NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
1
22
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
8 kip
Dx
30
4 kip
6 ft
4 ft
Ay
(a)
2 ft
Dy
Equation of Equilibrium: from FBD of (a)
∑M
∑F
∑F
D
=0
4 cos 30 (12) + 8(2) − Ay (6) = 0 Ay = 9.595kip
y
=0
D y + 9.595 − 4 cos 30 − 8 = 0 D y = 1.869kip
x
=0
D x + 9.595 − 4 sin 30 = 0 D x = 2.00kip
Ans
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
2
22
From FBD of (b):
∑M
∑F
∑F
C
=0
1.869(24) + 15 + 12(0.8) − B y (16) = 0 B y = 8.541kip
y
=0
C y + 8.541 − 1.869 − 12(0.8) = 0 C y = 2.93kip
x
=0
C x − 2.00 − 12(0.6) = 0 C x = 9.2kip
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.
Ans
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
Mechatronics Engineering
PAGE
3
22
SOLUTION 1-EXERCISE 6-91
FBD:
Ex
Ay
Ey
∑M
E
= 0; -Ay (3.5) + 400(2) + 300(3.5) + 300(1.5) = 0
Ay = 657 N
Ans
ANS
FBD of member CD:
400 N
1.5 m
Cx
2m
Dx
Cy
∑M
D
Dy
= 0; − C y (3.5) + 400(2) = 0 C y = 229 N
ANS
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
Mechatronics Engineering
FBD of member ABC:
229 N
∑M
∑F
Cx
2.5 m
FBD
7
7
5
2.5 m
5
∑F
B
= 0;
Cx = 0
x
= 0;
FBD = FBE
y
= 0; 657.1 − 229 − 2
5
FBD = 0
74
FBD = FBE = 368.7 N
B x = 0 ← Ans .
FBE
By =
5
(368.7)(2) = 429 N ← Ans.
74
PROBLEM 1-EXTRA PRACTICE 6-76
The three-hinged arch
support
the
loads
F1 = 8kN
and F2 = 5kN . Determine
the horizontal and vertical
components at the pin
supports A and B.
Take h = 2m .
PAGE
4
22
NAME & ID
Mechatronics Engineering
SOLUTION 1-EXTRA PRACTICE 6-76
Member AC:
∑M
∑F
∑F
A
=0
− 8(4) + C y (8) + C x (7) = 0
x
=0
8 − Ax − C x = 0
y
=0
− Ay + C y = 0 C y = Ay
Member BC:
DATE
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
5
22
NAME & ID
Mechatronics Engineering
∑M
∑F
∑F
B
=0
DATE
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
6
22
5(2) + C y (6) − C x (9) = 0
x
=0
C x − Bx = 0
y
=0
− C y + By − 5 = 0
Solving the above equations we have:
Ax = 5.6141 = 5.61kN
Ay = 1.9122 = 1.91kN
C x = 2.3859 = 2.39kN
C y = 1.9122 = 1.91kN
B x = 2.3859 = 2.39kN
B y = 6.9122 = 6.91kN
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.
Ans
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
Mechatronics Engineering
PAGE
7
22
SOLUTION 1-EXTRA PRACTICE 6-89
By
3 ft
1 ft
3 ft
Bx
80 lb
Ax
Fcd
80 lb
2
3
∑M
B
=0
FCD (
2
13
)(3) − 80(4) = 0
C x = Dx =
C y = Dy =
∑F
y
=0
− By +
2
13
3
13
2
13
FCD = 192.3lb
(192.3) = 160lb
Ans
(192.3) = 107lb
Ans
(192.3) − 80 = 0 B y = 26.7lb
Ey
Ans
3 ft
Ex
80 lb
Bx
26.7 lb
Ans
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
Mechatronics Engineering
∑M
∑F
∑F
E
=0
− B x (4) + 80(3) + 26.7(3) = 0 B x = 80.0lb
x
=0
y
=0
− E y + 26.7 = 0
x
=0
− Ax + 80 +
∑F
(192.3) − 80 = 0
Ans
Ax = 160lb
Ans
PROBLEM 1-EXERCISE 6-101
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 pinconnected at A, B, C and
D.
SOLUTION 1-EXERCISE 6-101
FBD of dismembered machine:
Ay
10 lb
Ax
Dx
Ax
FBC
Ay
F
22
Ans
E y = 26.7lb
13
8
Ans
E x + 80 − 80 = 0 E x = 0
3
PAGE
DY
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
∑ M = 0 : F (4) − 6(25) = 0 → F = 37.5lb
∑ F = 0 : − A + 6 = 0 → A = 6 lb
∑ F = 0 : − A + 37.5 = 0 → A =37.5b lb
∑ M = 0 : −5(6) − 37.5(9) + (39)( F ) = 0 → F = 9.42lb ← Ans.
A
BC
x
x
y
y
BC
x
y
D
PROBLEM 1-EXERCISE 6-120
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.
SOLUTION 1-EXERCISE 6-120
FBD
20 lb
Dy
Dx
Ax
Ay
PAGE
9
22
NAME & ID
DATE
Mechatronics Engineering
∑M
D
= 0;
− P (5.5) − Ax (0.5)+20(1)=0
5.5 P + 0.5 Ax = 0
∑F
∑F
x
= 0; Dx = A x
y
= 0; Dy − Ay − 20 = 0
Ay
3
FCB
2
Ax
∑M
B
= 0; Ay (0.75)+Ax (0.5) − 4.75 P = 0
3
FCB = 0
3
2
F
=
0;
A
+
P
F
=0
∑ y
y
CB
13
∑F
x
= 0; Ax −
Solving:
Ax = 13.3 lb, Ay = 6.46 lb
Dx = 13.3 lb, Dy = 28.9 lb
P = 2.42 lb
FCB = 16.0 lb
← Ans.
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
10
22
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PROBLEM 1-EXTRA PRACTICE 6-113
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.
SOLUTION 1-EXTRA PRACTICE 6-113
A) FBD
87.51 lb
87.51 lb
F/2
87.51 lb
175 lb
F/2
Nc
Bar:
∑F
y
= 0; 2( F / 2) − 2(87.5) = 0
F = 175 lb
ANS
= 0; N c − 175 − 2(87.5) = 0
N c = 350 lb
ANS
Man:
∑F
y
87.51 lb
PAGE
11
22
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
B) FBD
43.75 lb
43.75 lb
F/2
43.75 lb
175 lb
F/2
Nc
Bar:
∑F
y
= 0; 2(43.75) − 2( F ) = 0
F = 87.5 lb
ANS
Man:
∑F
y
= 0; N c − 175 − 2(43.75) = 0
N c = 87.5 lb
PROBLEM 1-EXTRA PRACTICE 6-117
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.
ANS
43.75 lb
PAGE
12
22
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
Mechatronics Engineering
PAGE
13
22
SOLUTION 1-EXTRA PRACTICE 6-117
From The bucket FBD:
∑M
∑F
∑F
E
=0
x
=0
y
=0
2452.5(0.1) − FAB (0.25) = 0 FAB = 981N
Ans
− E x + 981 = 0 E x = 981N
E y − 2452.5 = 0
E y = 2452.5 N
FE = (981) 2 + (2452.5) 2 = 2.64kN
Ans
Ans
(a) The Bucket FBD
(b) The Tractor Boom FBD
From the tractor boom FBD:
∑M
F
=0
2452.5(2.8) − FCD (cos 12.2 )(0.7) + FCD (sin 12.2 )(1.25) = 0
Ans
FCD = 16349 N = 16.3kN
∑F
∑F
x
=0
y
=0
Fx − 16349 sin 12.2 = 0 Fx = 3455 N
− Fy − 2452.5 + 16349 cos 12.2 = 0
FF = (3455) 2 + (13527) 2 = 14.0kN
Fy = 13527 N
Ans
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
Mechatronics Engineering
PROBLEM 3:
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.
SOLUTION 3
Free-body diagram of the entire beam:
∑F
y
= 0:
B + C − 750 = 0
∑M
pt . B
= 0:
0.375C − ( 0.25 )( 750 ) = 0
Solving: B = 250 N , C = 500 N
The distributed load is:
ω=
( 750 N ) = 7500 N
0.1m
m
From the equilibrium equations:
∑F
∑F
x
= 0 : PA = 0
y
= 0:
250 − VA − 0.05 ( 7500 ) = 0
∑M
right − end
=0:
PAGE
14
22
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
Mechatronics Engineering
PAGE
15
22
M A − ( 250 )( 0.25 ) +
+ ( 0.05 )( 7500 )( 0.025) = 0 Therefore, the internal forces at A are:
PA = 0 , VA = −125 N , M A = 53.1 N-m
Ans
PROBLEM 4-EXERCISE 7-5
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
(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
∑M
B
=0
− Ay (14) + 2500(20) + 900(8) + 3000(2) = 0
Ans
Ay = 4514lb
∑F
∑F
x
=0
Bx = 0
y
=0
B y + 4514 − 2500 − 900 − 3000 = 0
B y = 1886lb
Ans
NAME & ID
DATE
Mechatronics Engineering
∑M
∑F
∑F
C
=0
x
=0
y
=0
∑M
∑F
∑F
D
=0
x
=0
y
=0
2500(6) + M C = 0
VC = 2014lb
M D = 3771lb. ft
Ans
ND = 0
− VD − 3000 + 1886 = 0
Determine the normal
force, shear force, ad
moment at a section
passing through point
C. Take P = 8kN
1
Ans
− VC − 2500 + 4514 = 0
PROBLEM 4-EXERCISE 7-19
PAGE
16
22
M C = −15000lb. ft
NC = 0
1886(2) − M D = 0
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
VD = 1114lb
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
17
22
SOLUTION 4-EXERCISE 7-19
∑M
∑F
∑F
A
=0
− T (0.6) + 8(2.25) = 0
x
=0
Ax = 30kN
y
=0
Ay = 8kN
∑M
∑F
∑F
C
=0
x
=0
y
=0
8(0.75) − M C = 0
T = 30kN
Ans
M C = 6kN .m
N C = −30kN
VC + 8 = 0
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.
VC = −8kN
Ans
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
18
22
SOLUTION 4-EXTRA PRACTICE 7-13
Support Reactions: From FBD (a):
FBD (a)
∑M
A
=0
B y (8) + 800(2) − 2400(4) − 800(10) = 0
B y = 2000lb
Ans
Internal Forces: Applying the equations of equilibrium to segment ED (FBD (b)), we have:
FBD (b)
∑M
∑F
∑F
D
=0
x
=0
y
=0
− 800(2) − M D = 0
M D = −1600lb. ft
ND = 0
VD − 800 = 0
VD = 800lb
Ans
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
19
22
FBD (c)
Applying the equations of equilibrium to segment EC (FBD(c)), we have:
∑M
∑F
∑F
C
=0
x
=0
y
=0
− M C + 2000(4) − 1200(2) − 800(6) = 0
M C = 800lb. ft
NC = 0
VC + 2000 − 1200 − 800 = 0
Ans
VC = 0
PROBLEM 4-EXTRA PRACTICE 7-29
Determine the internal
force, shear force, and the
moment at point C and D.
SOLUTION 4-EXTRA PRACTICE 7-29
Support Reactions:
From FBD (a):
∑M
∑F
∑F
A
=0
x
=0
y
=0
B y (6 + 6 cos 45 ) − 12(3 + 6 cos 45 ) = 0
B y = 8.485kN
Ax = 0
Ay + 8.485 − 12.0 = 0
Ay = 3.515kN
Ans
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
FBD (a)
PAGE
20
22
FBD (b)
From FBD (b):
∑M
∑F
∑F
C
=0
M C − 3.515 cos 45 (2) = 0
M C = 4.97 kN .m
x
=0
3.515 sin 45 − N C = 0
N C = 2.49kN
y
=0
3.515 cos 45 (2) − VC = 0
VC = 2.49kN
Ans
FBD (c)
∑M
∑F
∑F
D
=0
x
=0
y
=0
8.485(3) − 6(1.5) − M D = 0
M D = 16.5kN .m
ND = 0
VD + 8.485 − 6 = 0
VD = −2.49kN
Ans
NAME & ID
Mechatronics Engineering
DATE
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PROBLEM 4-EXTRA PRACTICE 7-41
Determine the x, y, z
components of force and
moment at point C in the
pipe assembly. Neglect the
weight of the pipe. Take
F1 = {350iˆ − 400 ˆj}lb
and
F2 = {−300 ˆj + 150kˆ}lb
SOLUTION 4-EXTRA PRACTICE 7-41
Free Body Diagram: The support reactions need not be computed.
PAGE
21
22
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 6
SOLUTIONS
PAGE
22
22
Internal Forces: Applying the equations of equilibrium to segment BC. We have:
∑F
= 0;
N C + 350 = 0
y
= 0;
(VC ) y − 400 − 300 = 0 (VC ) y = 700lb
z
= 0; (VC ) z + 150 = 0
x
∑F
∑F
N C = −350lb
Ans
Ans
(VC ) z = −150lb
Ans
∑M
x
= 0; ( M C ) x + 400(3) = 0 ( M C ) x = −1200lb. ft = −1.2kip. ft
Ans
∑M
y
= 0; ( M C ) y + 350(3) − 150(2) = 0 ( M C ) y = −750lb. ft
Ans
∑M
z
= 0;
( M C ) z − 300(2) − 400(2) = 0 ( M C ) z = 1400lb. ft = 1.4kip. ft
Ans