NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
Mechatronics Engineering
PROBLEM 1
G
Systems 1 and 2 each consist of a couple. If they are equivalent, what is F ?
SOLUTION 1:
For couples, the sum of the forces vanish for both systems. For system 1, the two forces are
located at:
r11 = 4iˆ ,
r12 = +5 ˆj .
The forces are:
F1 = 200(iˆ cos 30 D + ˆj sin 30 D ) = 173.21iˆ + 100 ˆj
ˆj kˆ
M 1 = (r11 − r12 ) × F1 = 4
− 5 0 = 1266.05kˆ
173.21 100 0
iˆ
( N .m)
For system 2, the positions of the forces are r21 = 2iˆ and r22 = 5iˆ + 4 ˆj . The forces are:
F2 = F (iˆ cos(−20 D ) + ˆj sin(−20 D )) = F (0.9397iˆ − 0.3420 ˆj )
The moment of the couple in System 2 is:
ˆj
kˆ
M 2 = (r21 − r22 ) × F2 = F
−4
0 = 4.7848Fkˆ
0.9397 − 0.3420 0
iˆ
−3
from which, if the systems are to be equivalent,
F=
1266
= 264.6
4.7848
(N )
Å Ans
PAGE
1
14
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
Mechatronics Engineering
PROBLEM 2
Exercise Problem 4-106/107 (Textbook - page 175)
4-106: Goal: Replace the force and couple system by an equivalent force and couple
moment at point O.
4-107: Goal: Replace the force and couple system by an equivalent force and couple
moment at point P.
SOLUTION 2
Solution 4-106
The total forces acting on the x- axis are:
+
→ ∑ FRx = ∑ Fx
5
FRx = 6( ) − 4 cos 60 D = 0.30769 (kN )
13
The total forces acting on the y-axis are:
+
↑ ∑ FRy = ∑ Fy
12
FRy = 6( ) − 4 sin 60 D = 2.0744
13
FR = (0.30769) 2 + (2.0744) 2 = 2.10
(kN )
(kN ) Å-Ans
The angle of direction of total force (with respect to x-axis) is:
⎛ 2.0744 ⎞
D
⎟ = 81.6 ÅAns
0
.
30769
⎝
⎠
θ = − tan −1 ⎜
PAGE
2
14
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
Mechatronics Engineering
The total moment acting at point O is:
+
↓MO = ∑MO
M O = 8 − 6(
12
5
)(4) + 6( )(5) − 4 cos 60 D (4) = −10.62
13
13
M O = −10.62 (kN .m) Å Ans
Solution 4-107
Comparing to the previous problem, there would be no change in total force, but the
Changes will present in the moment which act at point P.
The total forces acting on the x- axis are:
+
→ ∑ FRx = ∑ Fx
5
FRx = 6( ) − 4 cos 60 D = 0.30769 (kN )
13
The total forces acting on the y-axis are:
+
↑ ∑ FRy = ∑ Fy
12
FRy = 6( ) − 4 sin 60 D = 2.0744
13
FR = (0.30769) 2 + (2.0744) 2 = 2.10
(kN )
(kN ) Å-Ans
The angle of direction of total force (with respect to x-axis) is:
⎛ 2.0744 ⎞
D
⎟ = 81.6 ÅAns
⎝ 0.30769 ⎠
θ = − tan −1 ⎜
The total moment acting at point P is:
+
↓ M P = ∑ M P M P = 8 − 6(
M P = −16.8 (kN .m) Å Ans
12
5
)(7) + 6( )(5) − 4 cos 60 D (4) + 4 sin 60 D (3)
13
13
PAGE
3
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
PROBLEM 3
Exercise Problem 4-113 (Textbook – page 175)
Goal: Replace the three forces acting on the shaft by a single resultant force.
Specify where the force acts, measured from end B.
SOLUTION 3
The total forces acting on the x- axis are:
+
→ ∑ FRx = ∑ Fx
4
5
FRx = −500( ) + 260( ) = −300(lb)
5
13
The total forces acting on the y-axis are:
+
↑ ∑ FRy = ∑ Fy
3
12
FRy = −500( ) − 200 − 260( ) = −740(lb)
5
13
FR = (−300) 2 + (−740) 2 = 798(lb)
Å-Ans
The angle of direction of total force (with respect to x-axis) is:
⎛ 740 ⎞
D
⎟ = 67.9 ÅAns
⎝ 300 ⎠
θ = − tan −1 ⎜
The total moment acting at point B is:
+
3
12
↓ M RB = ∑ M B 740( x) = 500( )(9) + 200(6) + 260( )(4)
5
13
PAGE
4
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
x = −16.8( ft ) Å Ans
PROBLEM 4
The tension in cable AB is 400N, and the tension in cable CD is 600N
G
a) If you represent the forces exerted on the left post by the cables by a force F
G
G
G
acting at the origin O and a couple M , what are F and M ?
G
b) If you represent the forces exerted on the post by the cables by the force F
alone, where does its line of action intersect the y-axis?
SOLUTION 4a
From the right triangle, the angle between the positive x axis and the cable AB is:
⎛ 400 ⎞
D
⎟ = −26.6
⎝ 800 ⎠
θ = − tan −1 ⎜
The tension in AB is:
T AB = 400(iˆ cos(−26.6 D ) + ˆj sin(−26.6 D )) = 357.77iˆ − 178.89 ˆj ( N )
The angle between the positive x axis and the cable CD is:
PAGE
5
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
⎛ 300 ⎞
D
⎟ = −20.6
800
⎝
⎠
α = − tan −1 ⎜
The tension in CD is:
TCD = 600(iˆ cos(−20.6 D ) + ˆj sin( −20.6 D )) = 561.8iˆ − 210.67 ˆj (N )
The equivalent force acting at the origin O is the sum of the forces acting on the left post:
F = (357.77 + 561.8)iˆ + (−178.89 − 210.67) ˆj
F = 919.61iˆ − 389.6 ˆj Å Ans
The sum of moments acting on the left post is the product of the moment arm and the xcomponent of the tensions:
∑ M = −0.7(357.77)kˆ − 0.3(561.8)kˆ = −419kˆ
( N .m)
Check: The position vectors at the point of application are rAB = 0.7 ˆj , and rCD = 0.3 ˆj .
The sum of the moments is:
ˆj
ˆj
iˆ
kˆ
iˆ
kˆ
0.7
0+ 0
0.3
0
∑ M = (rAB × TAB ) + (rCD × TCD ) = 0
357.77 − 178.89 0 561.8 − 210.67 0
∑ M = −0.7(357.77)kˆ − 0.3(561.8)kˆ = −419kˆ
Å Ans
SOLUTION 4b
Check: The equivalent single force retains the same scalar components, but must act at
appoint that duplicates the sum of the moments. The distance on the y axis is the ratio of the
sum of the moments to the x-component of the equivalent force. Thus:
D=
419
= 0.456
919.6
( m)
PAGE
6
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
Check: The moment is:
ˆj
iˆ
kˆ
0 = −919.6 Dkˆ = −419kˆ
M 2 = rF × F = 0
D
919.6 − 389.6 0
from which
D=
419
= 0.456
919.6
(m) , check.
PROBLEM 5
Exercise Problem 4-131 (Textbook – page 178)
Goal: Replace the system of forces acting on slings by an equivalent force and
couple moment at point O.
SOLUTION 5
Force vectors:
F = {6.00kˆ}(kN )
1
F2 = 5(− cos 45 D sin 30 D iˆ + cos 45 D cos 30 D ˆj + sin 45 D kˆ)
F2 = {−1.768iˆ + 3.062 ˆj + 3.536kˆ} (kN )
PAGE
7
14
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
Mechatronics Engineering
F2 = 4(cos 60 D iˆ + cos 60 D ˆj + cos 45 D kˆ)
F2 = {2.00iˆ + 2.00 ˆj + 2.828kˆ} (kN )
Equivalent force and couple moment at point O:
FR = F1 + F2 + F3 = (−1.768 + 2.00)iˆ + (3.06 + 2.00) ˆj + (6 + 3.536 + 2.828)kˆ
FR = {0.232iˆ + 5.06 ˆj + 12.4kˆ} (kN ) Å Ans
The position vectors are r1 = {2i + 6 j}(m) and r1 = {4i}( m)
M Ro = ∑ M 0 = (r1 × F2 ) + (r2 × F2 )
iˆ
ˆj
= 2 6
kˆ
iˆ
ˆj
kˆ
0 +
4
0
0
0 0 6.00
− 1.768 3.062 3.536
M Ro = {36iˆ − 26.1 ˆj + 12.2kˆ} (kN .m) Å Ans
PROBLEM 6
Exercise Problem 4-134 (Textbook – page 178)
Goal: Determine the
equivalent resultant
force and specify its
location (x, y) on the
slab.
F1 = 20kN , F2 = 50kN
PAGE
8
14
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
Mechatronics Engineering
SOLUTION 6
+
↓ ∑ FR = ∑ Fz
FR = 20 + 50 + 20 + 50
FR = 140(kN ) Å Ans
M Roy = ∑ M y
140( x) = (50)(4) + 20(10) + 50(10)
M Rox = ∑ M x
− 140( y ) = −(50)(3) − 20(11) − 50(13)
x = 6.43(m) Å Ans
y = 7.29(m) Å Ans
PROBLEM 7
Exercise Problem 4-135 (Textbook – page 179)
Goal: Replace the two
wrenches and forces,
acting on the pipe
assembly, by an
equivalent resultant
force and couple
moment at point O.
SOLUTION 7
Force and moment vectors:
F1 = {300kˆ}( N ) , F3 = {100 ˆj}( N )
F2 = 5(cos 45 D iˆ − sin 45 D kˆ)
F2 = {−1.41.42iˆ − 141.42kˆ} ( N )
M 1 = {100kˆ}( N .m)
M 2 = 180(cos 45 D iˆ − sin 45 D kˆ)
PAGE
9
14
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
Mechatronics Engineering
M 2 = {−127.28iˆ − 127.28kˆ} ( N .m)
Equivalent force and couple moment at point O:
FR = F1 + F2 + F3 = 141.42iˆ + 100 ˆj + (300 − 141.42)kˆ
FR = 141iˆ + 100 ˆj + 159kˆ
( N ) Å Ans
The position vectors are: r1 = {0.5 j}( m) and r2 = {1.1 j}( m)
M Ro = ∑ M 0 = (r1 × F2 ) + (r2 × F2 ) + M 1 + M 2
iˆ
ˆj
= 0 0 .5
0
0
kˆ
iˆ
ˆj
kˆ
0 +
0
1 .1
0
141.42
0
− 141.42
300
M Ro = {122iˆ − 183kˆ}
+ 100kˆ + 127.28iˆ − 127.28kˆ
( N .m) Å Ans
PROBLEM 8
Exercise Problem 4-137 (Textbook – page 179)
Goal: Replace the
three forces, acting on
the plate by a wrench.
Specify the magnitude
of the force and couple
moment for the
wrench and the point
P(x,y) where its line of
action intersect the
plate.
PAGE
10
14
NAME & ID
Mechatronics Engineering
SOLUTION 8
FR = 500iˆ + 300 ˆj + 800kˆ ( N )
FR = (−500) 2 + (300) 2 + (800) 2 = 990( N )
u FR = 0.5051iˆ + 0.3030 ˆj + 0.808kˆ
M Rx ' = ∑ M x '
;
M Rx ' = 800(4 − y )
M Ry ' = ∑ M y '
;
M Ry ' = 800 x
M Rz ' = ∑ M z '
;
M Rz ' = 500 y + 300(6 − x)
Since M R also acts in the direction of u FR ,
M R (0.5051) = 800(4 − y )
M R (0.3030) = 800 x
M R (0.8081) = 500 y + 300(6 − x)
M R = 3.07(kN .m)
x = 1.16(m)
Å Ans
Å Ans
y = 2.06(m) Å Ans
DATE
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
PAGE
11
14
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
Mechatronics Engineering
Extra Practice Problems:
For those of you interested in a little bit more of practice.
PROBLEM 9
Exercise Problem 4-121 (Textbook - page 177)
Goal: Replace the
loading on the frame
by a single resultant
force. Specify where
its line of action
intersects member CD,
measured from end C.
SOLUTION 9
The total forces acting on the x- axis are:
+
→ ∑ FRx = ∑ Fx
4
FRx = −250( ) − 500(cos 60 D ) = −450( N )
5
The total forces acting on the y-axis are:
+
↑ ∑ FRy = ∑ Fy
3
FRy = −300 − 250( ) − 500(sin 60 D ) = −883.0127( N )
5
FR = (−450) 2 + (−883.0127) 2 = 991( N )
Å-Ans
The angle of direction of total force (with respect to x-axis) is:
⎛ 883.0127 ⎞
D
⎟ = 63.0 ÅAns
⎝ 450 ⎠
θ = − tan −1 ⎜
The total moment acting at point B is:
PAGE
12
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
M RA = ∑ M C
3
883.0127( x) = −400 + 300(3) + 250( )(6) + 500 cos 60 D (2) + 500 sin 60 D (1)
5
x = 2.64(m) Å Ans
PROBLEM 10
The handpiece of a miniature industrial grinder weighs 2.4N, and its center of gravity
is located on the y-axis. The head of the handpiece is offset in the x-z plane in such
a way that line BC forms an angle of 25D with the x-direction. Show that the weight
G
G
of the handpiece and the two couples M 1 and M 2 can be replaced with a single
G
G
equivalent force. Further assuming that M1 = 0.068 N ⋅ m and M 2 = 0.065 N ⋅ m ,
determine:
(a) the magnitude and the direction of the equivalent force
(b) the point where its line of action intersects the x-z plane
SOLUTION 10a
First we find the total forces and moments acting on the present system. We know
that the only force acting on the system is the weight of the handpiece of grinder, so
the total equivalent force should have exactly the same amount of force, with the
same direction, but acting on a point in the space. We represent both the moments in
vector form:
PAGE
13
14
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 4
SOLUTIONS
M 1 = 0.068kˆ
M 2 = 0.065(− cos 25 D iˆ − sin 25 D ˆj ) = −0.0589iˆ − 0.0275kˆ
M 1 + M 2 = −0.0589iˆ + (0.068kˆ − 0.0275)kˆ = −0.0589iˆ + 0.0405kˆ
(1)
We can see that the moment does not have any component in “y” direction. We should expect
that the line of action of the resultant force intersect the x-z plane at point D with the
following position vector:
G
r = rx iˆ + ry ˆj
and since the only force in the system is W, the resultant force should have the following
magnitude and direction:
G
FR = FRx iˆ + FRy ˆj + FRz kˆ = W (− ˆj )
G
FR = −2.4 ˆj ( N ) Å Ans
SOLUTION 10b
G
G
We find the total moment with respect to point O, using FR = W (− ˆj ) and r = rx iˆ + ry ˆj :
iˆ ˆj
G G
M R = r × FR = 0 W
rx 0
kˆ
0 = (Wrz )iˆ − (Wrx )kˆ = (2.4rz )iˆ − (2.4rx )kˆ
rz
(2)
By equating the above resultant moment (Equation (2)) with the present moment in the
system (Equation (1)), the components of the position vector can be found:
(2.4rz )iˆ − (2.4rx )kˆ = −0.0589iˆ + 0.0405kˆ
rz =
− 0.0589
= −0.02454(m) = −24.54(mm)
2.4
rx =
− 0.0405
= −0.01688(m) = −16.88(mm)
2.4
G
r = −24.5iˆ − 16.9 ˆj (mm) Å Ans
PAGE
14
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