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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
PROBLEM 1-EXERCISE 6-73
The compound beam is pin-supported at C and supported by a roller A and B. There is a
hinge (pin) at D. Determine the reactions at the supports. Neglect the thickness of the beam.
SOLUTION 1-EXERCISE 6-73
6 ft 4 ft 2 ft
4 kip
8 kip
Dx
DyAy
30
(a)
Equation of Equilibrium: from FBD of (a)
∑ = 0DM 0)6()2(8)12(30cos4 =−+ yA�
kipAy 595.9=
∑ = 0yF 0830cos4595.9 =−−+�
yD kipDy 869.1=
∑ = 0xF 030sin4595.9 =−+�
xD kipDx 00.2=
1 22
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
From FBD of (b):
∑ = 0CM 0)16()8.0(1215)24(869.1 =−++ yB kipBy 541.8=
∑ = 0yF 0)8.0(12869.1541.8 =−−+yC kipC y 93.2=
∑ = 0xF 0)6.0(1200.2 =−−xC kipCx 2.9=
PROBLEM 1-EXERCISE 6-91
Determine the horizontal
and vertical components
of force which the pin at
A, B and C exert on
member ABC of the
frame.
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
SOLUTION 1-EXERCISE 6-91
FBD:
0; - (3.5) 400(2) 300(3.5) 300(1.5) 0
657 N ANS
E y
y
M A
A
= + + + =
=
∑
FBD of member CD:
0; (3.5) 400(2) 0 229 N ANSD y yM C C= − + = =∑
Cy Dy
Dx Cx
400 N
1.5 m 2 m
Ay
Ey
Ex
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
FBD of member ABC:
0; 0
0; F = F
50; 657.1 229 2 0
74
368.7 N
0 .
5(368.7)(2) 429 N Ans.
74
B x
x BD BE
y BD
BD BE
x
y
M C
F
F F
F F
B Ans
B
= =
=
= − − =
= =
= ←
= = ←
∑
∑
∑
PROBLEM 1-EXTRA PRACTICE 6-76
229 N
FBD
Cx
2.5 m
2.5 m
FBE
5
5
7 7
4
22
The three-hinged arch
support the loads
kNF 81 =
and kNF 52 = . Determine
the horizontal and vertical
components at the pin
supports A and B.
Take mh 2= .
Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
SOLUTION 1-EXTRA PRACTICE 6-76
Member AC:
∑ = 0AM 0)7()8()4(8 =++− xy CC
∑ = 0xF 08 =−− xx CA
∑ = 0yF 0=+− yy CA yy AC =
Member BC:
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
∑ = 0BM 0)9()6()2(5 =−+ xy CC
∑ = 0xF 0=− xx BC
∑ = 0yF 05 =−+− yy BC
Solving the above equations we have:
kNAx 61.56141.5 ==
kNAy 91.19122.1 ==
kNC x 39.23859.2 ==
kNC y 91.19122.1 ==
kNBx 39.23859.2 ==
kNBy 91.69122.6 ==
PROBLEM 1-EXTRA PRACTICE 6-89
Determine the horizontal
and vertical components
of force at each pin. The
suspended cylinder has a
weight of 80 lb.
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
SOLUTION 1-EXTRA PRACTICE 6-89
3 ft 3 ft 1 ft
80 lb
80 lb
By
Bx
Ax
Fcd3
2
∑ = 0BM 0)4(80)3)(13
2( =−CDF lbFCD 3.192=
lbDC xx 160)3.192(13
3===
lbDC yy 107)3.192(13
2===
∑ = 0yF 080)3.192(13
2=−+− yB lbBy 7.26=
Ey
Ex
Bx
80 lb
26.7 lb
3 ft
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
∑ = 0EM 0)3(7.26)3(80)4( =++− xB lbBx 0.80=
∑ = 0xF 08080 =−+xE 0=xE
∑ = 0yF 07.26 =+− yE lbE y 7.26=
∑ = 0xF 080)3.192(13
380 =−++− xA lbAx 160=
PROBLEM 1-EXERCISE 6-101
SOLUTION 1-EXERCISE 6-101
FBD of dismembered machine:
FBC Ax
Ay
10 lb
Ax
F DY
Dx
Ay
If a force P=6 lb is applied
perpendicular to the
handle of the mechanism,
determine the magnitude
of force F for equilibrium.
The members are pin-
connected at A, B, C and
D.
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
D
0 : (4) 6(25) 0 37.5
0 : 6 0 6 lb
0 : 37.5 0 =37.5b lb
0 : 5(6) 37.5(9) (39)( ) 0 9.42 .
A BC BC
x x x
y y y
M F F lb
F A A
F A A
M F F lb Ans
= − = → =
= − + = → =
= − + = →
= − − + = → = ←
∑
∑
∑
∑
PROBLEM 1-EXERCISE 6-120
SOLUTION 1-EXERCISE 6-120
FBD
Ax
Ay
Dy
Dx
20 lb
Determine the required
force P that must be
applied at the blade of the
pruning shears so that the
blade exerts a normal force
of 20 lb on the twig at E.
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
0; (5.5) (0.5)+20(1)=0
5.5 0.5 0
0; = A
0; 20 0
D x
x
x x x
y y y
M P A
P A
F D
F D A
= − −
+ =
=
= − − =
∑
∑
∑
0; (0.75)+ (0.5) 4.75 0
30; 0
3
20; + - =0
13
B y x
x x CB
y y CB
M A A P
F A F
F A P F
= − =
= − =
=
∑
∑
∑
Solving:
13.3 lb, 6.46 lb
13.3 lb, 28.9 lb
2.42 lb .
16.0 lb
x y
x y
CB
A A
D D
P Ans
F
= =
= =
= ←
=
2
3
Ay
Ax
FCB
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
PROBLEM 1-EXTRA PRACTICE 6-113
SOLUTION 1-EXTRA PRACTICE 6-113
A) FBD
Bar:
0; 2( / 2) 2(87.5) 0 175 lb ANSyF F F= − = =∑
Man:
0; 175 2(87.5) 0 350 lb ANSy c cF N N= − − = =∑
F/2 F/2
87.51 lb 87.51 lb
Nc
87.51 lb 87.51 lb 175 lb
A man having a weight of
175 lb attempts to lift
himself using one of the
two methods shown.
Determine the total force
he must exerts on bar AB
in each case and the
normal reaction he exerts
on the platform at C.
Neglect the weight of the
platform.
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
B) FBD
Bar:
0; 2(43.75) 2( ) 0 87.5 lb ANSyF F F= − = =∑
Man:
0; 175 2(43.75) 0 87.5 lb ANSy c cF N N= − − = =∑
PROBLEM 1-EXTRA PRACTICE 6-117
F/2 F/2
43.75 lb 43.75 lb
Nc
43.75 lb 43.75 lb 175 lb
1222
The tractor boom supports
the uniform mass of 500
kg in the bucket which
has a center of mass at G.
Determine the force in
each hydraulic cylinder
AB and CD and the
resultant force at pins E
and F. The load is
supported equally on each
side of the tractor by
similar mechanism.
Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
SOLUTION 1-EXTRA PRACTICE 6-117
From The bucket FBD:
∑ = 0EM 0)25.0()1.0(5.2452 =− ABF NFAB 981=
∑ = 0xF 0981 =+− xE NEx 981=
∑ = 0yF 05.2452 =−yE NE y 5.2452=
kNFE 64.2)5.2452()981( 22=+=
(a) The Bucket FBD (b) The Tractor Boom FBD
From the tractor boom FBD:
∑ = 0FM 0)25.1)(2.12(sin)7.0)(2.12(cos)8.2(5.2452 =+−��
CDCD FF
kNNFCD 3.1616349 ==
∑ = 0xF 02.12sin16349 =−�
xF NFx 3455=
∑ = 0yF 02.12cos163495.2452 =+−−�
yF NFy 13527=
kNFF 0.14)13527()3455( 22=+=
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
PROBLEM 3:
SOLUTION 3
Free-body diagram of the entire beam:
0yF =∑ :
750 0B C+ − =
. 0pt BM =∑ :
( )( )0.375 0.25 750 0C − =
Solving: 250B N= , 500C N=
The distributed load is:
( )7507500
0.1
N N
m mω = =
From the equilibrium equations:
0xF =∑ : 0A
P =
0yF =∑ :
( )250 0.05 7500 0AV− − =
0right endM−
=∑ :
Model the ladder rung as a simply
supported (pin supported) beam and
assume that the 750-N load exerted
by the person’s shoe is uniformly
distributed. Determine the internal
forces and moment at A.
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
( ) ( )250 0.25AM − +
( )( ) ( )0.05 7500 0.025 0+ = Therefore, the internal forces at A are:
0A
P = , 125A
V N= − , 53.1A
M = N-m
PROBLEM 4-EXERCISE 7-5
(a) Point C, which is just to the right of the bearing at A.
(b) Point D, which is just to the left of the 3000-lb force.
SOLUTION 4-EXERCISE 7-5
∑ = 0BM 0)2(3000)8(900)20(2500)14( =+++− yA
lbAy 4514=
∑ = 0xF 0=xB
∑ = 0yF 0300090025004514 =−−−+yB lbBy 1886=
The shaft is supported
by a journal bearing
at A and a thrust
bearing at B.
Determine the normal
force, shear force, and
moment at a section
passing through
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
∑ = 0CM 0)6(2500 =+ CM ftlbM C .15000−=
∑ = 0xF 0=CN
∑ = 0yF 045142500 =+−− CV lbVC 2014=
∑ = 0DM 0)2(1886 =− DM ftlbM D .3771=
∑ = 0xF 0=DN
∑ = 0yF 018863000 =+−− DV lbVD 1114=
PROBLEM 4-EXERCISE 7-19
Determine the normal
force, shear force, ad
moment at a section
passing through point
C. Take kNP 8=
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22
Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
SOLUTION 4-EXERCISE 7-19
∑ = 0AM 0)25.2(8)6.0( =+− T kNT 30=
∑ = 0xF kNAx 30=
∑ = 0yF kNAy 8=
∑ = 0CM 0)75.0(8 =− CM mkNM C .6=
∑ = 0xF kNNC 30−=
∑ = 0yF 08 =+CV kNVC 8−=
PROBLEM 4-EXTRA PRACTICE 7-13
Determine the internal
normal force, shear force,
and moment acting at point
C, and point D, which is
acted just to the right of the
roller support at B.
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
SOLUTION 4-EXTRA PRACTICE 7-13
Support Reactions: From FBD (a):
FBD (a)
∑ = 0AM 0)10(800)4(2400)2(800)8( =−−+yB lbBy 2000=
Internal Forces: Applying the equations of equilibrium to segment ED (FBD (b)), we have:
FBD (b)
∑ = 0DM 0)2(800 =−− DM ftlbM D .1600−=
∑ = 0xF 0=DN
∑ = 0yF 0800 =−DV lbVD 800=
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
FBD (c)
Applying the equations of equilibrium to segment EC (FBD(c)), we have:
∑ = 0CM 0)6(800)2(1200)4(2000 =−−+− CM ftlbM C .800=
∑ = 0xF 0=CN
∑ = 0yF 080012002000 =−−+CV 0=CV
PROBLEM 4-EXTRA PRACTICE 7-29
SOLUTION 4-EXTRA PRACTICE 7-29
Support Reactions:
From FBD (a):
∑ = 0AM 0)45cos63(12)45cos66( =+−+��
yB kNBy 485.8=
∑ = 0xF 0=xA
∑ = 0yF 00.12485.8 =−+yA kNAy 515.3=
Determine the internal
force, shear force, and the
moment at point C and D.
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
FBD (a) FBD (b)
From FBD (b):
∑ = 0CM 0)2(45cos515.3 =−�
CM mkNM C .97.4=
∑ = 0xF 045sin515.3 =− CN�
kNNC 49.2=
∑ = 0yF 0)2(45cos515.3 =− CV�
kNVC 49.2=
FBD (c)
∑ = 0DM 0)5.1(6)3(485.8 =−− DM mkNM D .5.16=
∑ = 0xF 0=DN
∑ = 0yF 06485.8 =−+DV kNVD 49.2−=
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
PROBLEM 4-EXTRA PRACTICE 7-41
SOLUTION 4-EXTRA PRACTICE 7-41
Free Body Diagram: The support reactions need not be computed.
Determine the x, y, z
components of force and
moment at point C in the
pipe assembly. Neglect the
weight of the pipe. Take
lbjiF }ˆ400ˆ350{1 −=
and
lbkjF }ˆ150ˆ300{2 +−=
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
Internal Forces: Applying the equations of equilibrium to segment BC. We have:
;0=∑ xF 0350 =+CN lbNC 350−=
;0=∑ yF 0300400)( =−−yCV lbV yC 700)( =
;0=∑ zF 0150)( =+zCV lbV zC 150)( −=
;0=∑ xM 0)3(400)( =+xCM ftkipftlbM xC .2.1.1200)( −=−=
;0=∑ yM 0)2(150)3(350)( =−+yCM ftlbM yC .750)( −=
;0=∑ zM 0)2(400)2(300)( =−−zCM ftkipftlbM zC .4.1.1400)( ==
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