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

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 + 138 – 257 FT = 119 N

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

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

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

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

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

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