Forces-Newton's Laws and Vectors

[joseg707]

Homework Statement

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?

Homework Equations

\sumF=ma

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.

 

[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.

 

[djeitnstine]

Try drawing a free body diagram

edit: jgens said it before me

 

[joseg707]

Ok. Would it be something like this?



\uparrow F_T
[ ]\leftarrow Acceleration?
\downarrow F_W

 

[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 = ...

 

[joseg707]

F_Tcos\theta+F_Tsin\theta+F_W.

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

 

[jgens]

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

 

[joseg707]

\sumF_x=F_Tsin\theta
\sumF_y=F_Tcos\theta+F_W

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

 

[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)?

 

[joseg707]

Oh! So,

\sumF_y=0

because it is not accelerating, and

F_Tcos\theta+mg=ma

Would gravity be negative?

 

[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.

 

[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.

 

[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.

 

[joseg707]

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

I substituted F_tinto \sumFx.

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

mg*tan\theta=ma

tan\theta=a/g

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

 

[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!

 

[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. :)