Explanation Given: The radius r is 3 ft . The weight of the bob is 5 lb . Write the equation for weight of the boy. m = 5 32.2 Write the equation of centripetal acceleration a n = υ 2 r (I). Here, centripetal acceleration is a , the velocity is υ and the radius is r . Write the expression for change in arc length, d s = r d θ (II). Here radius is r ., change in arc length is d s ad change in angle is d θ . Write the expression for tangential acceleration. a t = υ d υ d s υ d υ = a t d s (III). Here tangential acceleration is a , velocity is υ and rate of change of velocity with respect to distance is d υ d s . Draw the free body diagram of bob at angle θ is 0 ° as shown in Figure (1). Refer Figure (1) write the formula for equation of motion the normal direction. ∑ F n = m a n (IV). Here, summation of total force acting on normal direction is ∑ F y mass is m and acceleration in the normal direction is a n . Refer Figure (1) calculate the normal force in the vertical direction. ∑ F n = T (V). Draw the free body diagram of bob at angle θ is 45 ° as shown in Figure (2). Refer Figure (2) write the formula for equation of motion in the tangential direction. ∑ F t = m a t (VI). Here, summation of total force acting on tangential direction is ∑ F x mass is m and acceleration in the tangential direction is a t . Refer Figure (2) resolve the forces in the horizontal direction. ∑ F t = W cos θ (VII). Refer Figure (2) write the formula for equation of motion the normal direction. ∑ F n = m a n (VIII). Here, summation of total force acting on normal direction is ∑ F y mass is m and acceleration in the normal direction is a n . Refer Figure (2) calculate the normal force in the vertical direction. ∑ F n = T − W sin θ (IX). Conclusion: When bob at angle θ is 0 ° Velocity will be zero when angle θ is 0 ° Substitute 3 ft for r and υ is 0 in Equation (I). a n = υ 2 r = ( 0 ) 2 3 = 0 Substitute 0 for a n and 5 32.2 for m in Equation (IV). ∑ F n = 5 32.2 ( 0 ) = 0 Substitute 0 for ∑ F n in Equation (V). ∑ F n = T 0 = T Thus, the tension in string after the bob is released when angle θ is 0 ° is 0 . When bob at angle θ is 45 ° Substitute 5 32.2 for m in Equation (VI). ∑ F t = 5 32.2 a t Substitute 5 32

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 13.5, Problem 63P

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The tension in string after the bob is released when angle θ is 0°

The tension in string after the bob is released when angle θ is 45°

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Chapter 13.5, Problem 62P

Chapter 13.5, Problem 64P

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The tension in string after the bob is released when angle θ is 0 ° The tension in string after the bob is released when angle θ is 45 °

Solution Summary: The author explains the tension in string after the bob is released when angle theta is 45° and the equation for centripetal acceleration.

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The tension in string after the bob is released when angle θ is 0°

The tension in string after the bob is released when angle θ is 45°

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The tension in string after the bob is released when angle θ is 0 ° The tension in string after the bob is released when angle θ is 45 °

Solution Summary: The author explains the tension in string after the bob is released when angle theta is 45° and the equation for centripetal acceleration.

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The tension in string after the bob is released when angle θ is 0° The tension in string after the bob is released when angle θ is 45°

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Chapter 13 Solutions

The tension in string after the bob is released when angle θ is 0°

The tension in string after the bob is released when angle θ is 45°