Force diagram for integrated exercise

In summary, the wind-driven vibration can be characterized by the following equation: the tension in the wire is proportional to the square of the distance from the ground pivot.
  • #1
Engineering_stud
3
2
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.

Thank you in advance!
 

Attachments

  • wires2.PNG
    wires2.PNG
    14.3 KB · Views: 116
  • wires.PNG
    wires.PNG
    5.6 KB · Views: 112
  • Like
Likes Delta2
Physics news on Phys.org
  • #2
Engineering_stud said:
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.)
 
  • #3
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?
 
  • Like
Likes Delta2
  • #4
Engineering_stud said:
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.
 
  • Like
Likes Engineering_stud
  • #5
Thank you, I get it. It was a bit confusing since it was the first time I encountered such a 'perfectly horizontal' wire.
 
  • #6
Doc Al said:
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.
Engineering_stud said:
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:
  • Like
Likes Lnewqban
  • #7
Engineering_stud said:
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°.
 
  • Like
Likes haruspex
  • #8
Lnewqban said:
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.
 
  • Like
Likes Lnewqban
  • #9
If one worries about a (slightly) sagging wire, wouldn't one also have to consider that the poles oscillate as the wire vibrates? At least one would have to show that resonances may be ignored.
 
  • Like
Likes Lnewqban
  • #10
kuruman said:
If one worries about a (slightly) sagging wire, wouldn't one also have to consider that the poles oscillate as the wire vibrates? At least one would have to show that resonances may be ignored.
I was worried that the weight of the wire might disproportionately increase the tension, but I no longer think so.
The biggest flaw in the problem from a practical perspective was pointed out by @Lnewqban in post #7: the whole system is unstable and would collapse to one side. That makes an analysis of the oscillations of the poles somewhat irrelevant.
 

Related to Force diagram for integrated exercise

1. What is a force diagram for integrated exercise?

A force diagram for integrated exercise is a visual representation of the different forces acting on an object during an exercise. It includes all external forces such as gravity, friction, and applied forces, and helps to understand the overall movement and mechanics of the exercise.

2. How is a force diagram for integrated exercise useful?

A force diagram for integrated exercise is useful because it allows scientists to analyze and understand the different forces involved in an exercise, which can help in designing more effective and safe workout routines. It also helps in identifying potential areas of improvement for athletes or individuals performing the exercise.

3. How do you create a force diagram for integrated exercise?

To create a force diagram for integrated exercise, you first need to identify all the external forces acting on the object. Then, draw a diagram with arrows representing each force, making sure that the length and direction of the arrows are proportional to the magnitude and direction of the force. Finally, label each force and provide a key for easy understanding.

4. Can a force diagram for integrated exercise be used for any type of exercise?

Yes, a force diagram for integrated exercise can be used for any type of exercise as long as there are external forces acting on the object. It can be used for both simple and complex exercises, such as weightlifting, running, or even a simple push-up.

5. What are the benefits of using a force diagram for integrated exercise in research?

Using a force diagram for integrated exercise in research allows scientists to gather more accurate and detailed data on the forces involved in an exercise. This can help in developing new training methods, improving performance, and preventing injuries. It also allows for a better understanding of the biomechanics of different exercises, which can be applied to various fields such as sports science and rehabilitation.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
504
  • Introductory Physics Homework Help
Replies
5
Views
565
  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
29
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
219
  • Introductory Physics Homework Help
Replies
3
Views
504
  • Introductory Physics Homework Help
Replies
13
Views
4K
  • Introductory Physics Homework Help
Replies
12
Views
2K
Back
Top