P150_Exam3_F07sol

# P150_Exam3_F07sol - Physics 150 Exam#3 Solution 1 2 3 4 5 6...

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F h 657.854N = F h T 1 cos θ 1 () := Then to the horizontal forces equation: F v 1176N = F v m scaffold g m man g + T 1 sin θ 1 := Returning to the vertical forces equation: T 1 1023.439 N = T 1 L 2 m scaffold g x man m man g + L sin θ 1 := τ L 2 m scaffold g x man m man g + T 1 sin θ 1 L := 0 := Taking torques about the right end of the scaffold: T 1 cos θ 1 F h + 0N := Total horizontal forces; T 1 sin θ 1 F v + m scaffold g m man g := Total vertical forces: θ 1 50 deg := m man 120 kg := x man 4m := m scaffold 80 kg := L1 2 m := Problem #1: Physics 150 Exam #3 Solution 12/3/07
E 3.842 10 28 × J = EKU + := U 7.685 10 28 × J = U G M E m moon r := K 3.842 10 28 × J = K 1 2 m moon v 2 := Energies: D) v 1.022 km sec =

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P150_Exam3_F07sol - Physics 150 Exam#3 Solution 1 2 3 4 5 6...

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