(a) Explanation Given information: The force applied at the point A ( P ) is 0.75 lb . The weight of the rod AB ( W ) is 2 lb . The length of the rod AB ( L ) is 36 in . . The acceleration of the point B ( a B ) is 0. Calculation: Consider the acceleration due to gravity ( g ) as 32.2 ft / s 2 . Convert the unit of the length of the rod AB ( L ): L = 36 in . × 1 ft 12 in . = 3 ft Calculate the mass of the rod ( m ): m = W g Substitute 2 lb for W and 32.2 ft / s 2 for g . m = 2 32.2 lb ⋅ s 2 / ft Calculate the mass moment of inertia of the rod ( I ¯ ) : I ¯ = 1 12 m L 2 Substitute 2 32.2 lb ⋅ s 2 / ft for m and 3 ft for L . I ¯ = 1 12 × 2 32.2 × 3 2 = 46.5838 × 10 − 3 lb ⋅ s 2 ⋅ ft Show the free body diagram of the rod AB as in Figure 1. Here, h is the distance between gravity and force P , a is the acceleration of the rod AB , and α is the angular acceleration of rod AB . Refer to Figure 1. Calculate the moment about point G by applying the equation of equilibrium: ∑ M G = I G α ( P × h ) = I ¯ α P h = I ¯ α Substitute 0.75 lb for P and 46.5838 × 10 − 3 lb ⋅ s 2 ⋅ ft for I ¯ . ( 0.75 × h ) = 46.5838 × 10 − 3 × α 0.75 h = 46.5838 × 10 − 3 α α = 0 (b) To determine Find the acceleration of the point A ( a A ) .

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 16.1, Problem 16.49P

(a)

To determine

Find the location of the force P on the rod (zp).

(b)

To determine

Find the acceleration of the point A (aA).

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Chapter 16.1, Problem 16.48P

Chapter 16.1, Problem 16.50P

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Find the location of the force P on the rod ( z p ) .

Solution Summary: The author calculates the mass moment of inertia of the rod (z_p).