- #1
suspenc3
- 402
- 0
Homework Statement
The illustrated equilateral triangle is supported by two links. d = 0.5 m. At the illustrated position,[tex]\dot{\theta}= 9 rad/s[/tex] and [tex]\ddot{\theta}= 0 rad/s^2[/tex]. Find the magnitude of [tex]a_C[/tex].
http://img406.imageshack.us/img406/7264/tonguech66315yx0.th.gif
The Attempt at a Solution
First I found the angle [tex]\beta[/tex]. This is the angle between point B and the horizontal
[tex]\beta=30[/tex]
Next I found all the angular speeds I am going to need: [tex]\omega_{DA}, \omega_{EB}, \omega_{AB}[/tex]
[tex]\omega_{DA}=9 rad/s[/tex] (Given)
[tex]\omega_{EB}=\omega_{DA} \frac{AD}{EB}[/tex] (AD and EB are essentially given)
[tex]\omega_{EB}=18 rad/s[/tex]
[tex]\omega_{AB}=-\omega_{DA}DA+ \omega_{EB}EB[/tex]
[tex]\omega_{AB}=0[/tex]
Next I found all angular accelerations I will need by assuming PGM:[tex]a_B=a_A+a_{B/A}[/tex]
[tex]\omega_{EB}^2(EB)j-\alpha_{EB}(EB)i=\omega_{DA}^2(AD)j+\alpha_{AB}(AB)sin(90-\beta)j-\alpha_{AB}(AB)cos(90-\beta)i[/tex]
I arranged the [tex]i[/tex] and [tex]j[/tex] components and solved finding:
[tex]\alpha_{EB}=93.53[/tex]
[tex]\alpha_{AB}=93.53[/tex]
Now I can find the accelerations of C:
[tex]a_Ci+acj=\omega_{EB}^2(EB)j-\alpha_{EB}(EB)i+\alpha_{AB}(BC)i[/tex]
Solve resultant of [tex]a_C[/tex] to equal 168 whereas it should be .202.
I'm not good at this at all and could have made some pretty big mistakes so bear with me.
Any help would be greatly appreciated!
Last edited by a moderator: