(^) 112. CB OK, you know how. If you've done any
of the probltell you to draw theems in the book little dotted line, you know where t through whey here
the force would go, you know what I mean?
B
W^
- ME H114. CB And tmm. hen from, from this point, which Ak
what I mwould be theane orig, this linin. Youe, so it w'd drawould be a n the...you know inety
degwhat I'ree anm talkgle. Cing aban we do thout? at, anyhow? Know
W- ME WelIt would be 60 degrees. Bl that's...that's what I wut we'll just call it alpas thinking. ha.
Are you looking at this angle?
RQCl G^ - ME It'116. CB Yeahs going to be s. ixty degrees, because Ak^
60 an118. Sd 30 iT Ts ninety. his is if you count this, this is 30, this G^ - Cone's 30, tB Shis ono, see, te's 60, tau equhis is 90. als l times f. Cl C^
- S121. ME ItT Alpha, t would be this onhe se's alpha hame. Tere. his angle should Cl
be thworrye sam about it, I de as this oon't thne. Bink. Lut we don't have toet's just call it alp (^) ha
and sdown the see what wum ofe get th forcesere. So let'. s start writing
W
C^
W = th e angle is the same.
This is a nchallenges or mew claimodifies 118. since it neither
(^11) CBB 2
(^11) W CB 2
(^11) AkME (^3 11) W CB^4
RQCl^11 ME^5 ME^11 G^5
(^11) Ak CB (^6) ME^11 G^7
(^11) STC (^811) CBCl (^912) STCl 0
ME^12 C^1 ME^12 W^1
Figure C-7. Group 4D, Episode 15, lines 112-121