P150_Exam3_F06sol - Physics 150 1. 2. 3. 4. 5. 6. 7. 8. 9....

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Answer = A (One point given for answer E.) 12. Answer = B 13. Answer = C 14. Answer = D 15. Answer = C 16. Answer = A 17. Answer = D 18. Answer = C 19. Answer = A 20. Answer = E Physics 150 Exam #3 Solution 12/4/06 1. Answer = D 2. Answer = D 3. Answer = E 4. Answer = C 5. Answer = E 6. Answer = A 7. Answer = D 8. Answer = D 9. Answer = D 10. Answer = B 11.
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θ 2 65.818 deg = θ 2 atan T 2y T 2x := Angle: T 2 966.845N = T 2 T 2x 2 T 2y 2 + := Magnitude of T 2 : T 2x 396.062N = T 2x T 1 cos θ 1 () := Then to the horizontal forces equation: T 2y 882 N = T 2y m scaffold g m man g + T 1 sin θ 1 := Returning to the vertical forces equation: T 1 792.125N = 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 2 cos θ 2 T 1 cos θ 1 0N := Total horizontal forces; T 1 sin θ 1 T 2 sin θ 2 + m scaffold g m man g := Total vertical forces: θ 1 60 deg := m man 100 kg := x man 4m := m scaffold 60 kg := L1 0 m :=
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This note was uploaded on 04/10/2008 for the course PHYSICS 150 taught by Professor Naik during the Winter '08 term at University of Michigan.

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P150_Exam3_F06sol - Physics 150 1. 2. 3. 4. 5. 6. 7. 8. 9....

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