# Tension on a string attached to model airplane

1. Dec 6, 2003

### lovin_physics05 this is a harder problem then i thought
well, do u knowthose model airplanes that are attached to somethinglike a cieling an the go around in a circle?
i have to find the tension in the string
umm
the mass of the airplane is 13kg
i dontknow the velocity
the radius of the circle it is goinin in is 2.0m
what else do i need to know or how do i solve theproblem, been workin on it for a hile but i cant getit

2. 3. Dec 6, 2003

### Doc Al ### Staff: Mentor

Since the plane is traveling in a circle, it has a centripetal acceleration. To caculate that acceleration, you need to the the speed and the radius. (You need more information.)

Once you find the acceleration, then you can apply F=ma to find out what the centripetal force must be. That force is supplied by the tension in the string (the horizontal component, if it is angled).

Hope this helps a bit.

4. Dec 7, 2003

### lovin_physics05 well, ok that kindahelped. i do know howto do these centripital force problems, usin Fnet=mv^2/r. where the tension is Fnet. but u can only do that if it is horizontal to the surface. my problem is thatit has a slope. so i set up an imaginary problem and pluged numbers in for that. butsay that i have all the numbers, can anyone give mean equation to use or what to do. usually u have to take that angle and make it into Fx and Fy, two lines. so i did that and then i used vector addition to find thehypotenuse, i think thats theranswer. ionly realized this thismorning, thanks for the help. can anyone tell meif im right?

5. Dec 7, 2003

### Doc Al ### Staff: Mentor

I assume you mean that the string is angled down.

Then you must find the components of the string tension (T) in the horizontal and vertical directions. In the vertical direction, there is equilibrium: The forces on the plane sum to zero. In the horizontal direction, Thorizontal is the centripetal force.

6. Dec 7, 2003

### lovin_physics05 ok, so i dont know tension in the string, but i haveto make it into Tx and Ty, therefore i usethe equation Tcos(degree)= tx and Tsin(degree)= Ty. so to find tension in tx, i would use Fnet=mv^2/r

Tx=mv^2/r(cos(degree))
Ty=mv^2/r(sin(degree))

when i get those two
i use vector addition
r^2=Tx^2+Ty^2
want to plug in someimaginary numbers?
the thing is that he showed us the actual plane flying around, and wehave to find all the info our selves, butitsin school, so i have now way of figuring it out right now

7. Dec 7, 2003

### ceptimus Assuming the model is supported only by the string (it is not actually 'flying') then the vertical component of the tension in the string, must support the weight of the model.

If the model mass is 13kg, and gravitation acceleration is 9.81 m s ^-2, then the vertical component of the string tension is 13 * 9.81 newtons.

If the angle of the string to the vertical is &Delta; then the total string tension is 13* 9.81 / cos(&Delta;)

8. Dec 7, 2003

### lovin_physics05 ok, when i used my wn numbers, i set the mass to 13kg which would equal awieght of 130N

do u not need the velocityof the airplaeto determine thetension on the string since itisgoin around in a circle, i think i left that part out, that it is tied to the cieling by a string and it is movinaround in a cirle. say it takes 2.6s to completea rotation, where the radius is 3.0m
so velocity=2(3.14 pi)r/t
v=2(3.14)(3.0m)/2.6s
v=7.2m/s

so thats my velocity

9. Dec 7, 2003

### lovin_physics05 length of string is 3.6m

10. Dec 7, 2003

### ceptimus If we know the angle of the string and the weight of the model, we don't need to know anything else. The string could be any length (this would affect the speed and the time taken for each circle - we could work it out).

11. Dec 7, 2003

### Doc Al ### Staff: Mentor

Yes. (Assuming the angle is from the horizontal.) If you are given the angle, then finding the tension is easy. Just consider equilibrium in the y direction. For the y direction, what forces act on the plane? Those forces add to zero.
No, not Fnet. Only the x component of the tension is centripetal! Tx = m v2/r . But if you are given the angle, you can find the tension. (See above.) You will use this to solve for what you don't know, the speed.
No! See above.

12. Dec 7, 2003

### lovin_physics05 but i need to know the tension in the sring, notthe velocity
and yes, the anglegiven is fromthe horizon
wouldnt i find the tension of Tx and the tension of Ty and use vector addition to find T???

13. Dec 7, 2003

### ceptimus Are you sure? A three metre radius circle with only a 3.6 metre length string? But you said 2 metre radius earlier?

14. Dec 7, 2003

### ceptimus ok. with 3.6 metre radius string.

if circle is 2 metre radius, angle is 33.75 degrees, tension is 153.4 N
if circle is 3 metre radius, angle is 56.44 degrees, tension is 230.7 N

(the angle is between the string and the vertical)

15. Dec 7, 2003

### lovin_physics05 i may hav said it incorrectly, but let me clarify
if the radius was 3.0m and the distance from the object to the ceiling(where it is attached) is 2.0m
then that forms a right triangle, where the length of out string is the hypotenuse
so therefore
c^2=a^2+b^2
c^2= 3.0m^2+2.0m^2
= 13m
c = 3.6m
that is the length of the string
wouldntu agree?

16. Dec 7, 2003

### ceptimus Yes, and I think the answer I gave above is correct.

17. Dec 7, 2003

### Doc Al ### Staff: Mentor

You're overcomplicating the problem. If all you care about is the tension, then:

Ty = T sin&theta; = mg

Upward forces (string) must equal downward forces (weight). Solve for T. (This assumes, as ceptimus notes, that the plane isn't "flying".)

18. Dec 7, 2003

### lovin_physics05 my angle to the horizontal is 25 degrees, thats one that i made up just to get a sense of how to solve the real problem. now i allready had that degree, so how do i use itto solve tension again?
sorry for all this

19. Dec 7, 2003

### lovin_physics05 did u mean Ty = T sin&theta; + mg ????

20. Dec 7, 2003

### ceptimus for angle of string to the horizontal, divide by the sine of the angle.

Model weight = 13 kg. Tension if model is just hanging, not moving, = 13 * 9.81 = 127.5 Newtons.

If swinging in a circle with string at 25 degrees to the horizontal, divide 127.5 by sin(25) = 301.8 Newtons.

you can divide one figure by the other, if you want, to calculate the 'G' on the plane: this is just 1 / sin(25) which is 2.37 G

21. Dec 7, 2003

### Doc Al ### Staff: Mentor

No. Ty = T sin&theta;

Equilibrium implies:
Ty - mg = 0, or...
Ty = mg, thus...
T sin&theta; = mg

&theta;, in these equations, is measured from the horizontal.