Explanation Given: The weight is 3 lb . The radial distance ( r ) is 4 / ( 1 − cos θ ) . The transverse distance ( θ ) is 90 ° . The angular velocity ( θ ˙ ) is 4 rad/s . Write the expression for the radial distance ( r ) . r = 4 / ( 1 − cos θ ) (I). Write the expression for the transverse distance ( θ ) . θ = 45 ° = 45 ° π 180 ° = 0.785 rad Write the expression for the angular velocity ( θ ˙ ) . θ ˙ = 4 rad/s (II). Write the expression for angle ψ . tan ψ = r d r d θ (III). The radial co-ordinate is r and rate of change of radial distance with respect to angle. Write the expression for the radial acceleration. a r = r ¨ − r θ ˙ 2 (IV). Here, radial acceleration is a r and second time derivative of radius ( r ) is r ¨ . Write the expression for the transverse acceleration. a θ = r θ ¨ + 2 r ˙ θ ˙ (V). Here, transverse acceleration is a θ and second time derivative of θ is θ ¨ and first time derivative of radius ( r ) is r ˙ . Differentiate Equation (I). r = 4 / ( 1 − cos θ ) r ˙ = − 4 sin θ θ ˙ ( 1 − cos θ ) 2 (VI). Differentiate Equation (VI). r ˙ = − 4 sin θ θ ˙ ( 1 − cos θ ) 2 r ¨ = − 4 sin θ θ ¨ ( 1 − cos θ ) 2 + − 4 cos θ ( θ ˙ ) 2 ( 1 − cos θ ) 2 + 8 sin 2 θ ( θ ˙ ) 2 ( 1 − cos θ ) 3 (VII). Differentiate Equation (II). θ ˙ = 4 rad/s θ ¨ = 0 Draw the free body diagram of motorcycle as shown in Figure (1). Refer Figure (1) write the formula for equation of motion. ∑ F r = m a r (VIII). Here, summation of total force acting on radial direction is ∑ F r mass is m and acceleration radial direction is a r . Refer Figure (1) calculate the components of force. ∑ F r = P sin 45 ° − N cos 45 ° (IX). Refer Figure (1) write the formula for equation of motion. ∑ F θ = m a θ (X). Here, summation of total force acting on transverse direction is ∑ F θ mass is m and acceleration in the transverse direction is a θ . Refer Figure (1) calculate the components of force. ∑ F θ = − P cos 45 ° − N sin 45 ° (XI). Write the expression for mass. m = W 32.2 ft/s 2 (XII). Conclusion: Substitute 3 lb for W in Equation (XII). m = W 32.2 ft/s 2 = 3 lb 32.2 ft/s 2 Substitute 90 ° for θ in Equation (I). r = 4 / ( 1 − cos θ ) = 4 / ( 1 − cos 90 ° ) = 4 ft Substitute 90 ° for θ and 4 rad/s for θ ˙ in Equation (VI). r ˙ = − 4 sin θ θ ˙ ( 1 − cos θ ) 2 r ˙ = − 4 sin 90 ° ( 4 ) ( 1 − cos 90 ° ) 2 = − 16 ft/s Substitute 90 ° for θ , 0 for θ ¨ and 4 rad/s for θ ˙ in Equation (VII). r ¨ = − 4 sin θ θ ¨ ( 1 − cos θ ) 2 + − 4 cos θ ( θ ˙ ) 2 ( 1 − cos θ ) 2 + 8 sin 2 θ ( θ ˙ ) 2 ( 1 − cos θ ) 3 r ¨ = − 4 sin 90 ° θ ¨ ( 1 − cos 90 ° ) 2 + − 4 cos 90 ° ( 4 ) 2 ( 1 − cos 90 ° ) 2 + 8 sin 2 90 ° ( 4 ) 2 ( 1 − cos 90 ° ) 3 = 128 ft/s 2 Substitute 4 / ( 1 − cos θ ) for r in Equation (III)

Engineering Mechanics: Dynamics (14th Edition)

14th Edition

ISBN: 9780133915389

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 13.6, Problem 108P

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The tangential retarding force P and the normal force N acting on the collar when θ=90° .

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Chapter 13.6, Problem 107P

Chapter 13.6, Problem 109P

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Box A has a mass of 15 kilograms and is attached to the 20 kilogram Box B using the cord and pulley system shown. The coefficient of kinetic friction between the boxes and surface is 0.2 and the moment of inertia of the pulley is 0.5 kg * m^ 2. After 2 seconds, how far do the boxes move? A бро B

Chapter 13 Solutions

Engineering Mechanics: Dynamics (14th Edition)

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The tangential retarding force P and the normal force N acting on the collar when θ = 90 ° .

Solution Summary: The author calculates the tangential retarding force and normal force of the collar when theta = 45°.

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The tangential retarding force P and the normal force N acting on the collar when θ=90° .

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Chapter 13 Solutions