Hibbeler Statics 14e: Problem 2-132
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Problem 2-132
Determine the magnitude of the projected component of F along AC. Express this component as
a Cartesian vector.
Solution
Write the position vectors to the points A and C.
rA = h0, 0, 0i ft
rC = h7, 6, −4i ft
The unit vector going from A to C is then
ûAC =
rC − rA
h7, 6, −4i
=p
.
|rC − rA |
(7)2 + (6)2 + (−4)2
Take the dot product of F with ûAC to find the component of the force along AC.
h7, 6, −4i
260
Fk = F · ûAC = h30, −45, 50i · p
lb = − √
lb
101
(7)2 + (6)2 + (−4)2
Therefore, the magnitude of the component along AC is
260
|Fk | = √
lb ≈ 25.9 lb.
101
The component along AC as a vector is
260
h7, 6, −4i
p
Fk = Fk ûAC = − √
lb
2
101 (7) + (6)2 + (−4)2
1820 1560 1040
= −
,−
,
lb
101
101 101
≈ h−18.0, −15.4, 10.3i lb.
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