imageshack.us/img803/7161/physics.png [Broken]

You're saying that the tension in the x-dimension (the fictitious force?) is added to gravity? I don't fully understand.

Last edited by a moderator: May 5, 2017
4. Mar 26, 2012

### tiny-tim

have you done fictitious forces?

if not, there's no point in trying to do it that way, you must use the usual F = ma method instead

5. Mar 26, 2012

### mhz

Sort of, we've mentioned it as the force pulling the die backwards in this case (though it isn't really a force). So, I think the F = ma method is best for this case.

The real issue I'm having with this question is that I can't simply ignore the 14 degrees and solve for the angle normally then add the 14 degrees at the end because (at least, I think) the distance the die travel backwards is not fixed, it's based on the angle, so there are some ratios going on.

6. Mar 27, 2012

### tiny-tim

hi mhz!

(just got up :zzz:)
not really following you

the question doesn't ask for the tension, so call the equilibrium angle "θ", and do F = ma perpendicular to the string

7. Mar 27, 2012

### mhz

I believe that either my class is over analysing this question or you have some remarkable insight that the rest of us do not. This isn't a homework question, rather a question my teacher said was more difficult than usual and that he either cannot or has not yet tried to solve for whatever reason.

Could you please show me any calculations or equations you can come up with to answer the question?

note: I have a test tomorrow, and your solution would be helpful (although he guaranteed this question wouldn't be on the test lol).

Thanks.

8. Mar 27, 2012

### tiny-tim

no, you do it, and impress the class!

what is the component of acceleration perpendicular to the string?

what are the components of force perpendicular to the string?

9. Mar 27, 2012

### mhz

Alright, I'll play your game. ;)

Here's another image I made, showing the forces I believe are acting on the dice (the circle).

http://img718.imageshack.us/img718/6876/morephysics.png [Broken]

The red is what I'm not certain about.

Last edited by a moderator: May 5, 2017
10. Mar 27, 2012

### tiny-tim

i don't understand your diagram

why isn't Fgy parallel to Ty ?

and why is that red 14 - θ whe it looks like 90 - θ ?

anyway, you're not playing my game

my game was:
what is the component of acceleration perpendicular to the string?

what are the components of force perpendicular to the string?

11. Mar 27, 2012

### mhz

Because the car is going up a hill, not just cruising along a flat surface.

Alright, fair:

Would that be the acceleration of the car? 4.25 m/s^2 up the hill?

Is it not just the net force of the car travelling forward?

12. Mar 28, 2012

### tiny-tim

i meant, why are they both called "y" if they're not parallel?
i asked, what is the component of acceleration perpendicular to the string?

13. May 5, 2012

### mhz

This question is still unsolved if any of you are willing to help out. :)

14. May 5, 2012

### Steely Dan

This doesn't need to be a complicated problem. Call the angle $\theta$ that which the dice take with respect to the car. Then write down $F = ma$ for the directions perpendicular to and parallel to the plane. If you're having trouble with that, then that's just geometry, not a lack of understanding of the physics :)

15. May 5, 2012

### mhz

Does this make sense?

tan(θ - 14) = $\frac{macosθ}{mg}$

tan(θ - 14) = $\frac{4.23cos14}{9.8}$

θ ≈ 36°

Edit: No, I don't believe it does. :)

Edit 2: How about this!

tan(θ) = $\frac{ma}{mgcosθ}$

tan(θ) = $\frac{4.23}{9.8cos14}$

θ ≈ 24°

Last edited: May 6, 2012
16. May 6, 2012

### Steely Dan

Well your second attempt is on the right track but doesn't look quite right. I advise to stop skipping steps and make absolutely sure your two force equations are correct.

17. May 6, 2012

### mhz

Here's my thinking:

Since for these problems on a flat surface the solution is always done by doing tan(theta) is opposite over adjacent where adjacent is gravity (because tension in the y would be equal to gravity) and opposite is (and this is the only bit that confuses me) the net acceleration of the dice.

So, basically I do the same sort of setup but now I have an angle and we're still doing opposite is the net acceleration and adjacent becomes force of gravity in the y, do to the angle.

18. May 6, 2012

### Steely Dan

Your thinking is correct, but what you are missing is that gravity has components in both directions, so the "opposite" side has to include both the total acceleration up the plane and the (oppositely directed) gravitational acceleration tending to bring the car down the plane.

19. May 6, 2012

### mhz

Here's where I'm a bit sceptical. I considered this when making the equation, but I don't think I need to add it.

Here's why:

The right triangle I've created has smaller side Fgy and larger side Fnet with hypotenuse T. You're saying that the larger side should include Fgx but the thing is, Fnet is the total net force acting on the dice therefore gravity is already considered within the acceleration.

Also, if you add gravity again, the angle you solve for is less than 14° which is impossible.

20. May 6, 2012

### Steely Dan

It is incorrect to say that the larger side is "Fnet." All it is, is the component of the tension force parallel to the plane. Similarly, the smaller side is the component of the tension perpendicular to the plane. It is true that this latter term is equal in magnitude to the component of gravity pointing in the same direction. Why? Well, writing down Newton's second law perpendicular to the plane proves that they are. Similarly, if you want to solve for the component of the tension vector parallel to the plane, you need to write down Newton's second law and solve it for the magnitude of the tension as well. You will see that it must be equal to the net force minus the component of gravity that is along the plane. Doing so will get you an equation similar to your one with $\text{tan}(\theta)$, but with a term that you have missed.

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