Video: Finding the Tension in a String Loaded at Its Midpoint

A bird has a mass of 26 g and perches in the middle of a stretched telephone line. Assume that each half of the line is straight. Determine the tension when ๐œƒ = 5.0ยฐ. Determine the tension when ๐œƒ = 0.50ยฐ.

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Video Transcript

A bird has a mass of 26 grams and perches in the middle of a stretched telephone line. Assume that each half of the line is straight. Determine the tension when ๐œƒ equals 5.0 degrees. Determine the tension when ๐œƒ equals 0.50 degrees.

We have a multipart question here, where weโ€™re interested in the tension when ๐œƒ is two different angles. Weโ€™ve been given the mass of the bird, 26 grams, and weโ€™re told that each half of this line, this telephone line, is straight.

That means if we look at our diagram of the situation, if we have an angle ๐œƒ as labeled on the right side, then with each half of this power line as straight, then the left side will have the same angle of ๐œƒ. So we have a symmetrical setup that way.

What weโ€™re gonna do is solve this problem in general for any angle ๐œƒ, and then weโ€™ll be able to quickly work through the different ๐œƒ values weโ€™re to solve for. So letโ€™s start it on that now.

Now weโ€™re told that the mass of this bird is 26 grams and we want to solve for tension โ€” weโ€™ll call it capital ๐‘‡ โ€” for two different values of ๐œƒ, but what weโ€™d do first is solve for any value of ๐œƒ, well, that is, weโ€™ll give a general solution, and then in the end be able to plug in our particular given ๐œƒ values.

To make headway here, what weโ€™ll do is weโ€™ll draw a free body diagram of the forces acting on the lower point of the wire. Recall that a free body diagram is a drawing that shows all the forces acting on a body or a mass. So in this case, letโ€™s make a drawing of the bottom part of our telephone wire.

So here weโ€™ve drawn that lowest point of the wire that the bird is standing on, and now weโ€™re to draw in all the forces that are acting on that point. Well first, we know our bird is standing on that point, so thereโ€™s a gravity force or a weight force of the bird acting down, like that, and weโ€™ll call it ๐‘Š sub ๐‘ for weight of the bird.

Now we know that thatโ€™s not the only force acting on this part of the wire because the wire is still โ€” itโ€™s not in motion โ€” which means the forces on it balance out. Well what is the other force that acts on this wire? We know that itโ€™s under tension; that is a force that pulls on this lowest point in the diagram.

That tension force pulls to both ends of the wire and it goes along the wire, so weโ€™ll draw that in now. So these two arrows represent our tension in the wire, and weโ€™ll call those capital ๐‘‡ for tension.

Now we know again that all these forces balance out, which means that if we break our tension arrows into their horizontal and vertical components using dotted lines that the vertical parts of our tension force add up to perfectly counterbalance the weight force of the bird acting down. And the horizontal parts of our tension forces pull in opposite directions and perfectly cancel one another out so that the wire stays still.

What this means is, we can write out a force balance equation based on Newtonโ€™s second law. Remember that the second law says that the net force on an object is equal to the mass of that object multiplied by its acceleration. Now in this case, our acceleration is zero, which means that the forces acting on this slowest point in the wire balance out, just as we said earlier, and Newtonโ€™s second law is a way of mathematically showing why that is.

Now working on our free body diagram, you may wonder, where is ๐œƒ in this diagram? And thatโ€™s a great question. ๐œƒ does appear as the angle between our tension vectors and our horizontal dotted lines. Thatโ€™s the angle ๐œƒ, the same ๐œƒ, which if we look back to our original diagram appears on either side of our telephone wire.

So given that, letโ€™s write out the force balance equation in the vertical direction for our free body diagram. So we have a tension force, the vertical component of which acts up, which weโ€™ll call the positive direction. And that tension force that is vertical is equal to ๐‘‡, capital ๐‘‡, the tension of the wire, multiplied by the sin of ๐œƒ.

Now thatโ€™s the case for each one of the tension forces acting on our lowest point in the wire, but you see we have two. So in fact, the total vertical force that acts up is two times ๐‘‡ sin ๐œƒ.

And again thatโ€™s because we have tension forces that pull both up into the left and up into the right acting on this lowest point in the wire. Now the other vertical force at play here is the weight force of the bird, and that acts in the negative direction, so weโ€™ll write that in over a vertical force.

And now the question is, what does all this equal? Well looking back at the second law and again verifying that the bird and the lowest part of the wire are not accelerating, we know that these vertical forces cancel one another out; in other words, subtracting the weight force from two times the tension of sin ๐œƒ is equal to zero.

Now letโ€™s look back at our question, which asked about the tension for a given value of ๐œƒ. Now since thatโ€™s what weโ€™re after, letโ€™s rearrange this vertical forces equation to solve for capital ๐‘‡, the tension in the wire.

And to do that, letโ€™s first clear some space on the right side of our screen. Okay here we go. Weโ€™re again isolating the capital ๐‘‡ in our vertical forces equation because thatโ€™s we wanna solve for, so letโ€™s start by adding ๐‘Š sub ๐‘, the weight force of the bird, to both sides.

We see that that term cancels out in the left side of our equation. And if we now divide both sides by two times the sine of ๐œƒ, then as we look again on our left-hand side of our equation, we see that the twos cancel out as well as the sin ๐œƒ, leaving us with an equation that says capital ๐‘‡, the tension in the wire, is equal to the weight force of the bird, ๐‘Š sub ๐‘, divided by two times the sine of ๐œƒ.

Now we can go one step further because we know that ๐‘Š sub ๐‘, the weight force of the bird, equals the mass of the bird times ๐‘”, the acceleration due to gravity. So we see that the tension in the wire equals the mass of the bird times gravity divided by two times the sine of ๐œƒ.

Now weโ€™re ready to refer back to the ๐œƒ values for which we want to solve for ๐‘‡, the tension. Recall that those values were ๐œƒ equals 5.0 degrees and ๐œƒ equals 0.50 degrees. So first letโ€™s solve for the tension when ๐œƒ equals 5.0 degrees.

Since ๐‘‡ is a function of ๐œƒ, weโ€™ll write that as ๐‘‡ of 5.0 degrees like this. And thatโ€™s equal to the mass of our bird, 26 grams or 0.26 kilograms, multiplied by ๐‘”, 9.8 meters per second squared, all divided by two times the sine of 5.0 degrees.

Now looking at this equation, you might wonder, why did we bother converting our mass of 26 grams into kilograms? And the reason we did it is because weโ€™re looking for an answer in units of newtons, and the base units of newtons are indeed kilograms meters per second squared. So in order to get our answer in familiar units, thatโ€™s why we went ahead and converted our mass from grams to kilograms.

Now when we enter this equation in our calculator, we get a result of 1.5 newtons. We have two significant figures because thatโ€™s the number of significant figures of our original given mass. This is the tension in the wire when ๐œƒ equals 5.0 degrees.

Now letโ€™s move on to our next and final ๐œƒ value, 0.50 degrees. Now with this value of ๐œƒ, our equation for tension is again the mass of the bird in kilograms multiplied by ๐‘” divided by two times the sine of 0.50 degrees.

Now before we punch these numbers into our calculator, what do you think? Will this be more or less than the tension when ๐œƒ equals 5.0 degrees? Well looking at our diagram, when ๐œƒ gets less, it would seem that the wire is more taut, that the same mass of the bird pressing down on it would deflect it less, so we might expect that the tension in the wire will be higher for the smaller angle of ๐œƒ, so letโ€™s see.

When we type these values into our calculator, the result that we find is that the tension when ๐œƒ is 0.50 degrees is 15 newtons. So that indeed goes along with our expectation that the tension force is higher when ๐œƒ is smaller. So these are the values of the tension in the wire when ๐œƒ equals 5.0 degrees and 0.50 degrees, respectively.

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