© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
MA={–16.0i–32.1k} N #m
4–47.
SOLUTION
Ans.
Also,
Ans.MA=rC*F=
ijk
–0.400.2
–53.5 80.2 26.7
= – 16.0 i–32.1 kN#m
MA=rB*F=3ijk
0–0.6 0
–53.5 80.2 26.7 3=5–16.0 i–32.1 k6N#m
F=5–53.5 i+80.2 j+26.7 k6N
A force
F
having a magnitude of acts along the
diagonal of the parallelepiped. Determine the moment of F
about point A, using and MA=rC:F.MA=rB:F
F
=
100 N
F
z
y
x
C
200 mm
400 mm
r
C
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Ans:
MB={1.00i+0.750j–1.56k} kN #m
4–49.
F
orce
F
acts perpendicular to the inclined plane. Determine
the moment produced by Fabout point B. Express the
result as a Cartesian vector.
z
xy
3m
3m
4m
A
BC
F400 N
SOLUTION
is equal to the unit vector of the cross product, ,Fig. a. Here
Thus,
Then,
And finally
Vector Cross Product: The moment of Fabout point Bis
Ans.=[1.00i + 0.750j – 1.56k] kN # m
MB=rBC *F=3ijk
–34 0
249.88 187.41 249.88 3
=[249.88i+187.41j+249.88k]N
F=FuF=400(0.6247i+0.4685j+0.6247k)
uF=b
b=12i+9j+12k
2122+92+122
=0.6247i+0.4685j+0.6247k
b=rCA *rCB =3ijk
04–3
–34 03=[12i+9j+12k]m
2
rBC =(0 –3)i+(4 –0)j+(0 –0)k=[–3k+4j]m
rAC =(0 –0)i+(4 –0)j+(0 –3)k=[4j–3k]m
b=rAC *rBC
279
280
