3.34, 3.36, 3.53

# 3.34, 3.36, 3.53 - PHYS2626 Introductory Classical...

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Unformatted text preview: PHYS2626 Introductory Classical Mechanics Suggested Solutions for Assignment 1 September 2008 1. Problem 3.34 Refer to the figure for the symbols used. Only forces used to solve the problem are shown. The equations: N 1 + N 3 cos π 3 = ma. (1) ( N 2- N 3 ) cos π 3 = ma. (2) ( N 2 + N 3 ) sin π 3 = mg. (3) Using (2) and (3) to eliminate N 2 , 2 N 3 cos π 3 sin π 3 = mg cos π 3- ma sin π 3 . (4) For the case of minimum a , we have N 1 = 0, so (1) becomes N 3 cos π 3 = ma. (5) And finally eliminating N 3 with (4), we get 2 ma sin π 3 = mg cos π 3- ma sin π 3 3 a sin π 3 = g cos π 3 a = g cos π 3 3 sin π 3 = g 3 √ 3 . (6) 1 For the case of maximum a , we have N 3 = 0. From (4), 0 = mg cos π 3- ma sin π 3 a = g cos π 3 sin π 3 = g √ 3 . (7) Hence, all three cylinders will remain in contact with each other when g/ 3 √ 3 ≤ a ≤ g √ 3. 2. Problem 3.36 (a) Along the plane there are two forces acting on the block: friction μN and the component of gravitational force mg sin θ . Perpendicular to the plane there are the normal reaction N and mg cos θ . For the block to slide back....
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3.34, 3.36, 3.53 - PHYS2626 Introductory Classical...

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