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. Youcally, whereas the scalar approachNote that the vector approachyields sign information automati-which agrees with the sign conven-60 Chapter 2 Force Systems2m 5 m60 NDetermine the resultant of the four forces and one couple which act on theplate shown.50 NSAMPLE PROBLEM 2/945°140 N-mSolution. Point O is selected as a convenient reference point for the force-couplesystem which is to represent the given system.2 m80 N2 mO3040 N1 miR, == 40 + 80 cos 30- 60 cos 45° = 66.9 N(R, = EF)= 50 + 80 sin 30° + 60 cos 45= 132.4NR,R= (66.9)2 + (132,4) = 148.3 NAns.IR, = EF,)(R = R + R,R.Ans.R = 148.3 N- 63.266.9= tan-1A = tan-1 132.4Mo 60 sin 45 (7)- 140 - 50(5) + 60 cos 45°(4)-(a)e = 63.2O IM, = E(Fd))|Mol =237 N-m= - 237 N.mThe 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 representsthe original system.R = 148.3 NAns.(Rd = Mol]148.3d = 237d = 1.600 mHence, the resultant R may be applied at any point on the line which makes a63.2" angle with the x-axis and is tangent at point A to a circle of 1.600-m radiuswith center 0, as shown in part b of the figure. We apply the equation Rd = Mo inan 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 bto point C on the r-axis, Fig. c. With R, and R, acting through point C, only R,(b)63.21.600 mA---xB132.4x -- 66.9y =-237exerts a moment about O so that(c)yR,b = |Moland237= 1,792 m%3D132.4Alternatively, 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 usethe vector expressionR/ -b-Helpful Hints a1 We note that the choice of point O asa moment center eliminates any mo-ments due to the two forces whichpass through O. Had the clockwisesign convention been adopted, Mowould have been +237 N m, withthe plus signm indicating a sensewhich agrees with the sign conven-tion. Either sign convention, Ofcourse, leads to the conclusion of aclockwise moment Mo-rx R = M,where r = xi + yj is a position vector running from point O to any point on theline 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 = - 237kThus, the desired line of action, Fig. c, is given by132.4x - 66.9y = -237A By setting y = 0, we obtain x = -1.792 m, which agrees with our earlier calmletion of the distance b.yields sign information automatically, whereas the scalar approachshould master both methods.

Question

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. Youcally, whereas the scalar approachNote that the vector approachyields sign information automati-which agrees with the sign conven-60 Chapter 2 Force Systems2m 5 m60 NDetermine the resultant of the four forces and one couple which act on theplate shown.50 NSAMPLE PROBLEM 2/945°140 N-mSolution. Point O is selected as a convenient reference point for the force-couplesystem which is to represent the given system.2 m80 N2 mO3040 N1 miR, == 40 + 80 cos 30- 60 cos 45° = 66.9 N(R, = EF)= 50 + 80 sin 30° + 60 cos 45= 132.4NR,R= (66.9)2 + (132,4) = 148.3 NAns.IR, = EF,)(R = R + R,R.Ans.R = 148.3 N- 63.266.9= tan-1A = tan-1 132.4Mo 60 sin 45 (7)- 140 - 50(5) + 60 cos 45°(4)-(a)e = 63.2O IM, = E(Fd))|Mol =237 N-m= - 237 N.mThe 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 representsthe original system.R = 148.3 NAns.(Rd = Mol]148.3d = 237d = 1.600 mHence, the resultant R may be applied at any point on the line which makes a63.2&#34; angle with the x-axis and is tangent at point A to a circle of 1.600-m radiuswith center 0, as shown in part b of the figure. We apply the equation Rd = Mo inan 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 bto point C on the r-axis, Fig. c. With R, and R, acting through point C, only R,(b)63.21.600 mA---xB132.4x -- 66.9y =-237exerts a moment about O so that(c)yR,b = |Moland237= 1,792 m%3D132.4Alternatively, 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 usethe vector expressionR/ -b-Helpful Hints a1 We note that the choice of point O asa moment center eliminates any mo-ments due to the two forces whichpass through O. Had the clockwisesign convention been adopted, Mowould have been +237 N m, withthe plus signm indicating a sensewhich agrees with the sign conven-tion. Either sign convention, Ofcourse, leads to the conclusion of aclockwise moment Mo-rx R = M,where r = xi + yj is a position vector running from point O to any point on theline 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 = - 237kThus, the desired line of action, Fig. c, is given by132.4x - 66.9y = -237A By setting y = 0, we obtain x = -1.792 m, which agrees with our earlier calmletion of the distance b.yields sign information automatically, whereas the scalar approachshould master both methods.

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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 _ Nm. c)The resultant moment induced by all of the loads has a magnitude of ___ Nm.