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Rod structure

I have a problem understanding these basics. What is the tension in the ground rod due to the forces exerted by the slanted rods?

Please feel free to correct me at any point.

Rods are idealized and are long, light and thin, welded together to form a static structure in the form of an equilateral triangle. On the left, the weight of the load $F$ compresses the two standing slanted rods, with compression in both of them equalling $F / (2 sin 30^\circ)$. They in turn pull apart the ground rod with tension that equals the horizontal component of $F_1$ -- or is it two times that amount -- so $F / (2 ctg 30^\circ)$ or is it $F / (ctg 30^\circ)$?

On the right, it's no different for the slanted rods, but the ground rod here is not in equilibrium because of the non-existing horizontal force of the right rod but because of friction. Yet the tension in this rod is the same as it is in the example on the left?

Update

I can see my drawing isn't too helpful, yet I don't think it matters much how it's drawn; the rods are long, thin, light, and welded together. True, there should be any contact between the two standing rods and the ground (in the left case).

What's the tension then in the pink rod below. Is it the force exerted by each standing rod or is it twice that amount since each rod is pulling it apart?

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  • $\begingroup$ Is there any friction on the ground? $\endgroup$ Commented Jan 31, 2016 at 20:16
  • $\begingroup$ There could be, yes. It's static and still in any case. See update. $\endgroup$ Commented Jan 31, 2016 at 20:53

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Is the pink object a rod or not?
If it is a rod then the ground only needs to exert upward vertical forces.

My advice is that you draw the appropriate FBDs and then hopefully the setting up of equations becomes easier.

In the diagram on the right if friction provides the horizontal force at the bottom of the right hand rod why cannot that also happen for the left hand rod?

enter image description here

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  • $\begingroup$ Yes, the objects are all rods. Is the tension in the ground rod in your solution $x$ or $2 x$? $\endgroup$ Commented Jan 31, 2016 at 19:26
  • $\begingroup$ It is obvious from the diagram? $\endgroup$ Commented Jan 31, 2016 at 19:39
  • $\begingroup$ You need to add the forces (if any) on the bottom rod also. $\endgroup$ Commented Jan 31, 2016 at 20:01
  • $\begingroup$ @ja72 From the diagram I was not entirely sure what was going on at the bottom so I left those forces out. For example it is not clear whther the ground directly or the horizontal rod at the bottom provides the upward force $F$. $\endgroup$ Commented Jan 31, 2016 at 20:10
  • $\begingroup$ You can ignore the floor, and just assume the left and right corners are on rollers (with friction maybe). Or are they pinned? Hmm, difficult to ascertain from the diagram. $\endgroup$ Commented Jan 31, 2016 at 20:11
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(Ignoring friction) if the connections are welded, then the bottom rod is irrelevant to balancing the forces (also ignoring deflection effects). If you removed the bottom rod entirely the diagram would remain the same. With pinned connections the second figure is unstable and will collapse, and there will be tension in the bottom rod in the first figure. In the welded situation the forces on the second figure are balanced with torques that are not relevant in the first figure (due to symmetry).

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  • $\begingroup$ Left example, welded. The force exerted by the load on the two standing rods is obviously transmitted to the ground rod and it's pulling it apart. There is obviously some tension in it. $\endgroup$ Commented Feb 1, 2016 at 14:27
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    $\begingroup$ Nope, the tension in the bottom rod depends upon the torque absorbed in the top weld. The only way for there to be tension in the bottom rod is for thereto be deflection, which is not accounted for in standard mechanics. Think of it this way, take a coat hanger and clip off the bottom. The remaining top does not collapse. You may now suspend the clipped off section on the clipped ends and have no difference in the forces, except for minor deflections due to the weight of the suspended rod section. $\endgroup$ Commented Feb 2, 2016 at 18:59
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I used Force Effect to demonstrate the effect. One one case there is tension and on the other there isn't

pic

In the second case,there is nothing pulling on the rod to create tension. The forces from the side members are supported by the corners (and friction) of the triangle.

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