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Published byJohan Susman Modified over 6 years ago
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I can solve for a bunch more stuff Fg= 400 N θ = 40° Fg = mg
Given: I can solve for a bunch more stuff Fg= 400 N θ = 40° Fg = mg 400 = (m) (9.8) m = 40.8 kg μs = 0.45 μk = 0.35 F|| = Fg sinӨ fs ≤ μsFN fs ≤ (.45)(306) fs ≤ 138 N fk =μkFN fk = (.35)(306) fk = 107 N F|| = (400) sin(40º) = 257 N FN = Fg cosӨ For which questions are these values valid? FN = (400) cos(40º) = 306 N 1 through 5
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FT = 119 N 1. Draw the free body diagram: a = 0 F = ma FN FT
F = FT + fs – F|| ma = FT + fs – F|| FN FT F|| fs Fg 40.8(0) = FT – 257 FT = 119 N
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FT = 364 N 2. Draw the free body diagram: a = 0 F = ma FN FT
F = FT – fk – F|| ma = FT – fk – F|| FN FT F|| fk Fg 40.8(0) = FT – 107 – 257 FT = 364 N
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FT = 150 N 3. Draw the free body diagram: a = 0 F = ma FN FT
F = F|| – FT – fk ma = F|| – FT – fk FN FT F|| fk Fg 40.8(0) = 257 – FT – 107 FT = 150 N
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FT = 433 N 4. Draw the free body diagram: a F = ma FN FT
F = FT – F|| – fk ma = FT – F|| – fk FN FT F|| fk Fg 40.8(1.7) = FT – 257 – 107 FT = 433 N
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FT = 81 N 5. Draw the free body diagram: a F = ma FN FT
F = F|| – FT – fk ma = F|| – FT – fk FN FT F|| fk Fg 40.8(1.7) = 257 – FT – 107 FT = 81 N
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6/7. in # 6 fk = 0 N in # 7 fk = μkFN = 31.3 N
Draw the free body diagrams: F|| = (78.4) sin(37º) = 47.2 N a FT FN FT a FN = (78.4) cos(37°) = 62.6 N m2 F|| m1 fk Fg2 Fg1 F = m2a F = Fg2 - FT m2a = Fg2 - FT F = m1a F = FT - F|| - fk m1a = FT - F|| - fk in # 6 fk = 0 N in # 7 fk = μkFN = 31.3 N 6) a = 3.2 m/s2; FT = 73 N 7) a = 1.5 m/s2; FT = 91 N
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8. Draw the free body diagrams: F|| = (44.1) sin(37º) = 26.5 N a FT FN FT a FN = (44.1) cos(37°) = 35.3 N m2 F|| m1 fk Fg2 Fg1 fk = μkFN = 12.4 N F = m2a F = Fg2 - FT m2a = Fg2 - FT F = m1a F = FT - F|| - fk m1a = FT - F|| - fk 4m2 = 9.8m2 – 56.9 4.5(4) = FT – 26.5 – 12.4 FT = 56.9 N m2 = 9.8 kg
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