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Engineering Mechanics: Statics and Dynamics

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EXAMPLE 1 At any instant the horizontal position of the weather balloon in Fig. a is defined by where t is in seconds. If the equation of the path is determine the magnitude and direction of the v ... elocity and the acceleration when SOLUTION Velocity. The velocity component in the x direction is To find the relationship between the velocity components we will use the chain rule of calculus. vx = x # = d dt 18t2 = 8 ft>s : t = 2 s. y = x2 >10, x = 18t2 ft, vy = y # = d dt 1x2 >102 = 2xx# >10 = 21162182>10 = 25.6 ft>s c When the magnitude of velocity is therefore Ans. The direction is tangent to the path, Fig.b, where Ans. Acceleration. The relationship between the acceleration components is determined using the chain rule. We have Thus, Ans. The direction of a, as shown in Fig.c, is Ans. NOTE: It is also possible to obtain and by first expressing y = f1t2 = 18t2 and then taking successive time derivatives. 2 >10 = 6.4t 2 vy ay ua = tan-1 12.8 0 = 90° a = 4(02 2 + (12.822 = 12.8 ft>s 2 = 21822 >10 + 21162102>10 = 12.8 ft>s 2 c ay = v # y = d dt 12xx# >102 = 21x # 2x # >10 + 2x1x $ 2>10 ax = v # x = d dt 182 = 0 uv = tan-1 vy vx = tan-1 25.6 8 = 72.6° v = 4(8 ft>s22 + (25.6 ft>s22 = 26.8 ft>s t = 2 s, y A B x 16 ft (a) y x2 10 (b) B v 26.8 ft/s uv 72.6 (c) a 12.8 ft/s 2 B ua 90 EXAMPLE 2 For a short time, the path of the plane in Fig. a is described by If the plane is rising with a constant velocity of 10 , determine the magnitudes of the velocity and acceleration of the plane when it is at SOLUTION When , then or . Also, since , then ; Velocity. Using the chain rule to find the relationship between the velocity components, we have (1) Thus The magnitude of the velocity is therefore Ans. Acceleration. Using the chain rule, the time derivative of Eq. (1) gives the relation between the acceleration components. When The magnitude of the plane’s acceleration is therefore Ans. These results are shown in Fig.b. [Show More]

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