Question

GA The main section of this rocket has a mass of 12 Mg and a radius of gyration about its center of gravity kgA = 3 m. The two booster rockets each have a mass of 2 Mg and a radius of gyration about their center of gravity KGB = koc = 1.75 m. At the moment the booster rockets apply the unequal thrusts shown, determine the acceleration of the tip of the nosecone at A, which is 4 m above GA. 6 m B Gd GB Hint: start by finding the location of the center of gravity of the 3-piece rocket assembly. Tc = 20 KN Trn = 30 KN 1.5 m 1.5 m TA = 150 KN

Answer 1

d= Vortlog MA=12 Mg Код 2 3 » MaB=Muc=2 Mg Ki B z kuc=1.75m 6m Я т - N (%) (0) В lism ism TA= 150KN Т-2014 Тез 3 окN

the structure, man Calculation of unhe of f 1 ls Im; Emi ot 12X103 x6 + 0 - (2 + 12 + 2)x103 y 4.5m & we mbe me Knz dB z dc Moment of Inerlica of the body about A. (CA) = makĂ +2 2 (me kg + mode) 12x103 x 3 + 2 (2x103 13x1.75"+ 2x103x1671.54) (108 8 + 2(6.125+ 76.5)]x103 (10P +165] x103 273.25 Mg m 273.25(103) ка. z > m 2 + 2C1 (109)] 6.91) → +iz Fy z m (awy 150 (103) + 201109) + 30610) - [2(109) +126103) - (2010)+126100) +126109) +2010>)] + a

[150+20+30 - (2 +12 + 2) *9.81] (2 + 12+2) a > i 6 a = 43.04 (a a 2.69m/s 1 26103)(9.3) -ineg mas 2010(9.80) ÓA Now, + ECMWA ELINA & = 1.5 1,5 → 30 (10) 1.5-20(101.5 Tc=20UN Трг 30 = 273,25 (10) TA = 150KN n => dz 0.055 rad/s

the nos cone Now acceleration at the tip of (P P & GP = 10-4.5 4m 2 5.sm GA 5 6m 4,5 m uc AB

a a rola m i et Wzo as it starts, i.e t zo Now, au = 2.69 Î m/% 5,5 2 = 0.055 i rod/5 sos apa an+ 2x78/a - u rele au + 1xre/A 2,69 sî x 5.5 69 ŷ 0.3025 lapl = 2.7 m/s .. The acceleration of the top of the nose cone is ap = 2.699 -0.3025? and magnitude o 2.7m/s f + (0.055 2.69