Solving eqs 1 through eqs 4 simultaneously gives ans

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Solving Eqs. 1 through Eqs. 4 simultaneously gives Ans. A X = B X = 0 A Y = 71.6 N B Y = 26.5 N A Z = 0 © F Z = m 1 a G 2 Z ; A Y + B Y - 98.1 = 0 © F Y = m 1 a G 2 Y ; A X + B X = 0 © F X = m 1 a G 2 X ; A X = 1.25 B X A X 1 0.2 2 sin 10° - B X 1 0.25 2 sin 10° = 0 - 0 © M z = I z v # z - 1 I x - I y 2 v x v y A X = 1.25 B X A X 1 0.2 2 cos 10° - B X 1 0.25 2 cos 10° = 0 - 0 © M y = I y v # y - 1 I z - I x 2 v z v x - 0.2 A Y + 0.25 B Y = - 7.70 - 1 A Y 21 0.2 2 + 1 B Y 21 0.25 2 = 0 - 1 0.05 - 0.1 21 - 30 sin 10° 21 30 cos 10° 2 © M x = I x v # x - 1 I y - I z 2 v y v z 1 æ = V 2 1 a G 2 x = 1 a G 2 y = 1 a G 2 z = 0. v # x = v # y = v # z = 0. V # = 0 , 1 + Z 2 V V # = 1 V # 2 xyz æ = V , v # x = v # y = v # z = 0. v x = 0 v y = - 30 sin 10° v z = 30 cos 10° V æ = V . v = 30 rad > s, I x = I y = 0.05 kg # m 2 , I z = 0.1 kg # m 2 , 0.2 m 0.25 m Z Y X , x y z A B (a) 10 10 G v 30 rad / s 10 0.2 m 0.25 m x y z A B (b) 98.1 N B X A X A Y B Y 10 G A Z x y z A B G (c) 10 V 10 Fig. 21–12
606 C HAPTER 21 T HREE -D IMENSIONAL K INETICS OF A R IGID B ODY 21 The airplane shown in Fig. 21–13 a is in the process of making a steady horizontal turn at the rate of During this motion, the propeller is spinning at the rate of If the propeller has two blades, determine the moments which the propeller shaft exerts on the propeller at the instant the blades are in the vertical position. For simplicity, assume the blades to be a uniform slender bar having a moment of inertia I about an axis perpendicular to the blades passing through the center of the bar, and having zero moment of inertia about a longitudinal axis. v s . v p . EXAMPLE 21.5 V s V p (a) x y z G F R M R (b) Z , z ¿ , z X , x ¿ , x Y , y ¿ , y (c) V s V p Fig. 21–13 SOLUTION Free-Body Diagram. Fig. 21–13 b . The reactions of the connecting shaft on the propeller are indicated by the resultants and (The propeller’s weight is assumed to be negligible.) The x, y, z axes will be taken fixed to the propeller, since these axes always represent the principal axes of inertia for the propeller.Thus, The moments of inertia and are equal and Kinematics. The angular velocity of the propeller observed from the X, Y, Z axes, coincident with the x, y, z axes, Fig. 21–13 c , is so that the x, y, z components of are Since then To find , which is the time derivative with respect to the fixed X, Y, Z axes, we can use Eq. 20–6 since changes direction relative to X,Y, Z .The time rate of change of each of these components relative to the X,Y,Z axes can be obtained by introducing a third coordinate system which has an angular velocity and is coincident with the X,Y, Z axes at the instant shown.Thus æ¿ = V p z ¿ , y ¿ , x ¿ , V # = V # s + V # p V V # V # = 1 V # 2 xyz . æ = V , v x = v s v y = 0 v z = v p V V = V s + V p = v s i + v p k , I z = 0. 1 I x = I y = I 2 I y I x æ = V . M R . F R
21.4 E QUATIONS OF M OTION 607 21 Since the X,Y, Z axes are coincident with the x, y, z axes at the instant shown, the components of along x, y, z are therefore These same results can also be determined by direct calculation of however, this will involve a bit more work.To do this, it will be necessary to view the propeller (or the x, y, z axes) in some general position such as shown in Fig. 21–13 d . Here the plane has turned through an angle (phi) and the propeller has turned through an angle (psi) relative to the plane. Notice that is always directed along the fixed Z axis and follows the x axis. Thus the general

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