hmwk5 - / s 2 s = 31.4 N 6.0kg 9.8 m / s 2 = 0.5 7. F g x =...

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Wade Burelbach Homework 5 PHYS100A Kramer Chapter 4, Problems 18, 51, 78: 18. F N = mg , F N ' = 0.75 mg F = ma mg F N ' = ma F N ' = 0.75 mg = mg ma a = 0.25 g = 2.45m / s 2 (down) 51. DIAGRAM (SEE ANSWERS IN BOOK) F y = m B a B y F T m B g = m B a B y F T – m B g = m B a F x = m A a A x F T = m A a Ax F T = m A a a Ax =− a B y =− a m B a m B g = m A a m B m A a =− m B g a =− m B g m B m A , F T =− m A m B g m B m A 78. a. (Reference = helicopter) F = ma F lift F g = ma F lift = ma F g F lift = 7650 1250kg  0.80m / s 2 − 7650 1250 kg − 9.8m / s 2 = 94340N b. (Reference = construction frame) F = ma F T F g = ma F T = F g ma F T = 1250kg  9.8m / s 2  1250kg  0.8m / s 2 = 13250 N c. F T is the same throughout; therefore, 13250 N Chapter 5, Problems 2, 7: 2. a. F s max = s F N 35.0 N = s 6.0kg  9.8 m / s 2
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Wade Burelbach Homework 5 PHYS100A Kramer s = 35.0 N 6.0kg  9.8 m / s 2 = 0.6 b. F = ma 35.0 N F k = 6.0kg  0.60 m / s 2 F k = 35.0N − 6.0kg  0.60 m / s 2 = 31.4 N F k = k F N 31.4 N = k 6.0kg  9.8 m
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Unformatted text preview: / s 2 s = 31.4 N 6.0kg 9.8 m / s 2 = 0.5 7. F g x = 25.0kg 9.8m / s 2 sin 27 = 111N F x = ma x 111N F k = 25.0kg 0.30m / s 2 F k = 103.5N F y = ma y F N 113 N = F N = 113 N k = F k F N k = 37 N 113 N = 0.33 F g x = 117.6 sin 20 = 40 N F x = ma x 40 N F k = 12kg 0.3m / s 2 F k = 37 N F y = ma y F N 113 N = F N = 113 N k = F k F N k = 37 N 113 N = 0.33 K1. Diagram (see page) K2. Wade Burelbach Homework 5 PHYS100A Kramer Diagram? (see page) K3. Diagram (see page) K4. Diagram (see page)...
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This note was uploaded on 11/18/2011 for the course PHYS 100 taught by Professor Kramer during the Fall '08 term at Simons Rock.

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hmwk5 - / s 2 s = 31.4 N 6.0kg 9.8 m / s 2 = 0.5 7. F g x =...

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