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This is an off-level rigging problem. When solving for the tension forces in the red slings (lifting from point A), can the assumption be made that the y force vector at points B and C are each half of the weight of the object? Or must one calculate the force vectors at point A? In which case the y vectors are not symmetrical.

Off-level pick points

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No.

To convince yourself, why an equal distribution of the vertical component is wrong, try looking at the corresponding horizontal components (Kamran has calculated these values in his answer): The horizontal component at C is twice as large as the horizontal component at B, which results in a net horizontal load on the object, giving it a constant, horizontal acceleration off to the side. (And there is no moment equilibrium, so it would also rotate.)

I think, the easiest way to calculate the correct distribution of the vertical component is to extend the line A-C until it reaches the horizontal level of B. I call this point C'. The horizontal distance from C' to the center of gravity is $10'\cdot\frac{12'}{6'}=20'$, which is twice the distance from B to the center of gravity, so the vertical component of the reaction at B is twice the corresponding value at C or $\frac{2}{3}$ of the total weight.

Drawing the force vectors at point A and using horizontal and vertical equilibrium as you suggest in your question is also a good way to do it.

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No. The issue here is that the problem appears to be overconstrained at first glance. So you can't even assume that there is a physically valid solution at all. How many equations do you have? How many unknowns? Has it been constructed so that two of the equations are linear multiples of each other and one can be dropped?

Trying to take shortcuts like this will bite you in the butt someday.

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  • $\begingroup$ The reason the horizontal components are not on balance is because of the load's L shape there the vertical section of weight acting as a cantilever beam has horizontal reaction which has to add to sum of horizontal forces. We can't sum the horizontal forces at just one node. We add all nodes, A, B and C. The diagram indicates the location of CG, so if we add all horizontal forces at A,B,C the sum is equal to zero. $\endgroup$ – kamran Apr 10 '19 at 22:37
  • $\begingroup$ @kamran That works if there is a valid solution. We don't know if there is a valid solution. First, determine that there is. Then, justify the assumptions that apply when there is a solution. You can not just look at this diagram and assume anything. It might be an impossible arrangement. $\endgroup$ – Phil Sweet Apr 11 '19 at 2:21
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    $\begingroup$ Respectfully I don't understand what you mean by valid solution. I calculated the forces that will keep the load at equilibrium with these configuration. It is a very basic static problem. I don't know why many have difficulty with it. If one apply these forces to the cables the load will hang in equilibrium. If you still need help please clarify. $\endgroup$ – kamran Apr 11 '19 at 2:34
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Yes, we know the Y components of the cables are equal and each half of the weight because the axis of A passes through CG of the load.

The X components are as follows:

$$ \text{For B, }\quad X_B = \frac{10 \cdot W}{12\cdot 2} $$ $$ \text{For C,} \quad X_C =\frac{10 \cdot W}{6 \cdot 2} $$ Edit


After comments indicating my errors by @PhilSweet and @ingen0rd I thought best way is to calculate the tension in the cables by taking moment of forces about point B. $$ \Sigma M _B =0 \quad \therefore -W\cdot10 + F\ C_{vertical} \cdot 20 + F\ C_{ horizontal}\cdot6 =0 $$ $$\\ F \ C_{vertical} =6/10 \cdot F \ C_{horizontal} $$ So we have two equations and two unknowns which we can solve.

$$ F\ C_{horiz}\cdot6 +6/10*F\ C_{horiz}\cdot 20 -10W =0 $$

$$ 18 F\ C_{horiz}= 10W ,\quad F\ C_{horiz}= W/1.8 $$ $$F\ C_{vert}= 6W/18 = W/3$$

And so $F\ B_{ vert}=2W/3 $

Please check my numbers as always.

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  • $\begingroup$ Can you help me understand why summing the forces at A would not work? $\endgroup$ – Brice Edward Furr Apr 10 '19 at 15:30
  • $\begingroup$ @BriceEdwardFurr, the cables forces will be equal only when they are rigged symmetrically and the load is balanced with its CG right under A. But they always have Y component equal. the greater the angle of the cable with vertical axis, the greater is X component. Hence the greater the magnitude of the cable force. Ultimately if one cable approaches horizontal, its tension approaches infinity. $\endgroup$ – kamran Apr 10 '19 at 15:52
  • $\begingroup$ 6 FChoriz + 20*6/10 FChoriz should be 18 FChoriz, not 12. $\endgroup$ – ingenørd Apr 11 '19 at 7:10

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