Explanation Given: The weight of the crate lifted by the motor is 50 lb . The velocity of the crate v = 12 ft/s at s = 10 ft . The efficiency of the motor is ∈ = 0.74 . Initially the crate is at rest s 0 = 0 . Draw the free body diagram of the crate as shown in Figure (1). Refer Figure (1). Here, T is the tension acting on the cable. Write the formula for velocity of the crate at constant acceleration as the function of position. v 2 = v 0 2 + 2 a c ( s − s 0 ) (I) Here, v is the final velocity, v 0 is the initial velocity, s is the initial position, s 0 is the final position and a c is the constant acceleration. Write the formula for power. P o u t = T v (II) Write the formula forefficiency of the motor. ∈ = Power output ( P o u t ) Power input ( P i n ) (III) Conclusion: Calculate the acceleration of the crate. Substitute 12 ft/s for v , 0 for v 0 , s 0 and 10 ft for s in Equation (I). ( 12 ft/s ) 2 = ( 0 ) 2 + 2 a c ( 10 ft − 0 ) a c = ( 12 ) 2 20 a c = 7.2 ft/s 2 Refer Figure (1). Resolve the forces along y -axis. ∑ F y = m a y 2 T − 50 lb = 50 lb × 1 32 .2 ft/s 2 × a y ft/s 2 2 T = 1

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 14.5, Problem 5RP

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Thepower supplied to the motor at the instant of s=10 ft .

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Chapter 14 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

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Thepower supplied to the motor at the instant of s = 10 ft .

Solution Summary: The author calculates the power supplied to the motor at the instant of 1.80hp. The formula for velocity of the crate at constant acceleration as the function of position.

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