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Unformatted text preview: CIV100F
Section B/E Solutions
Fall 2011 Assignment 3
Due on Monday, October 3rd 2011 1 Force Q (Q=90kN) acts along a line which passes through H(2,2,5) and L(6,5,2). Force is in
direction H → L , determine vector Q as a Cartesian vector.
Solution:
H = (2,2,5) L = (6,5,−2) rHL = (4)i + (3) j − (7)k
rHL = 8.6023
Q=Q rHL
= (41.8i + 31.4 j − 73.2)kN
rHL 2 Consider the following points in space: A(1, 5, 3), B(0, 6, 2), C(7, 3, 3)
a) Find the angle formed by vectors AB and AC.
Solution:
r AB = B − A = (1,1,5) = 1i + 1 j + 5k
r AC = C − A = (8,−8,6) = 8i − 8 j + 6k
r AB = 12 + 12 + 52 = 27
r AC = 82 + (−8)2 + 6 2 = 164
cos θ = r AB ⋅ r AC
(1 × 8) − (1 × 8) + (5 × 6)
=
=0.45083
r AB r AC
27 × 164 θ = 63.20º
b) Find the magnitude of the component of AB in the direction of AC
Solution:
ABAC = rAB cos θ = 27 cos 63.20° = 2.34 Draw a FBD if necessary; Pay attention to significant figures; A ruler is always required. CIV100F
Section B/E Solutions
Fall 2011 Assignment 3
Due on Monday, October 3rd 2011 3 The 7m tree trunk is held stable by 2 cables BC
and BA. Knowing that TBA=555N, TBC=660N.
Express the two tensions as Cartesian vectors and
determine the angle formed by BA and BC. Solution:
First identify the coordinates of all points:
A (0.75, 0, 6)
B (0,7,0)
C (4.25, 0, 1)
Represent the forces in the cables as vectors:
rBA = A − B = (−0.75,−7,6)m = (−0.75i − 7 j + 6k )m
rBC = C − B = (4.25,−7,1)m = (4.25i − 7 j + 1k )m
rBA = (−0.75) 2 + (−7)2 + 6 2 = 9.25m
rBC = 4.252 + (−7) 2 + 12 = 8.25m u BA = rBA
= −0.081081i − 0.75675 j + 0.64865k
rBA u BC = rBC
= 0.51515i − 0.84848 j + 0.12121k
rBC T BA = T BA u BA = (−45i − 420 j + 360k )N
T BC = T BC u BC = (340i − 560 j + 80k )N
cos θ = rBA ⋅ rBC
(−0.75i − 7 j + 6k ) ⋅ (4.25i − 7 j + 1k )
=
= 0.67895
(9.25)(8.25)
rBA rBC θ = cos −1 (0.67895) = 47.24° Draw a FBD if necessary; Pay attention to significant figures; A ruler is always required. ...
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