Mechanical Engineering > EXAM > Lecture No.3_Engineering Mechanics II_ (All)

Lecture No.3_Engineering Mechanics II_

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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 velocity 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 [Show More]

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