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2–125.
Determine the magnitude of the projection of force
along the axis.uF =600 N
SOLUTION
Unit Vectors: The unit vectors and must be determined first.From Fig. a,
Thus, the force vectors is given by
Vector Dot Product: The magnitude of the projected component of along the
axis is
Ans.=246 N
=(–200)(sin30°) +400(cos 30°) +400(0)
Fu=F#uu=(–200i+400j+400k)#(sin30°i+cos 30°j)
uF
F=FuOA =600a– 1
3i–2
3j+2
3kb=5–200i+400j+400k6 N
F
uu=sin30°i+cos30°j
uOA =rOA
rOA
=(–2–0)i+(4 –0)j+(4 –0)k
3(–2–0)2+(4 –0)2+(4 –0)2
=-
1
3 i+2
3 j+2
3 k
uu
uOA
30
2 m
4 m
4 m
F 600 N
z
xu
O
y
A