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a)The problem sketch shows _____ loads. b)The moment load (the couple) has a magnitude of _____ N*m. c)The resultant moment induced by all of the loads has a magnitude of _____ N*m.

a)The problem sketch shows _____ loads. b)The moment load (the couple) has a magnitude of _____ N*m. c)The resultant moment induced by all of the loads has a magnitude of _____ N*m.

Elements Of Electromagnetics
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
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a)The problem sketch shows _____ loads.

b)The moment load (the couple) has a magnitude of _____ N*m.

c)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.
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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