Written Homework 6
Problem 1
∑M
C
= 0 ⇒ −6 ( P1 ) + 12 (T ) − 16 ( P2 ) + 24 (T sin 30˚) = 0
1
2
P1 + P2 = 4.20 kip
4
3
T=
∑F
= 0 ⇒ C x − T cos 30˚= 0
Cx =
3
T = 3.64 kip
2
∑F
= 0 ⇒ T + T sin 30˚+C y − P1 − P2 = 0
Cy =
5
P1 = 4.50 kip
8
x
y
Problem 2
û BC =
−5iˆ − 9 ĵ
2iˆ − 9 ĵ + 5 k̂
4 iˆ − 9 ĵ − 4 k̂
. û BD =
. û BE =
.
106
110
113
T = 6 kN. h = 90 m.
∑F
x
= 0 ⇒ FAx + (TBCx + TBDx + TBEx ) = 0
2
4 ⎞
⎛ −5
FAx = −T ⎜
+
+
⎟ = −0.488 kN
⎝ 106
110
113 ⎠
∑F
y
(
)
= 0 ⇒ FAy + TBCy + TBDy + TBEy = 0
9
9 ⎞
⎛ −9
FAy = −T ⎜
+
+
⎟ = 15.5 kN
⎝ 106
110
113 ⎠
∑ F = 0 ⇒ F + (T
z
Az
BCz
+ TBDz + TBEz ) = 0
−4 ⎞
⎛ 5
FAz = −T ⎜
+
⎟ = −0.603 kN
⎝ 110
113 ⎠
Vector method for moments:
!
!
!
!
!
! ∑ M A = 0 ⇒ M A + ∑ ⎡⎣ rAB × (Tû ) ⎤⎦ = 0 ⇒ M A = − rAB × ⎡⎣T ( û BC + û BD + û BE ) ⎤⎦
!
⎡
⎛
ˆ
⎢⎣
⎝
106
ˆ
⎞⎤
ˆ
! r!AB = hĵ. M A = −Th ⎡ ĵ × ( û BC + û BD + û BE ) ⎤ = −Th ⎢ ĵ × ⎜ −5i − 9 ĵ + 2i − 9 ĵ + 5 k̂ + 4 i − 9 ĵ − 4 k̂ ⎟ ⎥
⎣
⎦
110
113
⎠ ⎥⎦
!
4 ⎞ˆ ⎛ 5
2
4 ⎞ ⎤
⎡⎛ 5
ˆ
M A = −Th ⎢⎜
−
i +⎜
−
−
⎟
⎟⎠ k̂ ⎥ = −54.2i + 43.9 k̂ kNm
⎝
⎠
⎝
110
113
106
110
113
⎣
⎦
(
)
Scalar method for moments:
Sum moments around each axis.
⎛ 5
⎞
⎛ 4
⎞
! ∑ M x = 0 ⇒ M Ax + h ⎜
T ⎟ − h⎜
T⎟ = 0
⎝ 110 ⎠
⎝ 113 ⎠
5 ⎞
⎛ 4
−
! M Ax = Th ⎜
⎟ = −54.2 kNm
⎝ 113
110 ⎠
! ∑ M y = 0 ⇒ M Ay = 0
⎛ 2
⎞
⎛ 5
⎞
⎛ 4
⎞
! ∑ M z = 0 ⇒ M Az − h ⎜
T ⎟ + h⎜
T ⎟ − h⎜
T⎟ = 0
⎝ 110 ⎠
⎝ 106 ⎠
⎝ 113 ⎠
5
4 ⎞
⎛ 2
−
+
! M Az = Th ⎜
⎟ = 43.9 kNm
⎝ 110
106
113 ⎠
Problem 3
∑M
x
=0⇒−
3
7
TA ) − 1(TC ) + ( mg ) = 0
(
2
8
To get whole numbers, multiply by 8:
12TA + 8TC = 7mg (1)
∑M
z
=0⇒
9
5
5
TB ) + (TC ) − ( mg ) = 0
(
10
2
4
Multiply by 20: ! 18TB + 50TC = 25mg
∑F
y
(2)
= 0 ⇒ TA + TB + TC − mg = 0
TA + TB + TC = mg
(3)
Solve (1), (2) and (3) simultaneously to get ! TA = 1.09 kN. TB = 0.975 kN.
TC = 1.37 kN.
Problem 4
F1 = 3.6 kN.
∑F
x
F2 = 2.4 kN.
= 0 ⇒ Ox + F2 cos 50˚= 0
Ox = −F2 cos 50˚= −1.54 kN
∑F
y
= 0 ⇒ Oy − F2 sin 50˚+F1 cos 20˚= 0
Oy = F2 sin 50˚−F1 cos 20˚= −1.28 kN
∑F = 0 ⇒O
z
z
− F1 sin 30˚= 0
Oz = F1 sin 30˚= 1.80 kN
Scalar method for moments:
(∑ M )
!
O x
= 0 ⇒ M Ox + ( 0.15 m ) ( F1 cos 30˚) − ( 0.35 m ) ( F2 sin 50˚) = 0
! M Ox = − ( 0.15 m ) ( F1 cos 30˚) + ( 0.35 m ) ( F2 sin 50˚) = 0.176 kNm
(∑ M )
!
O y
= 0 ⇒ M Oy − ( 0.35 m ) ( F2 cos 50˚) = 0
! M Oy = ( 0.35 m ) ( F2 cos 50˚) = 0.540 kNm
(∑ M )
!
O z
= 0 ⇒ M Oz − ( 0.2 m ) ( F2 sin 50˚) = 0
! M Oz = ( 0.2 m ) ( F2 sin 50˚) = 0.368 kNm
Vector method for moments:
!
!
!
!
! ∑ M O = 0 ⇒ M O + −0.15 m k̂ × F1 + −0.35 m k̂ + 0.2 m iˆ × F2 = 0
(
!
(
)
)
(
(
)
) (
)
(
)
! M O = 0.15 m k̂ × ⎡ F1 cos 30˚ ĵ − sin 30˚k̂ ⎤ + 0.35 m k̂ − 0.2 m iˆ × ⎡ F2 cos 50˚iˆ − sin 50˚ ĵ ⎤ = 0
⎣
⎦
⎣
⎦
!
! M O = 0.176iˆ + 0.540 ĵ + 0.368 k̂ kNm
(
)