Figure 7 shows a transport helicopter carrying a load using four ropes which are

Question

Figure 7 shows a transport helicopter carrying a load using four ropes which are attached to the loading platform. Four ropes AO, BO, CO and DO are attached to the platform as shown in the plan, Figure 7(b). The tension in cables AO and BO are equal. The total mass of the and the platform is 1500kg and acting along line OG. Point G lies in the centre of the platform, vertically below point O, (Assume the helicopter is flying horizontally at a constant velocity.) Figure 7(a) (elevation view) Figure 7(b) (2 dimensional platform plan)

Explanation / Answer

First step is finding the length of all of the ropes:
Lag = sqrt(32^2+6^2) = 7.35m

Lbg = sqrt(32^2+6^2) = 7.35m

Ldg = sqrt(4.59^2+6^2) = 7.55m

Lcg = sqrt(4.42^2+6^2) = 7.454m

Now set up your relations:

Tag = Tbg

Since y direction-wise ag=bg, cg must = dg, therefore:

2.25/4.59 dg = 3/4.42 cg

dg = 1.385 cg

Now sum the x components of the tensions:

3/3sqrt2 ag + 3/3sqrt2 bg = 4/4.59 dg + 3.25/7.454 cg

ag = bg, so

6/3sqrt2 ag = 1.20665 cg + 3.25/7.454 cg

ag = 1.1615 cg

Now solve for sum of the forzes in the z-direction (up towards helicopter) = 0

1500*9.81=6/7.35 ag + 6/7.35 bg + 6/7.55 dg + 6/7.454 cg

plug in ; dg = 1.385 cg, ag = 1.1615 cg, bg = 1.1615 cg

1500*9.81 = 0.948cg + 0.948cg + 1.10066 cg + 0.8049 cg

cg = 3870 N

Back substitute into dg = 1.385 cg, ag = 1.1615 cg, bg = 1.1615 cg

dg = 1.385 * 3870 = 5361 N

ag = 1.1615 * 3870 = 4495 N

bg = 1.1615 * 3870 = 4495 N

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