The section of rope between Member-AD and Sheave-F is _____. options:1a two-force member2a multi-force member3a zero-force member4a torsional member i E E206 Chapter 4 Structures3 mSAMPLE PROBLEM 4/6The frame supports the 400-kg load in the manner shown. Neglect theweights of the members compared with the forces induced by the load and com-pute the horizontal and vertical components of all forces acting on each of the1.5 m0.5 m0.5 mmembers.1.5 mSolution. We observe first that the three supporting members which consti-O tute the frame form a rigid assembly that can be analyzed as a single unit. Wealso observe that the arrangement of the external supports makes the frame sta-tically determinate.From the free-body diagram of the entire frame we determine the external1.5 m400 kgHelpful Hintsreactions. Thus,1 We see that the frame corresponds tothe category illustrated in Fig. 4/14a,2 Without this observation, the prob-lem solution would be much longer,because the three equilibrium equa-tions for member BF would containfour unknowns: B, By, E, and E.Note that the direction of the linejoining the two points of force appli-cation, and not the shape of the mem-ber, determines the direction of theforces acting on a two-force member.[0 = "W3)5.5(0.4)(9.81) - 5D = 0D = 4.32 kNA- 4.32 = 0A, = 4.32 kN[o = "3)A, - 3.92 = 0A, = 3.92 kN[o = 13]Next we dismember the frame and draw a separate free-body diagram ofeach member. The diagrams are arranged in their approximate relative positionsto aid in keeping track of the common forces of interaction. The external reac-tions just obtained are entered onto the diagram for AD. Other known forces arethe 3.92-kN forces exerted by the shaft of the pulley on the member BF, as ob-tained from the free-body diagram of the pulley. The cable tension of 3.92 kN isalso shown acting on AD at its attachment point.Next, the components of all unknown forces are shown on the diagrams.2 Here we observe that CE is a two-force member. The force components on CEhave equal and opposite reactions, which are shown on BF at E and on AD at C.We may not recognize the actual sense of the components at B at first glance, sothey may be arbitrarily but consistently assigned.The solution may proceed by use of a moment equation about B or E formember BF, followed by the two force equations. Thus,3.92(5) – E,(3) = 0[0 = "W3)[0 = 3]B, + 3.92 – 13.08/2 = 0E, = 13.08 kNAns.%3D0.4(9.81)B, = 2.62 kNAns.N可268=B, + 3.92 – 13.08 = 0B = 9.15 kNAns.[0 = "3]Positive numerical values of the unknowns mean that we assumed their direc-tions correctly on the free-body diagrams. The value of C, = E, = 13.08 kN ob-tained by inspection of the free-body diagram of CE is now entered onto thediagram for AD, along with the values of B, and By just determined. The equa-tions of equilibrium may now be applied to member AD as a check, since all theforces acting on it have already been computed. The equations give3.92 kNA, = 3.92 kNNY 76'E= "v3.92 kN3.92 kN Y3.92 kN[0 = °W3)4.32(3.5) + 4.32(1.5) – 3.92(2) – 9.15(1.5) = 04.32 - 13.08 + 9.15 + 3.92 + 4.32 = 0[0 = "3)-13.08/2 + 2.62 + 3.92 = 0[0 = 3)4.32 kN

Question

The section of rope between Member-AD and Sheave-F is _____. options:1a two-force member2a multi-force member3a zero-force member4a torsional member i E E206 Chapter 4 Structures3 mSAMPLE PROBLEM 4/6The frame supports the 400-kg load in the manner shown. Neglect theweights of the members compared with the forces induced by the load and com-pute the horizontal and vertical components of all forces acting on each of the1.5 m0.5 m0.5 mmembers.1.5 mSolution. We observe first that the three supporting members which consti-O tute the frame form a rigid assembly that can be analyzed as a single unit. Wealso observe that the arrangement of the external supports makes the frame sta-tically determinate.From the free-body diagram of the entire frame we determine the external1.5 m400 kgHelpful Hintsreactions. Thus,1 We see that the frame corresponds tothe category illustrated in Fig. 4/14a,2 Without this observation, the prob-lem solution would be much longer,because the three equilibrium equa-tions for member BF would containfour unknowns: B, By, E, and E.Note that the direction of the linejoining the two points of force appli-cation, and not the shape of the mem-ber, determines the direction of theforces acting on a two-force member.[0 = &#34;W3)5.5(0.4)(9.81) - 5D = 0D = 4.32 kNA- 4.32 = 0A, = 4.32 kN[o = &#34;3)A, - 3.92 = 0A, = 3.92 kN[o = 13]Next we dismember the frame and draw a separate free-body diagram ofeach member. The diagrams are arranged in their approximate relative positionsto aid in keeping track of the common forces of interaction. The external reac-tions just obtained are entered onto the diagram for AD. Other known forces arethe 3.92-kN forces exerted by the shaft of the pulley on the member BF, as ob-tained from the free-body diagram of the pulley. The cable tension of 3.92 kN isalso shown acting on AD at its attachment point.Next, the components of all unknown forces are shown on the diagrams.2 Here we observe that CE is a two-force member. The force components on CEhave equal and opposite reactions, which are shown on BF at E and on AD at C.We may not recognize the actual sense of the components at B at first glance, sothey may be arbitrarily but consistently assigned.The solution may proceed by use of a moment equation about B or E formember BF, followed by the two force equations. Thus,3.92(5) – E,(3) = 0[0 = &#34;W3)[0 = 3]B, + 3.92 – 13.08/2 = 0E, = 13.08 kNAns.%3D0.4(9.81)B, = 2.62 kNAns.N可268=B, + 3.92 – 13.08 = 0B = 9.15 kNAns.[0 = &#34;3]Positive numerical values of the unknowns mean that we assumed their direc-tions correctly on the free-body diagrams. The value of C, = E, = 13.08 kN ob-tained by inspection of the free-body diagram of CE is now entered onto thediagram for AD, along with the values of B, and By just determined. The equa-tions of equilibrium may now be applied to member AD as a check, since all theforces acting on it have already been computed. The equations give3.92 kNA, = 3.92 kNNY 76'E= &#34;v3.92 kN3.92 kN Y3.92 kN[0 = °W3)4.32(3.5) + 4.32(1.5) – 3.92(2) – 9.15(1.5) = 04.32 - 13.08 + 9.15 + 3.92 + 4.32 = 0[0 = &#34;3)-13.08/2 + 2.62 + 3.92 = 0[0 = 3)4.32 kN

Accepted Answer

Section of rope between member AD and  Sheave F is a TWO FORCE MEMBER .

1	a two-force member
2	a multi-force member
3	a zero-force member
4	a torsional member

The section of rope between Member-AD and Sheave-F is _____. options: 1 a two-force member 2 a multi-force member 3 a zero-force member 4 a torsional member