
#1
Sep2610, 01:37 PM

P: 16

1. The problem statement, all variables and given/known data
The distance between two telephone poles is 46.0 m. When a 1.50 kg bird lands on the telephone wire midway between the poles, the wire sags 0.190 m. How much tension does the bird produce in the wire? Ignore the weight of the wire. 2. Relevant equations F=ma 3. The attempt at a solution I really just don't know where to start with this one...its confusing to me since the force of the bird (1.5 kg*9.8m/s^{2} = 14.7 N) is vertical, but the tension is sideways...so for F=ma, would I be looking at the sum of the forces in the x or y direction? Or is there some way to simplify it to one dimension? 



#3
Sep2610, 01:49 PM

P: 3,390

So the question is find the tension in the cable?




#4
Sep2610, 01:51 PM

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Tension problem
The tension is not entirely horizontal, because the rope sags. You'll have to resolve the forces into components. In reality, the rope would probably curve, but for simplicity, why don't you assume that the rope has two straight, sloping sections that meet at the midpoint (where the bird is sitting), to form a sort of "V" shape. The vertical position of the midpoint (bottom of the "V") is 0.190 m below where it was when the bird was absent and the rope was horizontal.
Does that help? 



#5
Sep2610, 01:52 PM

P: 16

Gah, somehow forgot to put the question in. Sorry.
How much tension does the bird produce in the wire? Ignore the weight of the wire. 



#6
Sep2610, 01:52 PM

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P: 1,999

Hello hm8.If by sideways you mean horizontal that's not exactly true.The wire sags so the wire on each side of the bird is inclined upwards at an angle to the horizontal.Start by sketching a diagram and marking in the forces at the point where the bird is.You will have the weight of the bird acting down and the tension in each side of the wire acting along the line of the wire.




#7
Sep2610, 02:01 PM

P: 16

Still not sure where to go from there...how do I use the length of the wire and the amount it sags? I guess I can figure out that the length of one half of this 'V' is 23.0007 m (square root of .19^{2} + (46/2)^{2}), but does that help me? 



#8
Sep2610, 02:05 PM

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Edit: okay maybe it's not so clear from his post as I thought. Use what you know of the geometry of the situation to figure out by what angle each half of the rope is inclined (relative to the horizontal). Then use that angle, and basic trigonometry, to resolve the tension force into horizontal and vertical components. Then use Newton's second law to figure out what conditions on those components must be true in order for the situation to be static (i.e. sums of forces in x and y directions are *each* equal to zero) . 



#9
Sep2610, 02:24 PM

P: 16

So the components of the tension vector would just be Cos(0.008)T and Sin(0.008)T So, in the Y direction... F = ma mg + Sin(0.008)T = 0 14.7 N + Sin(0.008)T = 0 T= 14.7 N/Sin(0.008) ...is that right so far? Or does the force of gravity on the bird not fit in there? EDIT: Sorry, make that .4733 degrees for the angle, my calculator was in radians mode. 



#10
Sep2610, 03:23 PM

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The method looks right with a few minor issues. Draw a free body diagram for the bird. What forces act on it? In particular, how many forces act on it due to the rope? What I'm hinting at is, are you sure that there should be only one Tsin(theta) term?



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