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Lecture 2 Force Systems Activity 1 Concept revision (Lecturer may explain and/or discuss with students)
- Step of finding a position vector
- Step of finding a force along a line.
- Step of find moments of a force in 3-D cases using a cross product method.
- The concepts of finding an equivalent force-couple system for a system of forces and couples. Activity 2 In class Questions
- Determine the position vector r directed from point A to point B and the length of cord AB.
- The door is held opened by two chains. If the tension in AB and CD is FA = 300 N and FC = 250 N, respectively, express each of these forces in Cartesian vector form.
- Determine the moment produced by the 4 - kN force about point A. At B āļ
13 āļ°^ -
āļĢāļĩ + āđāļāļĩ^ +2^ āļŦāļĩ m^ r^
6J t^ 4K āļŦāđ a B-_ -
tit āļģāļģāļģāļ āļđāđ
ljt (^2) āļāļĩ
2 f-
? 5 I^2 āļāđāđ to.7.sk
53 ra (^) B
= 2.3āļāļĩ -0.75 āļŦāļĩ āļ.^ āđ^ ra Bāđāļ° āļ°^42 m^ Fa^
āļ° (^300)
[t.tn ]^ āļ§āļš āļāļ°
Ξ 5 "" "āļģ^ "
"""^ "" "^ ""^ "^ ""
| = (^) """ ""āļŦāļķ "% āļāđ āļ āļŦāļĩ rap
āļ° āļ.^ - āļģāļāļąāđ? Fc^
āļ° (^25) o (^) [āļāļģ + (^) }% āļāļĩ - ° āļ§āļīāđ
]
āļ° 159.23 āļ^ +^ 183.12^ āļāļĩ -59.71 āļŦāļĩ y
ura.^ āļ°^ āļĢāļĩāđ + t -
(^) a à â | i. āđ. y <āļāļ^ nnng^ 9 Cos^95 āļ ro (^) āļāļšāļ°
-3āļāļĩ t 0.45āļ^ o . 9 Sz K
. | r F āļ (^). āļ° | ;āļāļ°āļāđāļ§āļ \5gtT.LT?T.s?s.nk.Iineso-2Sz
0 252 -^252 = 510052 t
N.MX (^0 0) 652k
Lecture 2 Force Systems Activity 1 Concept revision (Lecturer may explain and/or discuss with students)
- Step of finding a position vector
- Step of finding a force along a line.
- Step of find moments of a force in 3-D cases using a cross product method.
- The concepts of finding an equivalent force-couple system for a system of forces and couples. Activity 2 In class Questions
- Determine the position vector r directed from point A to point B and the length of cord AB.
- The door is held opened by two chains. If the tension in AB and CD is FA = 300 N and FC = 250 N, respectively, express each of these forces in Cartesian vector form.
- Determine the moment produced by the 4 - kN force about point A.
5A BifoB- 50A
Fo B =^6 āļ^ ÂĨ^ 4K
^ ro A^ =^3 āļē^ +2^ K ^ āļf FAB āļ°^ -3āļē^ +^ āđāļ^ +^
I
/ 5A^ Bl^ =^3 āļģāļ§āđ^ +^ āđ āđāļāļēāļ° 7 i
/ i
- 5
i.
÷:i
" āļ§āđāļē (^).
. ..^ ..si
XAB =^ 2.
^ _ 0.75āļŦāđ -2.
300
[
M =^ rxf āļ° (^) âĒ ^ -0.45āļ^ ) à (^) ( 4 cos 45 āļĄāļļāđ + 45in^45 ° āļ )
i.
10 KN Etilb ^ āļ§āļē rpcsosāđāļ§āđ^ cossJj â <-^8 rsi āļģ Ξ âĒ âĒ^ _^ _^ Âą^ _ _ _ nocos āđāļ§āđ 1 (^10) cos āđāļāļīāļ 535 ° i a) (^) 5A (^) B =^ roB- to (^) A o 1 1 ^
= [8C 0555 āļģ 8 Cos 3 Sj J - C 10 cos 6 āļ§āļī cos 35 ° itbcos 60 āļģ 0555 ° āļ 410 cos 3 āļ§āđāļāļļāđ )
^ (^) n āļ° (^) -9.59 I (^) -9,42 (^) j -8.66k lrABI āļ°^16 7A (^) B^ =^
_āļĢāļļāļĨāļđāđ
FAB =^1 °"^ AB āļ°^ -^9 āļŠāļļ
āļ§āļļ K N b) (^) 5A (^) āļ^ =^ tocos^ āļāļģ^0535 āļģ^ -10cos^6 āļ§āđ^ csss^ āļĢāļīāļ^ -10C^0530 ° I āļ° -4.1 I.^ 2.9 āļ - 8.66 Ii (^15) As | =^10
Fa; -5.66^ I^ -^ 5.65āļ^ -^ 5.04^ t
IF | =^ 9.^95 AB 1 āļ Uo (^) A āļ° 5Ao /^1 āļāļē l^ =^ -^ o.41i-o.zqj-o.gr^ t F A 0 = FAB ° UoA āļ° (^) 8.29 KN
C) thj -4.1^ I^ -2.9āļ^ -^ 8.āļ^61 Ã
^ Mo =^ 5Ao X^ FfB =^ -34.31^ I^ +28.35āļ^ t^ 6.7^51 t^ KN.^ M
āļ F. =^ 4J^ KN
f (^) fhsss.in.^.^ Fsi^ -6 āļŦāļĩ k^ N âĶ (^) 5C 0545 ° l (^) F }^ āļ°^ 3.5T - 3.5^51 āļāļļāđ^ k N
EFÃ āļ° 40 N^ āļāļļāđ =^ 4J EFy āļ°^ -^ 100N^ āļāļđāđ āļ°^ -6^ I āļāļĩāļ° 3.si^ - 3.āļĐāļļāđ tf (^) MRAāļ°^ - zoo - âĪ(^ āđ)^ -^ C^30 ](3) (^) h =^ 1.2Tt^ 1.2āļāđ^ t 4 I āļ° ' 200 - lronq (^0) Pz āļ° 2.9āļģ^ t 1.2J t (^) 2.8 E
āļĨ-470 M āļ° 53 āļ°^24 I^ -1.2J^ t^ 2.8^ E E (^) Mpo āļ° ro AXF āđ tbBXFzt roc XF^ } = (^) (4.8T -4.8āļ^ -0k (^) Jt C- 7.2Tt (^) 19.4 āļāļĩ tok ) t (^) C 9.8 āļ^ t (^) 9.8āļāļĩ -9.2 t^ ) āļ° (^) 7.4it (^) 23.4 āļāļĩ -4.2 āļŦāļĩ * KN. m
