Engineering Mechanics - Statics Episode 1 Part 9 pdf

Engineering Mechanics - Statics Chapter 4
Units Used:
kN 10
3
N=
Given:
w
1
1.5
kN
m
= a 3m=
w
2
1
kN
m
= b 3m=
w
3
2.5
kN
m
= c 1.5 m=
Solution:
F
R
w
1
aw
2

b+ w
3
c+= F
R
11.25 kN=
M
A
w
1
a
a
2
w
2
ba
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
+ w
3
ca b+
c
2
+

⎛
⎜
⎝
⎞
⎟
⎠
+=
M
A
45.563 kN m⋅= d
M
A
F
R
= d 4.05 m=
Problem 4-147
Determine the length b of the triangular load and its position a on the beam such that the
equivalent resultant force is zero and the resultant couple moment is M clockwise.
Units Used:
kN 10
3
N=
Given:
w
1
4
kN
m
= w
2

2.5
kN
m
=
M 8kNm⋅= c 9m=
Solution:
Initial Guesses:
a 1m= b 1m=
Given
1−
2
w
1
b
1
2
w
2
c+ 0=
321
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 4
1
2
w
1
ba
2b

3
+
⎛
⎜
⎝
⎞
⎟
⎠
1
2
w
2
c
2c
3
− M−=
a
b
⎛
⎜
⎝
⎞
⎟
⎠
Find ab,()=
a
b
⎛
⎜
⎝

⎞
⎟
⎠
1.539
5.625
⎛
⎜
⎝
⎞
⎟
⎠
m=
Problem 4-148
Replace the distributed loading by an equivalent resultant force and specify its location,
measured from point A.
Units Used:
kN 10
3
N=
Given:
w
1
800
N
m
=
w
2
200
N

m
=
a 2m=
b 3m=
Solution:
F
R
w
2
bw
1
a+
1
2
w
1
w
2
−
()
b+= F
R
3.10 kN=
xF
R
w
1
a
a
2

1
2
w
1
w
2
−
()
ba
b
3
+
⎛
⎜
⎝
⎞
⎟
⎠
+ w
2
ba
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠

+=
x
w
1
a
a
2
1
2
w
1
w
2
−
()
ba
b
3
+
⎛
⎜
⎝
⎞
⎟
⎠
+ w
2
ba
b
2

+
⎛
⎜
⎝
⎞
⎟
⎠
+
F
R
= x 2.06 m=
Problem 4-149
The distribution of soil loading on the bottom of a building slab is shown. Replace this loading
by an equivalent resultant force and specify its location, measured from point O.
Units Used:
322
© 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
kip 10
3
lb=
Given:
w
1
50
lb
ft
=

w
2
300
lb
ft
=
w
3
100
lb
ft
=
a 12 ft=
b 9ft=
Solution
:
F
R
w
1
a
1
2
w
2
w
1
−
()
a+

1
2
w
2
w
3
−
()
b+ w
3
b+= F
R
3.9 kip=
F
R
dw
1
a
a
2
1
2
w
2
w
1
−
()
a
2a

3
+
1
2
w
2
w
3
−
()
ba
b
3
+
⎛
⎜
⎝
⎞
⎟
⎠
+ w
3
ba
b
2
+
⎛
⎜
⎝
⎞

⎟
⎠
+=
d
3 w
3
ba 2 w
3
b
2
+ w
1
a
2
+ 2 a
2
w
2
+ 3 bw
2
a+ w
2
b
2
+
6F
R
= d 11.3 ft=
Problem 4-150
The beam is subjected to the distributed

loading. Determine the length b of the
uniform load and its position a on the beam
such that the resultant force and couple
moment acting on the beam are zero.
Given:
w
1
40
lb
ft
= c 10ft=
d 6ft=
w
2
60
lb
ft
=
Solution:
Initial Guesses
:
a 1ft= b 1ft=
323
© 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
1
2
w

2
dw
1
b− 0=
1
2
w
2
dc
d
3
+
⎛
⎜
⎝
⎞
⎟
⎠
w
1
ba
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠

− 0=
a
b
⎛
⎜
⎝
⎞
⎟
⎠
Find ab,()=
a
b
⎛
⎜
⎝
⎞
⎟
⎠
9.75
4.5
⎛
⎜
⎝
⎞
⎟
⎠
ft=
Problem 4-151
Replace the loading by an equivalent resultant force and specify its location on the beam,
measured from point B.

Units Used:
kip 10
3
lb=
Given:
w
1
800
lb
ft
=
w
2
500
lb
ft
=
a 12 ft=
b 9ft=
Solution:
F
R
1
2
aw
1
1
2
w
1

w
2
−
()
b+ w
2
b+= F
R
10.65 kip=
F
R
x
1
2
− aw
1
a
3
1
2
w
1
w
2
−
()
b
b
3
+ w

2
b
b
2
+=
x
1
2
− aw
1
a
3
1
2
w
1
w
2
−
()
b
b
3
+ w
2
b
b
2
+
F

R
= x 0.479ft=
( to the right of B )
Problem 4-152
Replace the distributed loading by an equivalent resultant force and specify where its line of action
intersects member AB, measured from A.
324
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:
w
1
200
N
m
=
w
2
100
N
m
=
w
3
200
N
m

=
a 5m=
b 6m=
Solution:
F
Rx
w
3
− a= F
Rx
1000− N=
F
Ry
1−
2
w
1
w
2
+
()
b= F
Ry
900− N=
y− F
Rx
w
3
a
a

2
w
2
b
b
2
−
1
2
w
1
w
2
−
()
b
b
3
−=
y
w
3
a
a
2
w
2
b
b
2

−
1
2
w
1
w
2
−
()
b
b
3
−
F
Rx
−
= y 0.1 m=
Problem 4-153
Replace the distributed loading by an equivalent resultant force and specify where its line of
action intersects member BC, measured from C.
Units Used:
kN 10
3
N=
325
© 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:

w
1
200
N
m
=
w
2
100
N
m
=
w
3
200
N
m
=
a 5m=
b 6m=
Solution
:
F
Rx
w
3
− a= F
Rx
1000− N=
F

Ry
1−
2
w
1
w
2
+
()
b= F
Ry
900− N=
x− F
Ry
w
3
− a
a
2
w
2
b
b
2
+
1
2
w
1
w

2
−
()
b
2b
3
+=
x
w
3
− a
a
2
w
2
b
b
2
+
1
2
w
1
w
2
−
()
b
2b
3

+
F
Ry
−
= x 0.556 m=
F
Rx
F
Ry
⎛
⎜
⎝
⎞
⎟
⎠
1.345 kN=
Problem 4-154
Replace the loading by an equivalent resultant force and couple moment acting at point O.
Units Used:
kN 10
3
N=
Given:
w
1
7.5
kN
m
=
326

© 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
w
2
20
kN
m
=
a 3m=
b 3m=
c 4.5 m=
Solution:
F
R
1
2
w
2
w
1
−
()
cw
1
c+ w
1
b+
1

2
w
1
a+=
F
R
95.6 kN=
M
Ro
1
2
− w
2
w
1
−
()
c
c
3
w
1
c
c
2
− w
1
bc
b
2

+
⎛
⎜
⎝
⎞
⎟
⎠
−
1
2
w
1
ab c+
a
3
+
⎛
⎜
⎝
⎞
⎟
⎠
−= M
Ro
349− kN m⋅=
Problem 4-155
Determine the equivalent resultant force
and couple moment at point O.
Units Used:
kN 10

3
N=
Given:
a 3m=
w
O
3
kN
m
=
wx() w
O
x
a
⎛
⎜
⎝
⎞
⎟
⎠
2
=
Solution:
F
R
0
a
xwx()
⌠
⎮

⌡
d= F
R
3kN=
M
O
0
a
xwx()ax−()
⌠
⎮
⌡
d= M
O
2.25 kN m⋅=
327
© 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
Problem 4-156
Wind has blown sand over a platform such that the intensity of the load can be approximated by
the function
ww
0
x
d
⎛
⎜
⎝

⎞
⎟
⎠
3
=
. Simplify this distributed loading to an equivalent resultant force and
specify the magnitude and location of the force, measured from A.
Units Used:
kN 10
3
N=
Given:
w
0
500
N
m
=
d 10 m=
wx() w
0
x
d
⎛
⎜
⎝
⎞
⎟
⎠
3

=
Solution:
F
R
0
d
xwx()
⌠
⎮
⌡
d= F
R
1.25 kN=
d
0
d
xxw x()
⌠
⎮
⌡
d
F
R
= d 8m=
Problem 4-157
Determine the equivalent resultant force and its location, measured from point O.
328
© 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
Solution:
F
R
0
L
xw
0
sin
π
x
L
⎛
⎜
⎝
⎞
⎟
⎠
⌠
⎮
⎮
⌡
d=
2w
0
L
π
=
d
0

L
xxw
0
sin
π
x
L
⎛
⎜
⎝
⎞
⎟
⎠
⌠
⎮
⎮
⌡
d
F
R
=
L
2
=
Problem 4-158
Determine the equivalent resultant force acting on the bottom of the wing due to air pressure and
specify where it acts, measured from point A.
Given:
a 3ft=
k 86

lb
ft
3
=
wx() kx
2
=
Solution:
F
R
0
a
xwx()
⌠
⎮
⌡
d= F
R
774lb=
x
0
a
xxw x()
⌠
⎮
⌡
d
F
R
= x 2.25ft=

Problem 4-159
Currentl
y
ei
g
ht
y
-five
p
ercent of all neck in
j
uries are caused b
y
rear-end car collisions. To
329
© 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
ygy p j y
alleviate this problem, an automobile seat restraint has been developed that provides additional
pressure contact with the cranium. During dynamic tests the distribution of load on the cranium
has been plotted and shown to be parabolic. Determine the equivalent resultant force and its
location, measured from point A.
Given:
a 0.5 ft=
w
0
12
lb

ft
=
k 24
lb
ft
3
=
wx() w
0
kx
2
+=
Solution:
F
R
0
a
xwx()
⌠
⎮
⌡
d= F
R
7lb=
x
0
a
xxw x()
⌠
⎮

⌡
d
F
R
= x 0.268ft=
Problem 4-160
Determine the equivalent resultant force of the distributed loading and its location, measured from
point A. Evaluate the integrals using Simpson's rule.
Units Used:
kN 10
3
N=
Given:
c
1
5=
c
2
16=
a 3=
330
© 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 1=
Solution:
F
R
0

ab+
xc
1
xc
2
x
2
++
⌠
⎮
⌡
d=
F
R
14.9=
d
0
ab+
xxc
1
xc
2
x
2
++
⌠
⎮
⌡
d
F

R
= d 2.27=
Problem 4-161
Determine the coordinate direction angles of F, which is applied to the end A of the pipe
assembly, so that the moment of F about O is zero.
Given:
F 20 lb=
a 8in=
b 6in=
c 6in=
d 10 in=
Solution:
Require M
o
= 0. This happens when force F is directed either towards or away from point O.
r
c
ab+
d
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= u
r
r

= u
0.329
0.768
0.549
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
If the force points away from O, then
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
acos u
()
=

α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
70.774
39.794
56.714
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
deg=
If the force points towards O, then
331
© 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
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
acos u−
()
=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠

109.226
140.206
123.286
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
deg=
Problem 4-162
Determine the moment of the force F about
point O. The force has coordinate direction
angles
α
,
β
,
γ
.
Express the result as a Cartesian
vector.
Given:
F 20 lb= a 8in=
α
60 deg= b 6in=
β
120 deg= c 6in=

γ
45 deg= d 10 in=
Solution:
r
c
ab+
d
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
v
F
cos
α
()
cos
β
()
cos
γ
()
⎛
⎜
⎜

⎜
⎝
⎞
⎟
⎟
⎟
⎠
= MrF
v
×= M
297.99
15.147
200−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb in⋅=
Problem 4-163
Replace the force at A by an equivalent resultant force and couple moment at point P. Express
the results in Cartesian vector form.
Units Used
:
332
© 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
kN 10
3
N=
Given:
a 4m=
b 6m=
c 8m=
d 4m=
F
300−
200
500−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
Solution
:
F
R
F= F
R
300−

200
500−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
M
P
a− c−
b
d
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F×= M
P
3.8−
7.2−
0.6−
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN m⋅=
Problem 4-164
Determine the moment of the force
F
C
about the door hinge at A. Express the result as a Cartesian
vector.
333
© 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 250 N=
b 1m=
c 2.5 m=
d 1.5 m=
e 0.5 m=
θ
30 deg=
Solution:
r
CB

ce−
bdcos
θ
()
+
d− sin
θ
()
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= r
AB
0
b
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F

v
F
r
CB
r
CB
=
M
A
r
AB
F
v
×= M
A
59.7−
0.0
159.3−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
Nm⋅=
Problem 4-165
Determine the magnitude of the moment of the force
F

C
about the hinged axis aa of the door.
334
© 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 250 N=
b 1m=
c 2.5 m=
d 1.5 m=
e 0.5 m=
θ
30 deg=
Solution:
r
CB
ce−
bdcos
θ
()
+
d− sin
θ
()
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
= r
AB
0
b
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
v
F
r
CB
r
CB
= u
a
1
0
0
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
M
aa
r
AB
F
v
×
()
u
a
⋅= M
aa
59.7− Nm⋅=
Problem 4-166
A force F
1
acts vertically downward on the Z-bracket. Determine the moment of this force
about the bolt axis (z axis), which is directed at angle
θ
from the vertical.
335
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Engineering Mechanics - Statics Chapter 4
Given:
F
1
80 N=
a 100 mm=
b 300 mm=
c 200 mm=
θ
15 deg=
Solution:
r
b−
ac+
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
F F
1
sin
θ

()
0
cos
θ
()
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
k
0
0
1
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
M
z

rF×
()
k=
M
z
6.212− Nm⋅=
Problem 4-167
Replace the force F having acting at point A by an equivalent force and couple moment at
point C.
Units Used:
kip 10
3
lb=
Given:
F 50 lb=
a 10 ft=
b 20 ft=
c 15 ft=
336
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 4
d 10 ft=
e 30 ft=
Solution
:
r
AB
d

c
e−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
F
v
F
r
AB
r
AB
=
r
CA
0
ab+
e
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
=
F
R
F
v
= F
R
14.286
21.429
42.857−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
M
R
r
CA
F
v
×= M
R
1.929−

0.429
0.429−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kip ft⋅=
Problem 4-168
The horizontal force F acts on the handle of the wrench. What is the magnitude of the moment
of this force about the z axis?
Given:
F 30 N=
a 50 mm=
b 200 mm=
c 10 mm=
θ
45 deg=
Solution:
F
v
F
sin
θ
()
cos
θ

()
−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= r
OA
c−
b
a
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= k
0
0
1
⎛
⎜

⎜
⎝
⎞
⎟
⎟
⎠
=
337
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 4
M
z
r
OA
F
v
×
()
k= M
z
4.03− Nm⋅=
Problem 4-169
The horizontal force
F

acts on the
handle of the wrench. Determine the
moment of this force about point O.

Specify the coordinate direction
angles
α
,
β
,
γ
of the moment axis.
Given:
F 30 N= c 10 mm=
a 50 mm=
θ
45 deg=
b 200 mm=
Solution:
F
v
F
sin
θ
()
cos
θ
()
−
0
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
= r
OA
c−
b
a
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
M
O
r
OA
F
v
×= M
O
1.06
1.06
4.03−
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
Nm⋅=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
acos
M
O
M
O
⎛
⎜
⎝
⎞

⎟
⎠
=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
75.7
75.7
159.6
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
deg=
Problem 4-170
If the resultant couple moment of the three couples acting on the triangular block is to be zero,

determine the magnitudes of forces
F
and
P.
338
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 4
Given:
F
1
10 lb=
a 3in=
b 4in=
c 6in=
d 3in=
θ
30 deg=
Solution:
Initial Guesses
:
F 1lb= P 1lb=
Given
0
F− c
0
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
0
0
P− c
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+
F
1
d
a
2
b
2
+
0
a
b
⎛
⎜

⎜
⎝
⎞
⎟
⎟
⎠
+ 0=
F
P
⎛
⎜
⎝
⎞
⎟
⎠
Find FP,()=
F
P
⎛
⎜
⎝
⎞
⎟
⎠
3
4
⎛
⎜
⎝
⎞

⎟
⎠
lb=
339
© 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 5
Problem 5-1
Draw the free-body diagram of the sphere of weight W resting between the smooth inclined
planes. Explain the significance of each force on the diagram.
Given:
W 10 lb=
θ
1
105 deg=
θ
2
45 deg=
Solution:
N
A
, N
B
force of plane on sphere.
W force of gravity on sphere.
Problem 5-2
Draw the free-body diagram of the hand punch, which is pinned at A and bears down on the
smooth surface at B.
Given:

F 8lb=
a 1.5 ft=
b 0.2 ft=
c 2ft=
340
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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 5
Problem 5-3
Draw the free-body diagram of the beam supported at A by a fixed support and at B by a roller.
Explain the significance of each force on the diagram.
Given:
w 40
lb
ft
=
a 3ft=
b 4ft=
θ
30 deg=
Solution:
A
x
, A
y
, M
A
effect of wall on beam.
N

B
force of roller on beam.
wa
2

resultant force of distributed load on beam.
Problem 5-4
Draw the free-body diagram of the jib crane AB, which is pin-connected at A and supported by
member (link) BC.
341
Solution:
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Units Used:
kN 10
3
N=
Given:

F 8kN=
a 3m=
b 4m=
c 0.4 m=
d 3=
e 4=
Solution:
Problem 5-5
Draw the free-body diagram of the

C-bracket supported at A, B, and C by
rollers. Explain the significance of each
forcce on the diagram.
Given:
a 3ft=
b 4ft=
θ
1
30 deg=
θ
2
20 deg=
F 200 lb=
342
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Solution:
N
A
, N
B
,

N
C
force of rollers on beam.
Problem 5-6
Draw the free-body diagram of the smooth

rod of mass M which rests inside the
glass. Explain the significance of each
force on the diagram.
Given:
M 20 gm=
a 75 mm=
b 200 mm=
θ
40 deg=
Solution:

A
x
,

A
y
, N
B
force of glass on rod.
M(g) N force of gravity on rod.
Problem 5-7
Draw the free-body diagram of the “spanner wrench” subjected to the force
F
. The support at
A
can be considered a
p
in, and the surface of contact at B is smooth. Ex
p

lain the si
g
nificance of
343
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Engineering Mechanics - Statics Chapter 5
p
pg
each force on the diagram.
Given:
F 20 lb=
a 1in=
b 6in=
Solution:
A
x
, A
y
, N
B
force of cylinder on wrench.
Problem 5-8
Draw the free-body diagram of the automobile, which is being towed at constant velocity up the
incline using the cable at C. The automobile has a mass M and center of mass at G. The tires are
free to roll. Explain the significance of each force on the diagram.
Units Used:
Mg 10
3

kg=
Given:
M 5Mg= d 1.50 m=
a 0.3 m= e 0.6 m=
b 0.75 m=
θ
1
20 deg=
c 1m=
θ
2
30 deg=
g 9.81
m
s
2
=
344
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 5
Solution:

N
A
,
N
B

force of road on car.
F
force of cable on car.
Mg
force of gravity on car.
Problem 5-9
Draw the free-body diagram of the uniform bar, which has mass M and center of mass at G. The
supports A, B, and C are smooth.
Given:
M 100 kg=
a 1.75 m=
b 1.25 m=
c 0.5 m=
d 0.2 m=
g 9.81
m
s
2
=
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.