590_Dynamics 11ed Manual

# 590_Dynamics 11ed Manual - ⎟ ⎠ N = 10.9 rad s 2 = Since...

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Engineering Mechanics - Dynamics Chapter 17 Given: m D 8kg = L 1m = m b 0kg = r 0.3 m = μ s 0.15 = θ 30 deg = k 0.1 = g 9.81 m s 2 = Solution: φ asin r L = Assume no slip Guesses N C 1N = F C 1N = α 1 rad s 2 = a A 1 m s 2 = F max 1N = Given N C L cos () m D gL cos θφ () m b g L 2 cos () 1 2 m D r 2 m D a A r m b a A r 2 = F C m D m b + () g sin () + m D m b + () a A = F C r 1 2 m D r 2 = a A r = F max s N C = N C F C a A F max Find N C F C , a A , , F max , () = N C F C F max 67.966 13.08 10.195
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Unformatted text preview: ⎟ ⎠ N = 10.9 rad s 2 = Since F C 13.08 N = > F max 10.195 N = then our no-slip assumption is wrong and we know that slipping does occur. Guesses N C 1 N = F C 1 N = 1 rad s 2 = a A 1 m s 2 = F max 1 N = 588...
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## This note was uploaded on 08/16/2011 for the course EGN 3321 taught by Professor Christianfeldt during the Spring '08 term at University of Central Florida.

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