Explanation Given information: The weight of the motor ( W ) is 360 lb. The weight of the rotor ( w ) is 0.9 oz. The spring constant ( k ) is 12.5 kip/ft. The axis of rotation ( r ) is 7.5 in. The fluctuating force exerted on the foundation is less than 5 lb. Calculation: Calculate the mass of the motor ( M ) using the formula: M = W g Here, g is the acceleration due to gravity. Substitute 360 lb for W and 32.2 ft / s 2 for g . M = 360 lb 32.2 ft / s 2 = 11.180 lb ⋅ s 2 / ft Convert the unit of unbalance weight ( w ) from oz to lb: w = 0.9 oz = 0.9 oz × 1 lb 16 oz = 0.05625 lb Calculate the mass of the unbalance weight ( m ) using the formula: m = w g Substitute 0.05625 lb for w , and 32.2 ft / s 2 for g . m = 0.05625 lb 32.2 ft / s 2 = 1.7469 × 10 − 3 lb ⋅ s 2 / ft Calculate the natural frequency ( ω n ) using the formula: ω n = k M Substitute 12.5 kips / ft for k and 11.180 lb ⋅ s 2 / ft for M . ω n = 12.5 kips / ft 11.180 lb ⋅ s 2 / ft = 12.5 × 1000 lb / ft 11.180 = 33.438 rad / s Write the expression for the amplitude of forced vibration as follows: x m = P m / k 1 − ( ω f ω n ) 2 k x m = P m 1 − ( ω f ω n ) 2 (1) Here, x m is the amplitude of forced vibration, P m is the unbalanced force and ω f is the frequency of fluctuating force. The amplitude of the fluctuating force must be less than ± 5 lb . This is given as the product of spring constant and deflection. The deflection is the amplitude of vibration and the spring constant is k . Calculate the fluctuating force using the relation: k x m < ± 5 lb (2) Substitute equation (1) in equation (2). k x m < ± 5 lb P m 1 − ( ω f ω n ) 2 < ± 5 (3) In the system given, the vibration is due to the unbalance of the rotor. Calculate the relation between the unbalanced force and centrifugal force due to eccentricity of the unbalanced weight using the relation: P m = m r ω f 2 (4) Here, r is the eccentricity of the unbalanced weight. Substitute equation (4) in equation (3). P m 1 − ( ω f ω n ) 2 < ± 5 m r ω f 2 1 − ( ω f ω n ) 2 < ± 5 (5) Substitute 1.7469 × 10 − 3 lb ⋅ s 2 / ft for m , 7.5 in. for r, and 33.4375 rad / s for ω n in Equation (5). P m 1 − ( ω f ω n ) 2 < ± 5 lb m r ω f 2 1 − ( ω f ω n ) 2 < ± 5 ( 1

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 19.4, Problem 19.120P

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Find the range of speeds of the motor (ωf) for which the amplitude of the fluctuating force exerted on the foundation is less than 5 lb.

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Chapter 19.4, Problem 19.119P

Chapter 19.4, Problem 19.121P

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Chapter 19 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

Ch. 19.1 - A particle moves in simple harmonic motion....Ch. 19.1 - Prob. 19.2P Ch. 19.1 - Prob. 19.3P Ch. 19.1 - Prob. 19.4P Ch. 19.1 - Prob. 19.5P Ch. 19.1 - Prob. 19.6P Ch. 19.1 - Prob. 19.7P Ch. 19.1 - Prob. 19.8P Ch. 19.1 - Prob. 19.9P Ch. 19.1 - Prob. 19.10P

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Find the range of speeds of the motor ( ω f ) for which the amplitude of the fluctuating force exerted on the foundation is less than 5 lb.

Find the range of speeds of the motor (ωf) for which the amplitude of the fluctuating force exerted on the foundation is less than 5 lb.

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Chapter 19 Solutions