ENGINEERING MECHANICS Â?? STATICS
ENGINEERING MECHANICS Â?? STATICS
15th Edition
ISBN: 9780137519132
Author: HIBBELER
Publisher: PEARSON
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Chapter 1.6, Problem 21P

If they are 800 mm apart, determine the force of gravity acting between them. Compare this result with the weight of each particle.

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Problem 5 (Optional, extra 6 points) A 6-lb homogeneous disk of radius 3 in. spins as shown at the constant rate w₁ = 60 rad/s. The disk is supported by the fork-ended rod AB, which is welded to the vertical shaft CBD. The system is at rest when a couple Mo= (0.25ft-lb)j is applied to the shaft for 2 s and then removed. Determine the dynamic reactions at C and D before and after the couple has been removed at 2 s. 4 in. C B Mo 5 in 4 in. Note: 2 rotating around CD induced by Mo is NOT constant before Mo is removed. and ₂ (two unknowns) are related by the equation: ₂ =0+ w₂t 3 in. Partial Answer (after Mo has been removed): C-7.81+7.43k lb D -7.81 7.43 lb
Problem 4. A homogeneous disk with radius and mass m is mounted on an axle OG with length L and a negligible mass. The axle is pivoted at the fixed-point O, and the disk is constrained to roll on a horizontal surface. The disk rotates counterclockwise at the constant rate o₁ about the axle. (mg must be included into your calculation) (a). Calculate the linear velocity of G and indicate it on the figure. (b). Calculate ₂ (constant), which is the angular velocity of the axle OG around the vertical axis. (c). Calculate the linear acceleration ā of G and indicate it on the figure. (d). Determine the force (assumed vertical) exerted by the floor on the disk (e). Determine the reaction at the pivot O. 1 Answers: N = mg +mr(r/L)² @² |j mr w IIG C R L i+ 2L =
Problem 2. The homogeneous disk of weight W = 6 lb rotates at the constant rate co₁ = 16 rad/s with respect to arm ABC, which is welded to a shaft DCE rotating at the constant rate 2 = 8 rad/s. Assume the rod weight is negligible compared to the disk. Determine the dynamic reactions at D and E (ignore mg). Answers: D=-7.12ĵ+4.47k lb r-8 in. 9 in. B D E=-1.822+4.47 lb 9 in. E 12 in. 12 in. x
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