Explanation Given: The weight of the collar is 2 lb . The initial speed of the collar is 5 ft/s . The unstretched length of the spring is 2 ft . The stiffness of the spring is k = 10 lb/ft . The equation of the path of the collar y = 4.5 − 1 2 x 2 Draw the free body diagram of collaras shown in Figure (1). Write the formula for kinetic energy ( T ) . T = 1 2 m v 2 (I) Here, m is the mass and v is the velocity. Write the formula for elastic potential energy ( V e ) . V e = 1 2 k s 2 (II) Here, k is the stiffness of the spring and s is the displacement. Write the formula for gravitational potential energy ( V g ) . V g = W h (III) Here, W is the weight and h is the displacement. Write the formula total potential energy ( V ) . V = V g + V e (IV) Here, V g is gravitational potential energy and V e is elastic potential energy. Write the formula for conservation of energy. T A + V A = T B + V B (V) Here, T A , T B are the initial and final kinetic energy and V A , V B are the initial and final potential energy. Resolve the forces along n − axis to determine the normal reaction force ( N ) . ∑ F n = m a n (VI) Here, m is the mass of skier and a n is the acceleration along axis − n . When the skier traveling along the path of y = 4.5 − 1 2 x 2 . The acceleration of the path is a n = v 2 ρ . Substitute v 2 ρ for a n in Equation (VI). ∑ F n = m ( v 2 ρ ) (VII) Here ρ is the radius of the curvature. Write the value of ρ . ρ = [ 1 + ( d y d x ) 2 ] 3 2 | d 2 y d x 2 | (VIII) Resolve the forces along t − axis to determine the tangential acceleration ( a t ) . ∑ F t = m a t (IX) Conclusion: At position A : Refer Figure (1). Calculate the extension length of the spring at position A . s A = 4.5 ft − 2 ft = 2 .5 ft Refer Figure (1). Calculate the kinetic energy of the collar. Substitute 2 lb for m and 5 ft/s for v in Equation (I). T A = 1 2 ( 2 lb × 1 32.2 ft/s 2 ) ( 5 ft/s ) 2 = 0.7764 ft ⋅ lb Calculate the elastic potential energy ( V e ) of the collar. Substitute 10 lb/ft for k and 2.5 ft for s in Equation (II) V e = 1 2 ( 10 lb/ft ) ( 2.5 ft ) 2 = 31 .25 ft ⋅ lb Calculate the gravitational potential energy ( V g ) of the collar. Substitute 2 lb for W and 4.5 ft for h in Equation (III). V g = 2 lb ( 4.5 ft ) = 9 ft ⋅ lb Calculate the total potential energy ( V A ) of the collar. Substitute V A for V , 31.25 ft ⋅ lb for V e and 9 ft ⋅ lb for V g in Equation (IV). V A = 31.25 ft ⋅ lb + 9 ft ⋅ lb = 40.25 ft ⋅ lb At position B : Refer Figure (1). Calculate the extension length of the spring at position B . s B = 3 ft − 2 ft = 1 ft Refer Figure (1). Calculate the kinetic energy of the collar. Substitute 2 lb for m and v B for v in Equation (I). T B = 1 2 ( 2 lb × 1 32.2 ft/s 2 ) ( v B ) 2 = 0.03106 v B 2 Calculate the elastic potential energy ( V e ) of the collar. Substitute 10 lb/ft for k and 1 ft for s in Equation (II) V e = 1 2 ( 10 lb/ft ) ( 1 ft ) 2 = 5 ft ⋅ lb When the collar is at the position of datum point, the gravitational potential energy becomes zero

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 14.5, Problem 88P

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The speed of the collar at point B , normal reaction, and rate of decrease in speed.

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Chapter 14.5, Problem 87P

Chapter 14.5, Problem 89P

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Chapter 14 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

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The speed of the collar at point B , normal reaction, and rate of decrease in speed.

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The speed of the collar at point B , normal reaction, and rate of decrease in speed.

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