Explanation Given information: The length of the ladder is L = 5 m . The mass of the ladder is m = 15 kg . The angle θ is 20 ° at rest. Calculation: Consider v A = v A ( ← ) ; v B = v B ( ↓ ) ; ω = ω ( clockwise ) . Draw the triangle ABC by making perpendicular line from v A . Show the free-body diagram of the triangle ABC as in Figure 1. The ladder is placed without any support. Therefore, the ladder slides freely without causing any friction at point A and B . The point C is instantaneous center and the triangle GCB is isosceles. The length of the points GA , GB , and GC is; G A = G B = G C = L 2 Find the velocity of the mass center ( G ) using the relation. v G = v ¯ = L 2 ω Find the mass moment of inertia ( I ¯ ) using the relation. I ¯ = m L 2 12 Find the total kinetic energy ( T ) using the equation. T = 1 2 m v ¯ 2 + 1 2 I ¯ ω 2 Substitute L 2 ω for v ¯ and m L 2 12 for I ¯ . T = 1 2 m ( L 2 ω ) 2 + 1 2 ( m L 2 12 ) ω 2 = m L 2 ω 2 8 + m L 2 ω 2 24 = m L 2 ω 2 6 (1) Show the inclined member AB as in Figure 2. Consider the datum is at point A . Refer to Figure 2. The elevation of point G from point A is; h = L 2 cos θ Find the total potential energy ( V ) using the relation. V = m g h Here, the acceleration due to gravity is g . The acceleration due to gravity is g = 9.8 1 m/s 2 . Substitute L 2 cos θ for h

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 17.1, Problem 17.37P

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Find the angular velocity of the ladder and the velocity of A when θ=45°.

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Chapter 17.1, Problem 17.36P

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Find the angular velocity of the ladder and the velocity of A when θ = 45 ° .

Solution Summary: The author shows the free-body diagram of the triangle ABC as in Figure 1. The ladder is placed without any support.

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Find the angular velocity of the ladder and the velocity of A when θ=45°.

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Chapter 17 Solutions