250
252
253
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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={–175i+5600j–900k} lb #ft
SOLUTION
4–30.
The force
F={400i–100j–700k} lb
acts at the end of
the beam. Determine the moment of this force about
pointA.
1.5 ft
8 ft
z
A
O
B
F
y
© 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:
MP={–24i+24j+8k} kN #m
SOLUTION
4–31.
Determine the moment of the force F about point P.
Express the result as a Cartesian vector.
2 m
1 m
3 m
3 m
3 m
2 m
A
O
P
y
z
© 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:
M
A
={–110i+70j–20k} N #m
SOLUTION
*4–32.
The pipe assembly is subjected to the force of
F={600i+800j–500k} N
. Determine the moment of
this force about point A.
y
0.5 m
0.4 m
0.3 m
x
z
B
A
© 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={–110i–180j–420k} N #m
SOLUTION
4–33.
The pipe assembly is subjected to the force of
F={600i+800j–500k} N
. Determine the moment of
this force about point B.
y
0.5 m
0.4 m
0.3 m
x
z
B
A
© 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={574i+350j+1385k} N #m
SOLUTION
4–34.
Determine the moment of the force of F
=
600 N about
point A.
4 m
4 m
z
x
6 m
6 m
A
C
B
F
45fi
© 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:
F=585 N
SOLUTION
4–35.
Determine the smallest force F that must be applied along
the rope in order to cause the curved rod, which has a radius
of 4 m, to fail at the support A. This requires a moment of
M=1500 N #m
to be developed at A.
4 m
4 m
z
x
6 m
6 m
A
C
B
F
45fi
© 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={–82.9i+41.5j+232k} lb #ft
SOLUTION
4–38.
Determine the moment of the force F about the door hinge
at A. Express the result as a Cartesian vector.
1.5 ft
7 ft 4 ft
z
C
A
BD
F fi 80 lb
45fi
© 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={–82.9i–96.8j–52.8k} lb #ft
SOLUTION
4–39.
Determine the moment of the force F about the door hinge
at B. Express the result as a Cartesian vector.
1.5 ft
7 ft 4 ft
z
C
A
BD
F fi 80 lb
45fi
© 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:
MC={–35.4i–128j–222k} lb #ft
*4–40.
The curved rod has a radius of 5 ft. If a force of 60 lb acts at
its end as shown, determine the moment of this force about
point C.
SOLUTION
Moment of Force About Point C: Applying Eq. 4–7, we have
Ans.=–35.4i–128j–222klb #ft
=
ij k
0 4.330 –2.50
51.231 22.797 –21.346
MC=rCA *FAB
FAB
=551.231i+22.797j–21.346k6lb
FAB =60
¢
16–02i+17–5 sin 60°2j+10–5 cos 60°2k
216–022+17–5 sin 60°22+10–5 cos 60°22
≤
lb
=54.330j–2.50k6m
5ft
5ft
60°
z
y
60 lb
A
C