Explanation Given: The weight of the load is W = 50 lb . The constant force exerted on the cable is F = 30 lb . The efficiency of the motor is ∈ = 0.76 . Draw the free body diagram of load as shown in Figure (1). Refer Figure (1). Resolve the forces along y − axis to determine the tensions in the cable. ∑ F y = m a y When, m = W 32.2 ft/s 2 = 50 lb 32.2 ft/s 2 and a y = a c . Substitute 50 lb 32.2 ft/s 2 for m and a c for a y . ∑ F y = 50 lb 32.2 ft/s 2 ( a c ) (I) Here, m is the mass of the load and a c is the constant acceleration. 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 ) (II) 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 ) . P o u t = F v (III) Here, F is the force and v is the velocity. Write the formula for efficiency ( ∈ ) . ∈ = Power output ( P o u t ) Power input ( P i n ) (IV) Conclusion: Refer Figure (1). Resolve the forces along y -axis and substitute in Equation (I). 2 ( 30 lb ) − 50 lb = 50 lb 32.2 ft/s 2 ( a c ) 10 lb=1

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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The area of science and mathematics concerned the motion of the objects, the relation between the force matter, and motion if specified. Forces exerted over the objects result in displacements or changes in the position of an object concerning its enviro…

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Chapter 14.5, Problem 6RP

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The power supplied to the motor when s = 10 ft .

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