How Is the Weight of a Bird Calculated When It Sits on a Kite?

[justinh8]

Homework Statement

A man is flying a kite. The string of the kite makes an angle of 50 degrees with respect to the horizontal and exerts a pull of 15 Newtons. A small bird decides to sit on the kite and causes the kite to reach a new equilibrium position with the sting at 30 degrees with respect to the horizontal. Assume that the force exerted by the wind on the kite alone is exactly the same as the force exerted by the wind on the kite and bird together. Calculate the weight of the bird.

The Attempt at a Solution

For this all i did was find the forces in the x direction for both cases and found the delta force so:
Fx1 = 15 N cos50 = 9.64
Fx2 = 15 N cos30 = 12.99

So delta Fx = 12.99-9.64 = 3.35N
So than 3.35N/9.81 = 0.3415kg

It just seems to simple and i think i did something wrong, any insights would be appreciated.

 

[PeterO]

Homework Statement

A man is flying a kite. The string of the kite makes an angle of 50 degrees with respect to the horizontal and exerts a pull of 15 Newtons. A small bird decides to sit on the kite and causes the kite to reach a new equilibrium position with the sting at 30 degrees with respect to the horizontal. Assume that the force exerted by the wind on the kite alone is exactly the same as the force exerted by the wind on the kite and bird together. Calculate the weight of the bird.

The Attempt at a Solution

For this all i did was find the forces in the x direction for both cases and found the delta force so:
Fx1 = 15 N cos50 = 9.64
Fx2 = 15 N cos30 = 12.99

So delta Fx = 12.99-9.64 = 3.35N
So than 3.35N/9.81 = 0.3415kg

It just seems to simple and i think i did something wrong, any insights would be appreciated.

In the first case, the weight of the kite + the tension in the string + the force of the wind = 0 since the kite is stationary - from that we can work out the size of the weight force and the wind force. Take particular notice of the direction of the sum of the Weight + wind forces.

Once the bird is added, the wind force stays the same, but the weight increases. This gives a change in angle of the sum of weight and wind. The tension in the string once the bird is added may well be different to 15N

 

[justinh8]

Ok so here's what i did for that one. For the scenario with know bird on it i did
mg - T1sin theta = 0
soo... m = (15Nsin50 degrees)/9.81 = 1.17kg = the mass of the kite at 50 degrees
Than for scenario 2 with the bird i did:
mg - T2sin theta = 0
soo... m = (15Nsin30)/9.81 = 0.76kg

And than i took difference of the two masses: 1.17 - 0.76 = 0.41 kg = 4.02 Newtons

Is this Correct??

 

[PeterO]

Ok so here's what i did for that one. For the scenario with know bird on it i did
mg - T1sin theta = 0
soo... m = (15Nsin50 degrees)/9.81 = 1.17kg = the mass of the kite at 50 degrees
Than for scenario 2 with the bird i did:
mg - T2sin theta = 0
soo... m = (15Nsin30)/9.81 = 0.76kg

And than i took difference of the two masses: 1.17 - 0.76 = 0.41 kg = 4.02 Newtons

Is this Correct??

Notice that you have a smaller mass for the kite+ bird, than for the kite alone; something is wrong.

The error is assuming that when the kite sinks to 30 degrees the Tension will still be 15N - it is a reaction force and can change. It is the force from the wind that remains the same.

 

[justinh8]

So basically, the tension in sequence number 2 is not 15 Newtons? Ya but in the question it states that the force exerted by the wind is the same both times doesn't it?

 

[PeterO]

So basically, the tension in sequence number 2 is not 15 Newtons? Ya but in the question it states that the force exerted by the wind is the same both times doesn't it?

That's right - the force exerted by the wind is the same. However, the weight involved changes, so the Tension may well change, not only in direction [they tell you that] but possibly magnitude as well.

Case 1.

Kite - weight down
Wind - Force in some direction
String - Tension at 50^o , so partially down and partially left.

The wind force have an up and a right component.

The right component balances the left component of the tension.
The up component balances the combined weight plus down component of the tension.

Case 2.

The wind force doesn't change. - so its up and right components are presumably unaltered.
The weight force increases.
So the tension has the same left component, but a smaller down component [since (increased) weight plus tension component are the same; as the wind force is the same]

Not surprisingly, the angle on the tension alters - and also its magnitude; Pythagoras will tell us that.

Presumably there is just enough information to get the actual weight of the bird. I am not convinced yet.

EDIT: Just re-read the question and there is enough information. You need to calculate the components of the Tension [left and down] the consider what I wrote above.

 

[Stanc]

So for the left direction in sequence 1 the left direction: T1 cos 50 = 15cos50 = 9.64 N and than for the y direction: T1sin50 = 15sin50 = 11.49N. So using pythogorean: hypotenuse = square root of 9.64^2 + 11.49^2 = 14.99832324N

The left direction in sequence 2 it is T2 cos 30 = 15cos30 = 12.99 N and than in the y direction for sequence 1: T2 sin 30 = 15sin30 = 7.5N
So using pythagorean u get the hypotenuse to equal square root of 12.99^2 + 7.5^2 = 14.99967N

so than delta = 14.99967 - 14.99832324 = 1.34 x10^-3N

Is that how i would do it?

 

[justinh8]

I did the exact same thing as stanc, however, i think that since the force of the wind stays the same for both situations, the force in the x would always 9.64N and the force in the y would always equal 11.49N
You would than have to say to find hypotenuse in situation 2 it is tan 30 = opp/adj = 11-x/9.64 where x represents the mass of the bird.
it turned out to be 5.93 Newtons

 

[azizlwl]

Before the lazy bird landed on the kite
$F_y=15Sin50^o$
$F_x=15Cos50^o$

Assuming the tension of the string remains.
$F_y-mg=15Sin30^o$
$mg=F_y-15Sin30^o$
mg = 15x0.2660=4N

If of unequal tension,
$mg=F_y-XSin30^o$ where X is the new tension.

 

[Stanc]

Before the lazy bird landed on the kite
$F_y=15Sin50^o$
$F_x=15Cos50^o$

Assuming the tension of the string remains.
$F_y-mg=15Sin30^o$
$mg=F_y-15Sin30^o$
mg = 15x0.2660=4N

If of unequal tension,
$mg=F_y-XSin30^o$ where X is the new tension.

Where did you get the 15x0.2660?

 

[azizlwl]

$F_x and F_y$ remain constant
$mg=15Sin50^o-15Sin30^o$
mg = 15x0.2660=4N

 

[justinh8]

$F_x and F_y$ remain constant
$mg=15Sin50^o-15Sin30^o$
mg = 15x0.2660=4N

But wouldn't that be saying 15sin 30 = 15sin50 -mg which is saying the second situation = the first situation subtract the force of the bird whereas the force of the bird would have to be added to the first situation to equal the second situation and also why would you only consider the y components?

 

[azizlwl]

But wouldn't that be saying 15sin 30 = 15sin50 -mg which is saying the second situation = the first situation subtract the force of the bird whereas the force of the bird would have to be added to the first situation to equal the second situation and also why would you only consider the y components?

The weight of the bird only acting downward. Assuming it a single point.
It has no horizontal component,mgCos90°=0

 

[justinh8]

Ok but can you explain to me why the equation mg = 15sin50 - 15sin30 is correct? instead of -mg = 15sin50 - 15sin 30?

 

[azizlwl]

Ok but can you explain to me why the equation mg = 15sin50 - 15sin30 is correct? instead of -mg = 15sin50 - 15sin 30?

I don't know how you get -mg = 15sin50 - 15sin 30.
Maybe you can show it.

 

[PeterO]

So for the left direction in sequence 1 the left direction: T1 cos 50 = 15cos50 = 9.64 N and than for the y direction: T1sin50 = 15sin50 = 11.49N. So using pythogorean: hypotenuse = square root of 9.64^2 + 11.49^2 = 14.99832324N

The left direction in sequence 2 it is T2 cos 30 = 15cos30 = 12.99 N and than in the y direction for sequence 1: T2 sin 30 = 15sin30 = 7.5N
So using pythagorean u get the hypotenuse to equal square root of 12.99^2 + 7.5^2 = 14.99967N

so than delta = 14.99967 - 14.99832324 = 1.34 x10^-3N

Is that how i would do it?

No THAT IS NOT THE WAY TO DO IT. [WHOOPS CAPS LOCK STUCK ON - still; that's better]

The first part of the calculation is correct,

the left direction component is indeed 9.64 N - indicating the wind force had a "9.64N to the right" component.

The vertical component was indeed 11.49N down, indicating the wind force had a [11.49N + Weight of kite] upward component.

Once the bird has landed, the horizontal component is STILL 9.64N since the wind force didn't change, and the combined downward component of Tension and Weight(kite + bird) must also still be the same, since the wind force didn't change.

Since we now have a new triangle with the Tension, we can see that tan30^o = [new downward component]/[old horizontal component].

That will give us a new, smaller downward component [it will also give us a new overall tension but who cares]

lets suppose that the new downward component was 8N, that would mean the component is 3.49N smaller, meaning kite + bird is 3.49N bigger than kite alone; so the bird weighs 3.49 N.

Now, I just invented that value of 8N. You must use your trig on the 30^o triangle to find what it is really equal to.

EDIT: If you want you can calculate what the new tension is - it is simply 9.64 / cos30^o

 

[justinh8]

Ok so than new downward component would be: tan 30 = downward component/9.64 so downward component would equal 5.56 Newtons and therefore the difference would be 11.49N - 5.56N = 5.924N. But when i find the distance do i not have to find the same downward component using the angle 50 degrees?? and also why can i not use mg = 15sin50 - 15sin30 to solve for the birds weight?

 

[PeterO]

Before the lazy bird landed on the kite
$F_y=15Sin50^o$
$F_x=15Cos50^o$

Assuming the tension of the string remains.
$F_y-mg=15Sin30^o$
$mg=F_y-15Sin30^o$
mg = 15x0.2660=4N

If of unequal tension,
$mg=F_y-XSin30^o$ where X is the new tension.

The problem is that highlited in red above. The tension in the string does not remain 15N

 

[PeterO]

Ok so than new downward component would be: tan 30 = downward component/9.64 so downward component would equal 5.56 Newtons and therefore the difference would be 11.49N - 5.56N = 5.924N. But when i find the distance do i not have to find the same downward component using the angle 50 degrees?? and also why can i not use mg = 15sin50 - 15sin30 to solve for the birds weight?

OK, you have the birds weight at 5.924N

if mg = 15sin50 - 15sin30 happens to give that same answer - then perhaps you can use it - but I don't expect it will.

I can understand why you would involve a 15sin50 factor, and/or a 15cos50 factor, because when the angle is 50 degrees, the tension is 15N - you were told that.

HOWEVER, you were told that the angle changed to 30 degrees, and the wind force remained the same. There was no suggestion that the tension remained at 15N. That is why a 15sin30 or a 15cos30, or even a 15tan30 term has no application to this problem.

Also, what distance were you trying to find, and why?

 

[justinh8]

Ok i understand, what i was wondering was since i found the new downward component using pythogorean with tan (angle) = downward component / 9.64N shouldn't I do the same for the first sequence by doing tan(50) = downward component / 9.64N? or is 15sin(50) the same thing?
EDIT: i didnt realize i wrote distance, i meant downward component, sorry..

 

[PeterO]

Ok i understand, what i was wondering was since i found the new downward component using pythogorean with tan (angle) = downward component / 9.64N shouldn't I do the same for the first sequence by doing tan(50) = downward component / 9.64N? or is 15sin(50) the same thing?

You don't have to - you already analysed that triangle - using the knowns of 50^o and hypotenuse 15 to get te other two sides.
You have now moved on to the second triangle using the new knowns - 9.64 and 30^o to find the one side you needed [the vertical component.
You don't bother to find the hypotenuse [the Tension] because no-one asked for its value.

 

[Rayquesto]

First, think about the y component of tension. What do you get?

 

[PeterO]

First, think about the y component of tension. What do you get?

Are you joining late here?

Do you mean y-component when the string is at 50^o or the y-component when the string is at 30^o

 

[Rayquesto]

You know what doesn't make sense though...If the y-component of the tension decreases with smaller angles, how does the rope deceleration in the y direction when the bird decides to sit down. We know this y component decelerated from 50 to 30 degrees

 

[Rayquesto]

How would we set up a way to find the wind force? Using differential equations?

 

[justinh8]

Soo.. what should I go with than, you mean find the y component of the second situation and don't subtract it from the first?

 

[PeterO]

Soo.. what should I go with than, you mean find the y component of the second situation and don't subtract it from the first?

I would recommend you go with the answer you got in your post #17, and ignore the posts from Rayquesto.

 

[PeterO]

How would we set up a way to find the wind force? Using differential equations?

You do it the way they did here - but know the mass of the kite.
ie: measure it.

When a real bird lands on a real kite, the real wind force may well alter - but in this problem, it was stated that the wind force remained the same, so it turned into a semi theoretical problem.

As for the kite accelerating when the bird sat on it - in the real world this is true, but the kite would oscillate a few times then assume a new stable position - and who knows, it may even change from 50 degrees to 30 degrees.
rm my experience with kites as a child, the string won't be straight anyway, but will have a droop, and while it may make an angle of 30 degrees when it leaves my hand, it may well be 60 degrees to the horizontal where it connects to the kite.

This is a theoretical question with a real-world practical basis.

 

[PeterO]

You know what doesn't make sense though...If the y-component of the tension decreases with smaller angles, how does the rope deceleration in the y direction when the bird decides to sit down. We know this y component decelerated from 50 to 30 degrees

Suppose you were flying this kite from the top of a tower. If the bird, kite and wind were just right, the string could be horizontal; there would be no y component at all!

A description of yours "decelerated from 50 to 30 degrees" is a curious one, and I am not sure which word in the incorrectly used one - though I suspect it is "decelerated" ?

 

[azizlwl]

I've made a mistake not calculating for the the new tension of the string.
Since the body in equilibrium, and forces produced by wind on kite remain.

Taking x direction
$F_2Cos30^o=F_1Cos50^0$

Taking y direction
From this we can calculate the weight by sum 3 forces, weight, string and kite which is equal 0

http://img39.imageshack.us/img39/7465/lazybird.jpg

 

[Rayquesto]

I say decelerated, because, initially, the kite has a tension and in equilibrium, but when the weight exerts a force on the kite, the kite moves with some force that will accelerate the kite to move down y axis, yet somehow the kite stops. I suspect that there then would exist a deceleration sometime between the weight exerted by the bird on the kite and when the new equilibrium position is reached. I think there is a change in acceleration with respect to time in this sense, because the wind force is such a weird thing to deal with where its force increases with increasing velocity. So, I mean, what do you think about the idea of the wind force in this situation?

Also, I want to note that the components of the tension do in fact change with a change in angle. As the angle gets shorter, the tension in the y direction decreases and that's also partially why this is a strange situation to me.

 

[PeterO]

I say decelerated, because, initially, the kite has a tension and in equilibrium, but when the weight exerts a force on the kite, the kite moves with some force that will accelerate the kite to move down y axis, yet somehow the kite stops. I suspect that there then would exist a deceleration sometime between the weight exerted by the bird on the kite and when the new equilibrium position is reached. I think there is a change in acceleration with respect to time in this sense, because the wind force is such a weird thing to deal with where its force increases with increasing velocity. So, I mean, what do you think about the idea of the wind force in this situation?

Also, I want to note that the components of the tension do in fact change with a change in angle. As the angle gets shorter, the tension in the y direction decreases and that's also partially why this is a strange situation to me.

What you say about a real kite in a real wind with a real string and a real bird is all correct. However an important part of the original post/question was:
"Assume that the force exerted by the wind on the kite alone is exactly the same as the force exerted by the wind on the kite and bird together"

This immediately made the problem a simplified, idealised situation, so real world logic is no longer that important.

As you obviously realize, the behaviour of wind, with the kite tipping at different angles is extremely complex - that is why, in this case, a greatly simplified situation is to be assumed.

Many problems we are presented with in school/college physics are simplified/idealised presentations of an every day occurrence, and we have to be ready to go with that.

When was the last time you had to deal with a satellite in elliptical orbit, and allow for added air resistance at the points where it is nearer the Earth?
We mostly idealise to perfectly circular orbit around a perfectly spherical Earth of uniform density.
We mostly assume air resistance is insignificant [ignored], or constant regardless of speed.
Surfaces are magically frictionless etc. etc.
Even when launching a rocket that achieves enormous speed, we still ignore the air resistance.

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