# Statics Quiz 1 1 - Tum Wax-x15 *D SORVC 1 O. DO NOT EM...

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Unformatted text preview:Tum Wax-x15 *'D SORVC 1 O. DO NOT EM 306 quiz 1 Name: i \( EID: Session Number: A boomerang mathematically represented by the parabola y = x2 — 4; is pinned to the ground at point "A ".- An inextensibleostring of length 2 m is fixed to the rim of the boomerang at point C. The string 80 makes an angle ofjéfi6 with the x-axis (as shown in the figure) and an take a maximum load of km u v\ a g . pounds before it breaks. Find the range of anglesnat which a load "P"; pounds can be imposed on the boomerang at "B", such that the string does not break. Hint: a pin cannot support moments, therefore, if the system is in equilibrium {which is the case in our problem), the resultant moments at A (or any point to that matter will be zero. Therefore the moments LX' : O /\ due load P will be counteracted by.... ?! W jigs 9 .6 \oo/H'f ~> X ((9:0) MOM ioqolino shim? :{oo' TY:TS*Y'\}O 3 OST'V'QO Tr: T005707 9'1: [cog-IND PX : i055 036' '3 Ciro} , '9th Com-dico'fbﬂ OF C); sz_lco330° (~,:-ts'mit°= 4 :> CMLWC'lflL—l) fl) gum-k dex P his (om'ioneqk of Show" above ' ~-'_ .\ J .— .L1 3) M°MU\'\ Nomi A - camel?" _ AWAV I 9,, 1 \/——\ - \_/— \ gt ,Pb', at Z "A: T7 (1-1-13) + T2303 — Pt, 01) -— o SubS'iilde 7, ,Tgpt, Ts'm'so" (p.17) +Tmao°m — '1 U0935iﬁ9 '— 0 Li) Consider (awe; MWC'. T10 H: r, BY 9:6, 6 'N'U' Lama T= (m is ﬂy as.) oil ————____ Thus, 0°: esp—(Hf vat/va o 53 CoASiLLM (Miro-w I ° 6 offer "Aﬂumnfwdv ' hawk T W: m-Forto. vedcrﬁv} cm Le MQJCA +0 +k¢ MGM. 2)Now +0401 VMW'M/Vf 0499* A. (04%"? jk'AﬁLA—'tiﬂx : EMA: T703 — Hum to SJUY'WH'? Ty .Py 'Tsin}o°(-v§ — L1 (\W)S'Vn8=0 33 (paddy T110 w! 810' r (ow H3 (mam 13mm, bx, 3 «9': MW" e Aw bovm,' Thug I we CKH'KI! «it "a. ftm(_\ Cpﬁc'v)'!on: O- '5 e g