Explanation Given information: The mass of the compressor ( m C ) is 250 kg. The speed of the rotor ( N ) is 2,000 rpm. The force transmitted to the ground ( P m ) is m r ω f 2 . The density of concrete ( ρ ) is 2400 kg/m 3 . The spring constant ( k ) is 80 × 10 6 N / m . The area of the concrete block ( A ) is 1.5 m × 1.5 m . Calculation: Calculate the frequency of the force transmitted ( ω f ) to the ground using the relation: ω f = 2 π N 60 Substitute 2000 rpm for N . ω f = 2 π ( 2000 rpm ) 60 = 209.44 rad / s Consider the system as a spring mass system, where the compressor with the concrete block acts as mass unit and the soil as spring. Thus, the force transmitted to the ground is the product of spring constant of the spring and amplitude of the vibration. The expression for the force transmitted to the ground ( P ) as follows: P = k x m (1) Here, x m is the amplitude of the forced vibration. The expression for the force ( P m ) due to the unbalance as follows: P m = m r ω f 2 Here, mr is the force due to unbalance is, and the unbalance. The amplitude of the forced vibration is out-of-phase with the unbalance of the compressor. Calculate the amplitude of forced vibration ( x m ) using the formula: x m = − P m / k 1 − ( ω f ω n ) 2 Here, ω n is the natural frequency of vibration of the soil. Substitute m r ω f 2 for P m . x m = − m r ω f 2 / k 1 − ( ω f ω n ) 2 k x m = − m r ω f 2 1 − ( ω f ω n ) 2 k x m = m r ω f 2 ( ω f ω n ) 2 − 1 . (2) Substitute Equation (1) in Equation (2). k x m = m r ω f 2 ( ω f ω n ) 2 − 1 P = m r ω f 2 ( ω f ω n ) 2 − 1 (3) The force transmitted to the ground is the centripetal force due to the unbalance of the compressor. The objective is to reduce the force transmitted to the ground by 75%. The expression for the force transmitted to the ground as follows: P = m r ω f 2 − 75 % ( m r ω f 2 ) = ( 1 − 0.75 ) m r ω f 2 = 0.25 m r ω f 2 (4) Substitute equation (4) in equation (3). P = m r ω f 2 ( ω f ω n ) 2 − 1 0

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259638091

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, Problem 19.167RP

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The depth (h) that will reduce the force transmitted to the ground by 75 percent.

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Chapter 19, Problem 19.166RP

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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 - A particle moves in simple harmonic motion....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 - A simple pendulum consisting of a bob attached to...Ch. 19.1 - Prob. 19.9P Ch. 19.1 - Prob. 19.10P

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The depth ( h ) that will reduce the force transmitted to the ground by 75 percent.

Solution Summary: The author calculates the force transmitted to the ground using the relation comega_f=2pi N60.

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The depth (h) that will reduce the force transmitted to the ground by 75 percent.

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Chapter 19 Solutions