{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

P150_Exam3_F06sol - Physics 150 1 2 3 4 5 6 7 8 9 10 11 12...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 0 N := Total vertical forces: θ 1 60 deg := m man 100 kg := x man 4 m := m scaffold 60 kg := L 10 m :=
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}