# Force diagram for integrated exercise

Engineering_stud
Homework Statement:
A 6.00 m, 0.732 kg wire is used to support two uniform 235 N posts of equal length (the figure (Figure 1)). Assume that the wire is essentially horizontal and that the speed of sound is 344 m/s. A strong wind is blowing, causing the wire to vibrate in its 5th overtone.
Relevant Equations:
\sum tau = 0
Hello,

Can someone help me out with this one. I solved the exercise but didn't get the right answer so I went searching on internet for solutions. It seems something was wrong with my force-diagram. In my force diagram I had another force included excerted by the wire on the post, since the wire has weight that must be carried by the posts there has to be a reaction force hasn't it? Although in the solutions I added to the post the only force excerted by the wire on the post is horizontal tension force that cannot cancel out the weight of the wire.

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Delta2

Mentor
It seems something was wrong with my force-diagram. In my force diagram I had another force included excerted by the wire on the post, since the wire has weight that must be carried by the posts there has to be a reaction force hasn't it?
Since you are assuming the wire to be horizontal, you can ignore the weight of the wire (at least as far as the tension goes). (Compare the weight of the wire to that of the posts.)

Engineering_stud
Hi Doc Al

The weight of the wire of 7N seems (although much less) not insignificant however. I understand that since the posts are pulling the wire horizontally they cannot carry the weight of the wire, but at the mean time they are the only possibility to carry it. Is there still a mistake in my thinking?

Delta2
Mentor
Hi Doc Al

The weight of the wire of 7N seems (although much less) not insignificant however. I understand that since the posts are pulling the wire horizontally they cannot carry the weight of the wire, but at the mean time they are the only possibility to carry it. Is there still a mistake in my thinking?
Your thinking makes perfect sense. In "real life" the wire cannot be perfectly horizontal if it has weight and the posts must support the weight of the wire. But, for the purpose of this textbook problem, you can make the simplifying assumption that the wire is horizontal and its weight can be neglected in your force diagram.

Engineering_stud
Engineering_stud
Thank you, I get it. It was a bit confusing since it was the first time I encountered such a 'perfectly horizontal' wire.

Homework Helper
Gold Member
2022 Award
Since you are assuming the wire to be horizontal, you can ignore the weight of the wire (at least as far as the tension goes).
I'd say that differently. The justification for ignoring the wire's weight in the force diagram, if it is justified, is that it is small compared with the weights of the posts, and the simplification that the wire is straight follows from that.
I worry that although W is smallish compared with the weight of the post, it does not follow that its effect on the tension is correspondingly small.
The weight of the wire of 7N seems (although much less) not insignificant however. I understand that since the posts are pulling the wire horizontally they cannot carry the weight of the wire, but at the mean time they are the only possibility to carry it. Is there still a mistake in my thinking?
In reality, the wire will form a catenary. The horizontal component of tension will be constant along it, but the vertical component will be zero in the centre and nonzero at the ends.
Let one half of the wire have weight Ww, (horizontal) length Lw, horizontal tension Th, and vertical tension at the top end Tv. If the low point is distance h below the supported end then taking moments about that upper end, and pretending the weight is uniformly distributed horizontally:
##\frac 12W_wL_w=T_hh##
##T_v=W_w##.
A catenary takes the general form ##y=a\cosh(\frac xa)##, where the origin is at the lowest point. Hence here ##\frac{T_v}{T_h}=\sinh(\lambda)##, where λ satisfies ##\lambda\frac h{L_w}=\cosh(\lambda)##.
Further equations can be had from considering forces on the post. In principle, this allows us to find the tensions, but I don't think there's an analytic solution.
It is possible that doing the torque correctly, instead of assuming uniform horizontal distribution of weight, actually simplifies matters enough to allow an analytic solution.

Last edited:
Lnewqban
Homework Helper
Gold Member
Hi Doc Al

The weight of the wire of 7N seems (although much less) not insignificant however. I understand that since the posts are pulling the wire horizontally they cannot carry the weight of the wire, but at the mean time they are the only possibility to carry it. Is there still a mistake in my thinking?
You could add half of the weight of the wire to the weight of each post, and then calculate the moment each of those vertical forces create about the ground pivot.

Those two combined moments will induce a tension on the wire, which is the information needed to calculate the characteristics of the wind-driven vibration.

As it is represented, the mechanism is non-stable, unless the rotation of each pivot is limited in such way that the angle between each pole and ground is not much less than 57°.

haruspex
Homework Helper
Gold Member
2022 Award
You could add half of the weight of the wire to the weight of each post, and then calculate the moment each of those vertical forces create about the ground pivot.
Yes, that'd be close. The small inaccuracy is that the sag in the wire reduces the torque arm of the tension, increasing the tension needed to balance the weights.

Lnewqban