Engineering Mechanics - Statics Episode 1 Part 8 docx

Engineering Mechanics - Statics Chapter 4
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
acos
M
M
⎛
⎜
⎝
⎞
⎟
⎠
=
α
β
γ
⎛
⎜
⎜
⎜

⎝
⎞
⎟
⎟
⎟
⎠
155.496
114.504
90
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
deg=
Problem 4-96
Express the moment of the couple acting on the pipe in Cartesian vector form. What is the magnitude
of the couple moment?
Given:
F 125 N=
a 150 mm=
b 150 mm=
c 200 mm=
d 600 mm=
Solution:
M
c

ab+
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
0
F
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
×= M
37.5
25−
0
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
Nm⋅= M 45.1 N m⋅=
Problem 4-97
If the couple moment acting on the pipe has a magnitude M, determine the magnitude F of the
forces applied to the wrenches.
281
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
Given:
M 300 N m⋅=
a 150 mm=
b 150 mm=
c 200 mm=
d 600 mm=
Solution:
Initial guess:
F 1N=
Given
c
ab+
0
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
0
0
F
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
× M= F Find F()= F 832.1 N=
Problem 4-98
Replace the force at A by an equivalent force and couple moment at point O.
Given:
F 375 N=
a 2m=
b 4m=
c 2m=
d 1m=
θ
30 deg=
282
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4

F
v
F
sin
θ
()
cos
θ
()
−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
v
187.5
324.76−
0
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
N=
M
O
a−
b
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
v
×= M
O
0
0
100.481−
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
Nm⋅=
Problem 4-99
Replace the force at A by an equivalent force and couple moment at point P
.
Given:
F 375 N=
a 2m=
b 4m=
c 2m=
d 1m=
θ
30 deg=
Solution:
F
v
F
sin
θ
()
cos
θ
()
−
0
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
= F
v
187.5
324.76−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
M
P
a− c−
bd−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

F
v
×= M
P
0
0
736.538
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
Nm⋅=
Problem 4-100
Replace the force system by an equivalent resultant force and couple moment at point O.
283
Solution:
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
F
1
60 lb= a 2ft=
F
2
85 lb= b 3ft=

F
3
25 lb= c 6ft=
θ
45 deg= d 4ft=
e 3=
f 4=
Solution:
F
0
F
1
−
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
F
2
e
2
f
2

+
e−
f−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+ F
3
cos
θ
()
sin
θ
()
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+=

F
33.322−
110.322−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb= F 115.245 lb=
M
O
c−
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
F
1
−

0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×
0
a
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
e
2
f
2
+

e−
f−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+
d
b
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
F
3
cos
θ
()
sin
θ
()
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+=
M
O
0

0
480
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb ft⋅= M
O
480lb ft⋅=
Problem 4-101
Replace the force system by an equivalent resultant force and couple moment at point P.
284
Given:
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
Given:
F
1
60 lb= a 2ft=
F
2
85 lb= b 3ft=
F
3

25 lb= c 6ft=
θ
45 deg= d 4ft=
e 3= f 4=
Solution:
F
0
F
1
−
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
F
2
e
2
f
2
+
e−
f−

0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+ F
3
cos
θ
()
sin
θ
()
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+=
F
33.322−
110.322−

0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb= F 115.245 lb=
M
P
c− d−
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
F
1
−
0
⎛
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×
d−
a
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
e
2
f
2
+
e−
f−
0

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+
0
b
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
3

cos
θ
()
sin
θ
()
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+=
M
P
0
0
921
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb ft⋅= M
P
921lb ft⋅=
Problem 4-102
Replace the force system by an equivalent
force and couple moment at point O.
Units Used:
kip 10
3
lb=
Given:
F
1
430 lb= F
2
260 lb=
a 2ft= e 5ft=
285
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
b 8ft= f 12=

c 3ft= g 5=
da=
θ
60 deg=
Solution:
F
R
F
1
sin
θ
()
−
cos
θ
()
−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
g
2

f
2
+
g
f
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+= F
R
272−
25
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb= F
R
274lb=

M
O
d−
b
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
1
sin
θ
()
−
cos
θ
()
−
0
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×
e
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
g
2
f
2
+
g

f
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+= M
O
0
0
4.609
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
kip ft⋅=
Problem 4-103
Replace the force system by an equivalent force and couple moment at point P.
Units Used:
kip 10
3
lb=
Given:
F
1
430 lb= F
2
260 lb=
a 2ft= e 5ft=
b 8ft= f 12=
c 3ft= g 5=
da=
θ
60 deg=
Solution:
F
R
F
1
sin
θ
()
−
cos

θ
()
−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
g
2
f
2
+
g
f
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

+= F
R
272−
25
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb= F
R
274lb=
286
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
M
P
0
bc+
0
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
F
1
sin
θ
()
−
cos
θ
()
−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥

⎦
×
de+
c
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
g
2
f
2
+
g
f
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+= M
P
0
0
5.476
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kip ft⋅=
Problem 4-104
Replace the loading system acting on the post by an equivalent resultant force and couple
moment at point O.
Given:
F
1
30 lb= a 1ft= d 3=

F
2
40 lb= b 3ft= e 4=
F
3
60 lb= c 2ft=
Solution:
F
R
F
1
0
1−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
1
0
0
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
+
F
3
d
2
e
2
+
d−
e−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+=
F
R
4
78−
0

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb= F
R
78.1 lb=
M
O
0
ab+ c+
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
1
0
1−
0
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×
0
c
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
1

0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+
0
bc+
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

F
3
d
2
e
2
+
d−
e−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+=
M
O
0

0
100
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb ft⋅=
Problem 4-105
Replace the loading system acting on the post by an equivalent resultant force and couple
moment at point P.
287
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
Given:
F
1
30 lb=
F
2
40 lb=
F
3
60 lb=
a 1ft=

b 3ft=
c 2ft=
d 3=
e 4=
Solution:
F
R
F
1
0
1−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
1
0
0
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
+
F
3
d
2
e
2
+
d−
e−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+=
F
R
4
78−
0
⎛
⎜

⎜
⎝
⎞
⎟
⎟
⎠
lb= F
R
78.1 lb=
M
P
0
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
ft
⎡
⎢
⎢
⎣
⎤
⎥
⎥

⎦
F
1
0
1−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×
0
a− b−
0
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
F
2
1
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+
0
a−
0
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
3
d
2
e
2
+
d−
e−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤

⎥
⎥
⎦
×+=
M
P
0
0
124
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb ft⋅=
Problem 4-106
Replace the force and couple system by an equivalent force and couple moment at point O.
Units Used:
kN 10
3
N=
288
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
Given:

M 8kNm=
θ
60 deg=
a 3m= f 12=
b 3m= g 5=
c 4m= F
1
6kN=
d 4m= F
2
4kN=
e 5m=
Solution:
F
R
F
1
f
2
g
2
+
g
f
0
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
F
2
cos
θ
()
−
sin
θ
()
−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+=
F
R
0.308
2.074
0
⎛
⎜

⎜
⎝
⎞
⎟
⎟
⎠
kN= F
R
2.097 kN=
M
O
0
0
M
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
c−
e−
0
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
F
1
f
2
g
2
+
g
f
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+

0
d−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
cos
θ
()
−
sin
θ
()
−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+=
M
O
0
0
10.615−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN m⋅=
Problem 4-107
Replace the force and couple system by an equivalent force and couple moment at point P.
Units Used:
kN 10
3
N=
289

© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
Given:
M 8kNm⋅=
θ
60 deg=
a 3m= f 12=
b 3m= g 5=
c 4m= F
1
6kN=
d 4m= F
2
4kN=
e 5m=
Solution:
F
R
F
1
f
2
g
2
+
g
f
0

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
cos
θ
()
−
sin
θ
()
−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+=
F
R

0.308
2.074
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN= F
R
2.097 kN=
M
P
0
0
M
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
c− b−
e−
0

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
1
f
2
g
2
+
g
f
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣

⎤
⎥
⎥
⎦
×+
b−
d−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
cos
θ
()
−
sin
θ
()
−
0
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+=
M
P
0
0
16.838−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN m⋅=
Problem 4-108
Replace the force system by a single force resultant and specify its point of application, measured

along the x axis from point O.
Given:
F
1
125 lb=
F
2
350 lb=
F
3
850 lb=
290
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
a 2ft=
b 6ft=
c 3ft=
d 4ft=
Solution:
F
Ry
F
3
F
2
− F
1
−= F

Ry
375lb=
F
Ry
xF
3
bc+()F
2
b()− F
1
a()+=
x
F
3
bc+()F
2
b()− F
1
a+
F
Ry
= x 15.5 ft=
Problem 4-109
Replace the force system by a single force resultant and specify its point of application, measured
along the x axis from point P.
Given:
F
1
125 lb= a 2ft=
F

2
350 lb= b 6ft=
F
3
850 lb= c 3ft=
d 4ft=
Solution:
F
Ry
F
3
F
2
− F
1
−= F
Ry
375lb=
F
Ry
xF
2
dc+()F
3
d()− F
1
ab+ c+ d+()+=
x
F
2

dF
2
cF
3
d−+ F
1
a+ F
1
b+ F
1
c+ F
1
d+
F
Ry
=
x 2.47 ft=
(to the right of P)
291
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
The forces and couple moments which are exerted on the toe and heel plates of a snow ski are
F
t
,
M
t
, and

F
h
,
M
h
, respectively. Replace this system by an equivalent force and couple moment
acting at point O. Express the results in Cartesian vector form.
Given:
a 120 mm=
b 800 mm=
Solution:
F
t
50−
80
158−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N= F
h
20−
60
250−
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
N= M
t
6−
4
2
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
Nm⋅= M
h
20−
8
3
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
Nm⋅=
F
R
F
t
F
h
+=
F
R
70−
140
408−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
r
0Ft
a
0
0

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
M
RP
r
0Ft
F
t
×
()
M
t
+ M
h
+=
M
RP
26−
31
14.6
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
Nm⋅=
Problem 4-111
The forces and couple moments which are exerted on the toe and heel plates of a snow ski are
F
t
,
M
t
, and
F
h
,
M
h
, respectively. Replace this system by an equivalent force and couple moment
292
Problem 4-110
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
acting at point P. Express the results in Cartesian vector form.
Given:
a 120 mm=
b 800 mm=

F
t
50−
80
158−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
M
t
6−
4
2
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
Nm⋅=
F
h

20−
60
250−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
M
h
20−
8
3
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
Nm⋅=
Solution:
F
R
F

t
F
h
+= F
R
70−
140
408−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
M
P
M
t
M
h
+
b
0
0
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
F
h
×+
ab+
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
t
×+= M
P
26−
357.4
126.6
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
Nm⋅=
Problem 4-112
Replace the three forces acting on the shaft by a single resultant force. Specify where the force
acts, measured from end B.
293
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
Given:
F
1
500 lb=
F
2
200 lb=
F
3
260 lb=
a 5ft= e 3=
b 3ft= f 4=
c 2ft= g 12=
d 4ft= h 5=
Solution:
F
R

F
1
e
2
f
2
+
f−
e−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
0
1−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
+
F
3
g
2
h
2
+
h
g−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+= F
R
300−
740−
0
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
lb= F
R
798lb=
Initial guess:
x 1ft=
Given
a
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
1
e
2
f
2
+
f−
e−
0

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×
ab+
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2

0
1−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+
ab+ c+
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
F
3
g
2
h
2
+
h
g−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+
x−
0

0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
R
×=
x Find x()= x 7.432− ft=
Problem 4-113
Replace the three forces acting on the shaft by a single resultant force. Specify where the force
acts, measured from end B.
Given:
F
1
500 lb=
F
2
200 lb=
F
3
260 lb=
a 5ft= e 3=
294
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
b 3ft= f 4=
c 2ft= g 12=
d 4ft= h 5=
Solution:
F
R
F
1
e
2
f
2
+
f−
e−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
0
1−

0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+
F
3
g
2
h
2
+
h
g−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+= F
R

300−
740−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb= F
R
798lb=
Initial guess:
x 1ft=
Given
b− c− d−
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
1

e
2
f
2
+
f−
e−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×
c− d−
0
0
⎛
⎜

⎜
⎝
⎞
⎟
⎟
⎠
F
2
0
1−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+
d−
0

0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
3
g
2
h
2
+
h
g−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢

⎣
⎤
⎥
⎥
⎦
×+
x−
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
×=
x Find x()= x 6.568 ft=
measured to the left of B
Problem 4-114
Replace the loading on the frame by a single resultant force. Specify where its line of action
intersects member AB, measured from A.
Given:
F
1
300 lb= M 600 lb ft⋅=
a 3ft=
F

2
200 lb=
b 4ft=
F
3
400 lb=
c 2ft=
F
4
200 lb=
d 7ft=
Solution:
F
Rx
F
4
−= F
Rx
200− lb=
F
Ry
F
1
− F
2
− F
3
−= F
Ry
900− lb=

295
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
FF
Rx
2
F
Ry
2
+= F 922lb=
θ
atan
F
Ry
F
Rx
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
77.5 deg=
F
Ry
xF

2
− aF
3
ab+()− F
4
c− M+=
x
F
2
a() F
3
ab+()+ F
4
c+ M−
F
Ry
−= x 3.556 ft=
Problem 4-115
Replace the loading on the frame by a single resultant force. Specify where the force acts,
measured from end A.
Given:
F
1
450 N= a 2m=
F
2
300 N= b 4m=
F
3
700 N= c 3m=

θ
60 deg= M 1500 N m⋅=
φ
30 deg=
Solution:
F
Rx
F
1
cos
θ
()
F
3
sin
φ
()
−= F
Rx
125− N=
F
Ry
F
1
− sin
θ
()
F
3
cos

φ
()
− F
2
−= F
Ry
1.296− 10
3
× N=
FF
Rx
2
F
Ry
2
+= F 1.302 10
3
× N=
θ
1
atan
F
Ry
F
Rx
⎛
⎜
⎝
⎞
⎟

⎠
=
θ
1
84.5 deg=
F
Ry
x() F
1
− sin
θ
()
aF
2
ab+()− F
3
cos
φ
()
ab+ c+()− M−=
x
F
1
− sin
θ
()
aF
2
ab+()− F
3

cos
φ
()
ab+ c+()− M−
F
Ry
= x 7.36 m=
Problem 4-116
Replace the loading on the frame by a single resultant force. Specify where the force acts,
measured from end B.
296
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
Given:
F
1
450 N= a 2m=
F
2
300 N= b 4m=
F
3
700 N= c 3m=
θ
60 deg= M 1500 N m⋅=
φ
30deg=
Solution:

F
Rx
F
1
cos
θ
()
F
3
sin
φ
()
−= F
Rx
125− N=
F
Ry
F
1
− sin
θ
()
F
3
cos
φ
()
− F
2
−= F

Ry
1.296− 10
3
× N=
FF
Rx
2
F
Ry
2
+= F 1.302 10
3
× N=
θ
1
atan
F
Ry
F
Rx
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
1
84.5 deg=

F
Ry
xF
1
sin
θ
()
bF
3
cos
φ
()
c− M−=
x
F
1
sin
θ
()
bF
3
cos
φ
()
c− M−
F
Ry
= x 1.36 m=
(to the right)
Problem 4-117

Replace the loading system acting on
the beam by an equivalent resultant
force and couple moment at point O.
Given:
F
1
200 N=
F
2
450 N=
M 200 N m⋅=
a 0.2 m=
b 1.5 m=
c 2m=
d 1.5 m=
297
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
θ
30 deg=
Solution:
F
R
F
1
0
1
0

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
sin
θ
()
−
cos
θ
()
−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+=
F
R

225−
190−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N= F
R
294 N=
M
O
bc+
a
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
1
0

1
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×
b
a
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

F
2
sin
θ
()
−
cos
θ
()
−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+ M
0
0

1−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+=
M
O
0
0
39.6−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
Nm⋅=
Problem 4-118
Determine the magnitude and direction
θ
of force
F
and its placement d on the

beam so that the loading system is
equivalent to a resultant force
F
R
acting
vertically downward at point A and a
clockwise couple moment M.
Units Used:
kN 10
3
N=
Given:
F
1
5kN= a 3m=
F
2
3kN= b 4m=
F
R
12 kN= c 6m=
M 50 kN m⋅= e 7= f 24=
Solution:
Initial guesses:
F 1kN=
θ
30 deg= d 2m=
298
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
Given
e−
e
2
f
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
1
F cos
θ
()
+ 0=
f−
e
2
f
2
+
⎛
⎜
⎝

⎞
⎟
⎠
F
1
F sin
θ
()
− F
2
− F
R
−=
f
e
2
f
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
1
aFsin
θ
()

ab+ d−()+ F
2
ab+()+ M=
F
θ
d
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
Find F
θ
, d,
()
= F 4.427 kN=
θ
71.565 deg= d 3.524 m=
Problem 4-119
Determine the magnitude and direction
θ

of force
F
and its placement d on the beam so that the
loading system is equivalent to a resultant force
F

R
acting vertically downward at point A and a
clockwise couple moment M.
Units Used:
kN 10
3
N=
Given:
F
1
5kN= a 3m=
F
2
3kN= b 4m=
F
R
10 kN= c 6m=
M 45 kN m⋅= e 7=
f 24=
Solution:
Initial guesses:
F 1kN=
θ
30 deg= d 1m=
Given
e−
e
2
f
2

+
⎛
⎜
⎝
⎞
⎟
⎠
F
1
F cos
θ
()
+ 0=
f−
e
2
f
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
1
F sin
θ
()

− F
2
− F
R
−=
299
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
f
e
2
f
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
1
aFsin
θ
()
ab+ d−()+ F
2
ab+()+ M=

F
θ
d
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
Find F
θ
, d,
()
= F 2.608 kN=
θ
57.529 deg= d 2.636 m=
Problem 4-120
Replace the loading on the frame by a single resultant force. Specify where its line of action
intersects member AB, measured from A.
Given:
F
1
500 N= a 3m=
b 2m=
F
2
300 N=
c 1m=

F
3
250 N=
d 2m=
M 400 N m⋅=
e 3m=
θ
60 deg= f 3=
g 4=
Solution:
F
Rx
F
3
−
g
g
2
f
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
1
cos

θ
()()
−= F
Rx
450− N=
F
Ry
F
2
− F
3
f
f
2
g
2
+
⎛
⎜
⎝
⎞
⎟
⎠
− F
1
sin
θ
()
−= F
Ry

883.0127− N=
F
R
F
Rx
2
F
Ry
2
+= F
R
991.066 N=
θ
1
atan
F
Ry
F
Rx
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
1
62.996 deg=
300

© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
F
Rx
− y() MF
1
cos
θ
()
a+ F
3
g
g
2
f
2
+
ba+()+ F
2
d()− F
3
f
g
2
f
2
+
⎛

⎜
⎝
⎞
⎟
⎠
de+()−=
y
MF
1
cos
θ
()
a+ F
3
g
g
2
f
2
+
ba+()+ F
2
d()− F
3
f
g
2
f
2
+

⎛
⎜
⎝
⎞
⎟
⎠
de+()−
F
Rx
−
=
y 1.78 m=
Problem 4-121
Replace the loading on the frame by a single resultant force. Specify where its line of action
intersects member CD, measured from end C.
Given:
F
1
500 N= a 3m=
b 2m=
F
2
300 N=
c 1m=
F
3
250 N=
d 2m=
M 400 N m⋅=
e 3m=

θ
60 deg= f 3=
g 4=
Solution:
F
Rx
F
3
−
g
g
2
f
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
1
cos
θ
()()
−= F
Rx
450− N=
F

Ry
F
2
− F
3
f
f
2
g
2
+
⎛
⎜
⎝
⎞
⎟
⎠
− F
1
sin
θ
()
−= F
Ry
883.0127− N=
F
R
F
Rx
2

F
Ry
2
+= F
R
991.066 N=
301
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
θ
1
atan
F
Ry
F
Rx
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
1
62.996 deg=
F
Ry

x() MF
2
dc+()− F
3
f
g
2
f
2
+
⎛
⎜
⎝
⎞
⎟
⎠
cd+ e+()− F
1
b( ) cos
θ
()
− F
1
csin
θ
()
−=
x
MF
2

dc+()− F
3
f
g
2
f
2
+
⎛
⎜
⎝
⎞
⎟
⎠
cd+ e+()− F
1
b( ) cos
θ
()
− F
1
csin
θ
()
−
F
Ry
=
x 2.64 m=
Problem 4-122

Replace the force system acting on the frame
by an equivalent resultant force and specify
where the resultant's line of action intersects
member AB, measured from point A.
Given:
F
1
35 lb= a 2ft=
F
2
20 lb= b 4ft=
F
3
25 lb= c 3ft=
θ
30 deg= d 2ft=
Solution:
F
Rx
F
1
sin
θ
()
F
3
+= F
Rx
42.5 lb=
F

Ry
F
1
− cos
θ
()
F
2
−= F
Ry
50.311− lb=
F
R
F
Rx
2
F
Ry
2
+= F
R
65.9 lb=
θ
1
atan
F
Ry
F
Rx
⎛

⎜
⎝
⎞
⎟
⎠
=
θ
1
49.8− deg=
F
Ry
xF
1
− cos
θ
()
aF
2
ab+()− F
3
c()+=
302
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
x
F
1
− cos

θ
()
aF
2
ab+()− F
3
c()+
F
Ry
= x 2.099 ft=
Problem 4-123
Replace the force system acting on the frame by an equivalent resultant force and specify where the
resultant's line of action intersects member BC, measured from point B.

Given:
F
1
35 lb=
F
2
20 lb=
F
3
25 lb=
θ
30 deg=
a 2ft=
b 4ft=
c 3ft=
d 2ft=

Solution:
F
Rx
F
1
sin
θ
()
F
3
+= F
Rx
42.5 lb=
F
Ry
F
1
− cos
θ
()
F
2
−= F
Ry
50.311− lb=
F
R
F
Rx
2

F
Ry
2
+= F
R
65.9 lb=
θ
1
atan
F
Ry
F
Rx
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
1
49.8− deg=
F
Rx
yF
1
cos
θ
()

bF
3
c()+=
303
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
y
F
1
cos
θ
()
bF
3
c()+
F
Rx
= y 4.617 ft=
(Below point B)
Problem 4-124
Replace the force system acting on the frame by an equivalent resultant force and couple moment
acting at point A.
Given:
F
1
35 lb= a 2ft=
F
2

20 lb= b 4ft=
F
3
25 lb= c 3ft=
θ
30 deg= d 2ft=
Solution:
F
Rx
F
1
sin
θ
()
F
3
+= F
Rx
42.5 lb=
F
Ry
F
1
cos
θ
()
F
2
+= F
Ry

50.311 lb=
F
R
F
Rx
2
F
Ry
2
+= F
R
65.9 lb=
θ
1
atan
F
Ry
F
Rx
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
1
49.8 deg=
M

RA
F
1
− cos
θ
()
aF
2
ab+()− F
3
c()+= M
RA
106− lb ft⋅=
Problem 4-125
Replace the force and couple-moment system by an equivalent resultant force and couple
moment at point O. Express the results in Cartesian vector form.
304
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 4
Units Used:
kN 10
3
N=
Given:
F
8
6
8

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN= a 3m=
e 5m=
b 3m=
f 6m=
c 4m=
g 5m=
M
20−
70−
20
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN m⋅=
d 6m=
Solution:
F

R
F= M
R
M
f−
e
g
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F×+= F
R
8
6
8
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN= M
R

10−
18
56−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN m⋅=
Problem 4-126
Replace the force and couple-moment system by an equivalent resultant force and couple
moment at point P. Express the results in Cartesian vector form.
Units Used:
kN 10
3
N=
Given:
F
8
6
8
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
kN=
M
20−
70−
20
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN m⋅=
a 3m=
b 3m= e 5m=
c 4m= f 6m=
305
© 2007 R. C. Hibbeler. Published by 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.