# Forces-Newton's Laws and Vectors

1. Feb 15, 2009

### joseg707

1. The problem statement, all variables and given/known data
A pair of fuzzy dice is hanging by a string from your rearview mirror. While you are accelerating from a stoplight to 28 m/s in 6.0 s, what angle $$\vartheta$$ does the string make with the vertical?

2. Relevant equations
$$\sum$$F=ma

3. The attempt at a solution
I found the acceleration to be about 4.667, but I honestly have no clue where to go from here. I've looked for other examples in the book that are similar but I haven't found any. The angle part is really throwing me off. I'm completely lost and I just want to be pushed in the right direction.

2. Feb 15, 2009

### jgens

Your value for the acceleration seems alright. Have you drawn a free-body diagram yet and carefully labeled all the forces acting on the dice? If you have, write it out here.

3. Feb 15, 2009

### djeitnstine

Try drawing a free body diagram

edit: jgens said it before me

4. Feb 15, 2009

### joseg707

Ok. Would it be something like this?

$$\uparrow$$ FT
[ ]$$\leftarrow$$ Acceleration?
$$\downarrow$$ FW

5. Feb 15, 2009

### jgens

Not quite. Since, the dice are going to make an angle, Ft should not be positioned directly above mg; hence, your tension has two components, one in the x direction and one in the y. Additionally, your diagram should not show the acceleration as a force.

If it's easier you could just write out the results of your free-body diagram as:
∑Fx = ...
∑Fx = ...

6. Feb 15, 2009

### joseg707

FTcos$$\theta$$+FTsin$$\theta$$+FW.

Is that correct? Would i have to break it into two equations because of the x and y components?

7. Feb 15, 2009

### jgens

Yes, it should be broken into two x and y components.

8. Feb 15, 2009

### joseg707

$$\sum$$Fx=FTsin$$\theta$$
$$\sum$$Fy=FTcos$$\theta$$+FW

Is that correct? I can substitute FW for mass times gravity right?

9. Feb 15, 2009

### jgens

Your equations look correct and Fw can be replaced with mg (just make sure you use g < 0). However, you can extrapolate more still. Are the dice accelerating in the vertical component and what does this suggest about ∑Fy? Eventually, the dice must be accelerating at the same rate as the car, what does this suggest about ∑Fx (think about Newton's second law)?

10. Feb 15, 2009

### joseg707

Oh! So,

$$\sum$$Fy=0

because it is not accelerating, and

FTcos$$\theta$$+mg=ma

Would gravity be negative?

11. Feb 15, 2009

### jgens

Well, it seems you've gotten your forces in the x and y directions mixed up. Your equations should read:

∑Fx = Ftsin(theta) = ma
∑Fy = Ftcos(theta) - abs(g)m = 0.

12. Feb 15, 2009

### joseg707

Oh yeah, sorry about that. But that still gives me three variables in both equations. I still only know that acceleration is 4.667m/s2 and gravity is 9.8m/s2.

13. Feb 15, 2009

### jgens

True, but you can easily reduce it. Divide ∑Fy by ∑Fx and solve, or using ∑Fy come up with an expression for Ft and substitute it into ∑Fx and solve. Either method should work.

14. Feb 15, 2009

### joseg707

Ok, I got answer, 25 degrees, but I just want to make sure my math is correct.

I substituted Ftinto $$\sum$$Fx.

(mg/cos$$\theta$$)sin$$\theta$$=ma

mg*tan$$\theta$$=ma

tan$$\theta$$=a/g

tan-1(a/g)=$$\theta$$

15. Feb 15, 2009

### jgens

Your math it correct. I can't check your angle measure because I don't have a calculator or table of values at this point in time but I would presume it is correct. Good job!

16. Feb 15, 2009

### joseg707

Thanks so much for your help! I've been having trouble understanding force and just going through that problem with you helped me a lot. Thanks so much for your time. :)