Question

exercise 10.12
tq very much

Exercise 10.12 Member BD is welded onto rigid body ABC at B as shown. The mass of the result- ing rigid body is 2 kg/m. If the body is rotating clockwise with an angular velocity a = 5 rad/s, determine the components of the reaction at A for that instant. 1.0 m 0.5 m В 0.5 m (Answer: A = 83.64 N Ay = 14.40 NT)

hello sir,
I have already check the answer and it is still the same answer sir. this is because I take this question straight from my book.

it's okay sir tq

Answer 1

step :- free body diagram Compare and kinematic diagram Ay B 0 FBD 'nggg d=angular acceleration of rod m BDO BC no normal acceleration Es tangential acceleration BABY (malabn B (maloon A IBOA malast (ma), (maloot BC IBCA с mas Bet EMA) FED (EMA)KO step@:- (Moc 9 x1 + IBOC meg)x1.25 I ABX + IBC X → (Magy) Ź + X 1.25 + mabast Ž + malept +mal x JIR +0.25?. Bet Here, mag = 2 ks ) Cım) = 2kg meo - 2X0.5 = = 1 kg 9 = 9.81 mis? e - 14.0360 tan' (0.25) Sae men 2X 12 '16 T2 IxGS IBC = IBO ४ 12

0.52 (angle = @BO) - 1.252 (ABC) = JP +0.25? a (OAB) = ² w ² = 1 (5)² = 12.5 m/s2 (ABC), Jeto.152 x 25 = 25.769 @Bo). 1.25*25 = 31.25 (2x9.81) Ź 1x 9.81x1 = S2'1 x 186 x + C + 1 48 + 4 is). + 2 x ois? 1X CR +0.28²) a + 1x 1.25² & + 31.8825 3.333 & 9.58475 rad/g2 CW step®:- 09323) C&Fx) KD -Ax -ma), ma), (ma) coso -ma) But ODD sino ABD =) At - 2 X 12.5 +4 X 31.25 1 x 25.769 x COS14.036 + 1*25.969 sinly; 034 = 87.5 N : Ax=87.50 )

(& Fy) FED Fy) Step :- h) → Ay - MABI Bot -Macg mong = {malage + (malech sino -mar (maject coso 9.56476 * Ay - 39.24 = -9.56475 +6.24979 -11.99994 A Ay 14.4N(D) think NOTE:- text book is I Asc in wrong. Imay be mispointed.