3. Replace the two forces by an equivalent resultant force and couple moment at point O if F 20 lb. (Hibbeler, 2010) Ans. Fa-29.9 lb at 0-78.4° from the positive x-axis, Ma-155.6 lb-in S
3. Replace the two forces by an equivalent resultant force and couple moment at point O if F 20 lb. (Hibbeler, 2010) Ans. Fa-29.9 lb at 0-78.4° from the positive x-axis, Ma-155.6 lb-in S
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
How to answer this number 3 number 3 in lesson
![R = R + R and 0 = tan
R.
The location of the line of action of the resultant with respect to an arbitrary reference point, say O, can be
determined by applying the principle of moments. Hence,
M =EM
SAMPLE PROBLEMS
1. Three forces are applied to a beam as shown in the figure. Replace the three forces by an equivalent force-
couple system at point C. (Riley, 1993)
80 Ib
75 lb
90 lb
14 in
16 in
To in
20 in
2. A couple of magnitude M = 540 mN-m and three forces shown are applied to an angle bracket. Find the
resultant of this system of forces and locate the points where the line of action of the resultant intersects the
line AB and line BC. (Beer, 1998)
10N
30 N
120 mm
80 mm
45 N
3. Determine the magnitudes of F, and F, and the direction of F, so that the loading creates a zero resultant
force and couple on the wheel. (Hibbeler, 1986)
D
F2
0.75 f
80 Ib-ft
60 lb
B.
30 Ib
4. Determine the magnitude and direction 0 of force F and the couple moment M such that the loading system
is equivalent to a resultant force of 600 N, acting vertically downward at O, and a clockwise couple moment
of 4000 N-m. (Hibbeler, 1986)
800 N
Im
300 N
EXERCISES
1. Replace the loading system by an equivalent resultant force and specify where the resultant's line of action
intersects the beam measured from O. (Hibbeler, 2010) Ans. Fg 1250 lb; x 6 ft
ITI
500 Ib
S00 ib
250 ib
3 ft-3t--3n-3t-
2. Replace the loading system acting on the post by a single resultant force and specify its point of application
on the post measured from point O. (Hibbeler, 1986) Ans. 69.255 lb : 0= 18,96: 3.22 t
20 Ib
70 th
60 Ib
45
3. Replace the two forces by an equivalent resultant force and couple moment at point O if F = 20 lb.
(Hibbeler, 2010) Ans. FR-29,9 lb at 0= 78.4° from the positive x-axis, M = 155.6 lb-in)
15 in
40
-2 in](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffc69e191-1797-46b9-8a0a-b557825582f3%2Fa0eaa705-b8d0-488a-8b0e-860c141c195d%2Fnbfg1o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:R = R + R and 0 = tan
R.
The location of the line of action of the resultant with respect to an arbitrary reference point, say O, can be
determined by applying the principle of moments. Hence,
M =EM
SAMPLE PROBLEMS
1. Three forces are applied to a beam as shown in the figure. Replace the three forces by an equivalent force-
couple system at point C. (Riley, 1993)
80 Ib
75 lb
90 lb
14 in
16 in
To in
20 in
2. A couple of magnitude M = 540 mN-m and three forces shown are applied to an angle bracket. Find the
resultant of this system of forces and locate the points where the line of action of the resultant intersects the
line AB and line BC. (Beer, 1998)
10N
30 N
120 mm
80 mm
45 N
3. Determine the magnitudes of F, and F, and the direction of F, so that the loading creates a zero resultant
force and couple on the wheel. (Hibbeler, 1986)
D
F2
0.75 f
80 Ib-ft
60 lb
B.
30 Ib
4. Determine the magnitude and direction 0 of force F and the couple moment M such that the loading system
is equivalent to a resultant force of 600 N, acting vertically downward at O, and a clockwise couple moment
of 4000 N-m. (Hibbeler, 1986)
800 N
Im
300 N
EXERCISES
1. Replace the loading system by an equivalent resultant force and specify where the resultant's line of action
intersects the beam measured from O. (Hibbeler, 2010) Ans. Fg 1250 lb; x 6 ft
ITI
500 Ib
S00 ib
250 ib
3 ft-3t--3n-3t-
2. Replace the loading system acting on the post by a single resultant force and specify its point of application
on the post measured from point O. (Hibbeler, 1986) Ans. 69.255 lb : 0= 18,96: 3.22 t
20 Ib
70 th
60 Ib
45
3. Replace the two forces by an equivalent resultant force and couple moment at point O if F = 20 lb.
(Hibbeler, 2010) Ans. FR-29,9 lb at 0= 78.4° from the positive x-axis, M = 155.6 lb-in)
15 in
40
-2 in
![R = R + R and 0 = tan
R.
The location of the line of action of the resultant with respect to an arbitrary reference point, say O, can be
determined by applying the principle of moments. Hence,
M =EM
SAMPLE PROBLEMS
1. Three forces are applied to a beam as shown in the figure. Replace the three forces by an equivalent force-
couple system at point C. (Riley, 1993)
80 Ib
75 lb
90 lb
14 in
16 in
To in
20 in
2. A couple of magnitude M = 540 mN-m and three forces shown are applied to an angle bracket. Find the
resultant of this system of forces and locate the points where the line of action of the resultant intersects the
line AB and line BC. (Beer, 1998)
10N
30 N
120 mm
80 mm
45 N
3. Determine the magnitudes of F, and F, and the direction of F, so that the loading creates a zero resultant
force and couple on the wheel. (Hibbeler, 1986)
D
F2
0.75 f
80 Ib-ft
60 lb
B.
30 Ib
4. Determine the magnitude and direction 0 of force F and the couple moment M such that the loading system
is equivalent to a resultant force of 600 N, acting vertically downward at O, and a clockwise couple moment
of 4000 N-m. (Hibbeler, 1986)
800 N
Im
300 N
EXERCISES
1. Replace the loading system by an equivalent resultant force and specify where the resultant's line of action
intersects the beam measured from O. (Hibbeler, 2010) Ans. Fg 1250 lb; x 6 ft
ITI
500 Ib
S00 ib
250 ib
3 ft-3t--3n-3t-
2. Replace the loading system acting on the post by a single resultant force and specify its point of application
on the post measured from point O. (Hibbeler, 1986) Ans. 69.255 lb : 0= 18,96: 3.22 t
20 Ib
70 th
60 Ib
45
3. Replace the two forces by an equivalent resultant force and couple moment at point O if F = 20 lb.
(Hibbeler, 2010) Ans. FR-29,9 lb at 0= 78.4° from the positive x-axis, M = 155.6 lb-in)
15 in
40
-2 in](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffc69e191-1797-46b9-8a0a-b557825582f3%2Fa0eaa705-b8d0-488a-8b0e-860c141c195d%2Ftn2vex_processed.jpeg&w=3840&q=75)
Transcribed Image Text:R = R + R and 0 = tan
R.
The location of the line of action of the resultant with respect to an arbitrary reference point, say O, can be
determined by applying the principle of moments. Hence,
M =EM
SAMPLE PROBLEMS
1. Three forces are applied to a beam as shown in the figure. Replace the three forces by an equivalent force-
couple system at point C. (Riley, 1993)
80 Ib
75 lb
90 lb
14 in
16 in
To in
20 in
2. A couple of magnitude M = 540 mN-m and three forces shown are applied to an angle bracket. Find the
resultant of this system of forces and locate the points where the line of action of the resultant intersects the
line AB and line BC. (Beer, 1998)
10N
30 N
120 mm
80 mm
45 N
3. Determine the magnitudes of F, and F, and the direction of F, so that the loading creates a zero resultant
force and couple on the wheel. (Hibbeler, 1986)
D
F2
0.75 f
80 Ib-ft
60 lb
B.
30 Ib
4. Determine the magnitude and direction 0 of force F and the couple moment M such that the loading system
is equivalent to a resultant force of 600 N, acting vertically downward at O, and a clockwise couple moment
of 4000 N-m. (Hibbeler, 1986)
800 N
Im
300 N
EXERCISES
1. Replace the loading system by an equivalent resultant force and specify where the resultant's line of action
intersects the beam measured from O. (Hibbeler, 2010) Ans. Fg 1250 lb; x 6 ft
ITI
500 Ib
S00 ib
250 ib
3 ft-3t--3n-3t-
2. Replace the loading system acting on the post by a single resultant force and specify its point of application
on the post measured from point O. (Hibbeler, 1986) Ans. 69.255 lb : 0= 18,96: 3.22 t
20 Ib
70 th
60 Ib
45
3. Replace the two forces by an equivalent resultant force and couple moment at point O if F = 20 lb.
(Hibbeler, 2010) Ans. FR-29,9 lb at 0= 78.4° from the positive x-axis, M = 155.6 lb-in)
15 in
40
-2 in
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