The resultant moment induced by all of the loads has a magnitude of _____ N*m.
The resultant moment induced by all of the loads has a magnitude of _____ N*m.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
The resultant moment induced by all of the loads has a magnitude of _____ N*m.
![is more physically oriented. You
cally, whereas the scalar approach
Note that the vector approach
yields sign information automati-
which agrees with the sign conven-
60 Chapter 2 Force Systems
2m 5 m
60 N
Determine the resultant of the four forces and one couple which act on the
plate shown.
50 N
SAMPLE PROBLEM 2/9
45°
140 N-m
Solution. Point O is selected as a convenient reference point for the force-couple
system which is to represent the given system.
2 m
80 N
2 m
O30
40 N
1 mi
R, =
= 40 + 80 cos 30- 60 cos 45° = 66.9 N
(R, = EF)
= 50 + 80 sin 30° + 60 cos 45= 132.4N
R,
R= (66.9)2 + (132,4) = 148.3 N
Ans.
IR, = EF,)
(R = R + R,
R.
Ans.
R = 148.3 N
- 63.2
66.9
= tan-1
A = tan-1 132.4
Mo 60 sin 45 (7)
- 140 - 50(5) + 60 cos 45°(4)-
(a)
e = 63.2
O IM, = E(Fd))
|Mol =
237 N-m
= - 237 N.m
The force-couple system consisting of R and Mo is shown in Fig. a.
We now determine the final line of action of R such that R alone represents
the original system.
R = 148.3 N
Ans.
(Rd = Mol]
148.3d = 237
d = 1.600 m
Hence, the resultant R may be applied at any point on the line which makes a
63.2" angle with the x-axis and is tangent at point A to a circle of 1.600-m radius
with center 0, as shown in part b of the figure. We apply the equation Rd = Mo in
an absolute-value sense (ignoring any sign of Mo) and let the physics of the situa-
tion, as depicted in Fig, a, dictate the final placement of R. Had Mo been counter-
clockwise, the correct line of action of R would have been the tangent at point B.
The resultant R may also be located by determining its intercept distance b
to point C on the r-axis, Fig. c. With R, and R, acting through point C, only R,
(b)
63.2
1.600 m
A
---x
B
132.4x -- 66.9y =
-237
exerts a moment about O so that
(c)
y
R,b = |Mol
and
237
= 1,792 m
%3D
132.4
Alternatively, the y-intercept could have been obtained by noting that the mo-
ment about O would be due to R, only.
A more formal approach in determining the final line of action of R is to use
the vector expression
R/ -b-
Helpful Hints a
1 We note that the choice of point O as
a moment center eliminates any mo-
ments due to the two forces which
pass through O. Had the clockwise
sign convention been adopted, Mo
would have been +237 N m, with
the plus signm indicating a sense
which agrees with the sign conven-
tion. Either sign convention, Of
course, leads to the conclusion of a
clockwise moment Mo-
rx R = M,
where r = xi + yj is a position vector running from point O to any point on the
line of action of R. Substituting the vector expressions for r, R, and M, and car-
rying out the cross product result in
(xi + yj) x (66.9i - 132.4j) = -237k
!!
(132.4x - 66.9y)k = - 237k
Thus, the desired line of action, Fig. c, is given by
132.4x - 66.9y = -237
A By setting y = 0, we obtain x = -1.792 m, which agrees with our earlier calmle
tion of the distance b.
yields sign information automati
cally, whereas the scalar approach
should master both methods.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6cf78908-a1f9-4330-9ece-f4021df10613%2F4d685774-9484-4a26-ad12-cbb5d1e45808%2Fbwizlku_processed.jpeg&w=3840&q=75)
Transcribed Image Text:is more physically oriented. You
cally, whereas the scalar approach
Note that the vector approach
yields sign information automati-
which agrees with the sign conven-
60 Chapter 2 Force Systems
2m 5 m
60 N
Determine the resultant of the four forces and one couple which act on the
plate shown.
50 N
SAMPLE PROBLEM 2/9
45°
140 N-m
Solution. Point O is selected as a convenient reference point for the force-couple
system which is to represent the given system.
2 m
80 N
2 m
O30
40 N
1 mi
R, =
= 40 + 80 cos 30- 60 cos 45° = 66.9 N
(R, = EF)
= 50 + 80 sin 30° + 60 cos 45= 132.4N
R,
R= (66.9)2 + (132,4) = 148.3 N
Ans.
IR, = EF,)
(R = R + R,
R.
Ans.
R = 148.3 N
- 63.2
66.9
= tan-1
A = tan-1 132.4
Mo 60 sin 45 (7)
- 140 - 50(5) + 60 cos 45°(4)-
(a)
e = 63.2
O IM, = E(Fd))
|Mol =
237 N-m
= - 237 N.m
The force-couple system consisting of R and Mo is shown in Fig. a.
We now determine the final line of action of R such that R alone represents
the original system.
R = 148.3 N
Ans.
(Rd = Mol]
148.3d = 237
d = 1.600 m
Hence, the resultant R may be applied at any point on the line which makes a
63.2" angle with the x-axis and is tangent at point A to a circle of 1.600-m radius
with center 0, as shown in part b of the figure. We apply the equation Rd = Mo in
an absolute-value sense (ignoring any sign of Mo) and let the physics of the situa-
tion, as depicted in Fig, a, dictate the final placement of R. Had Mo been counter-
clockwise, the correct line of action of R would have been the tangent at point B.
The resultant R may also be located by determining its intercept distance b
to point C on the r-axis, Fig. c. With R, and R, acting through point C, only R,
(b)
63.2
1.600 m
A
---x
B
132.4x -- 66.9y =
-237
exerts a moment about O so that
(c)
y
R,b = |Mol
and
237
= 1,792 m
%3D
132.4
Alternatively, the y-intercept could have been obtained by noting that the mo-
ment about O would be due to R, only.
A more formal approach in determining the final line of action of R is to use
the vector expression
R/ -b-
Helpful Hints a
1 We note that the choice of point O as
a moment center eliminates any mo-
ments due to the two forces which
pass through O. Had the clockwise
sign convention been adopted, Mo
would have been +237 N m, with
the plus signm indicating a sense
which agrees with the sign conven-
tion. Either sign convention, Of
course, leads to the conclusion of a
clockwise moment Mo-
rx R = M,
where r = xi + yj is a position vector running from point O to any point on the
line of action of R. Substituting the vector expressions for r, R, and M, and car-
rying out the cross product result in
(xi + yj) x (66.9i - 132.4j) = -237k
!!
(132.4x - 66.9y)k = - 237k
Thus, the desired line of action, Fig. c, is given by
132.4x - 66.9y = -237
A By setting y = 0, we obtain x = -1.792 m, which agrees with our earlier calmle
tion of the distance b.
yields sign information automati
cally, whereas the scalar approach
should master both methods.
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