Tension & Magnitude

• April 28th 2009, 01:24 PM
jmoran
Tension & Magnitude
A tetherball weighing 1 lb is pulled outward from the pole by a horizontal force u until the rope makes an angle of theta degrees with the pole

1. Determine the resulting tension in the rope and magnitude of u when theta = 30 degrees
2. Write the tension T in the rope and the magnitude of u as functions of theta
3. Determine the domains of the functions
4. Compare T and ||u|| as theta increases

I know that ||u|| = sqrt a^2+b^2 but I just don't know what the values of a and b are.

Jennifer
• April 28th 2009, 01:47 PM
earboth
Quote:

Originally Posted by jmoran
A tetherball weighing 1 lb is pulled outward from the pole by a horizontal force u until the rope makes an angle of theta degrees with the pole

1. Determine the resulting tension in the rope and magnitude of u when theta = 30 degrees
2. Write the tension T in the rope and the magnitude of u as functions of theta
3. Determine the domains of the functions
4. Compare T and ||u|| as theta increases

I know that ||u|| = sqrt a^2+b^2 but I just don't know what the values of a and b are.

Jennifer

Draw a sketch.

You deal with a right triangle with the hypotenuse $\vec t$ and the legs $\vec w$ and $\vec u$.

You know: $\cos(\theta) = \dfrac{|\vec w|}{|\vec t|}$

Since you know the magnitude of w and the angle $\theta$ you can determine the magnitude of t

$\tan(\theta) = \dfrac{|\vec u|}{|\vec w|}$

For a specific value of $\theta$ you can calculate the magnitude of u.
• April 30th 2009, 08:51 AM
jmoran
I don't know if I understand.

I am assuming the magnitude of w is = 1??

If that is the case then in order to get the value of t (which is the tension) when angle is 30 degrees, I would use

t = 1/cos(30)

So now I have the t = 1.15.

Next -

Is this the formula to determine the magnitude of u??

http://www.mathhelpforum.com/math-he...d0f1d82d-1.gif

If so, again assuming the magnitude of w is = 1 and the angle is 30 degrees, would it be

i * tan(30) = .58

so |u| = .58 ??

I think I am missing something but I'm not sure what....

Thanks,

Jennifer
• April 30th 2009, 09:39 AM
pflo
The fact that w's magnitude is 1 is given in the problem (the ball is 1 pound). Look again at the diagram posted by earboth. The only assumption is that we are disregarding the weight of the rope itself.

Other than that, your work is correct. It is all right-triangle ratios (SOH, CAH, TOA).
• May 1st 2009, 06:28 AM
jmoran
So if the question was

1. Determine the resulting tension in the rope and magnitude of u when theta = 30 degrees

tension = 1/cos(30) = 1.15
|u| = 1 * tan(30) = .58

Is that right?

Jennifer
• May 1st 2009, 07:59 AM
earboth
Quote:

Originally Posted by jmoran
So if the question was

1. Determine the resulting tension in the rope and magnitude of u when theta = 30 degrees

tension = 1/cos(30) = 1.15
|u| = 1 * tan(30) = .58

Is that right?

Jennifer

Yes (Clapping)

(Keep in mind that the forces are measured in pounds (very unusual) and if you have to use Newtons you must transform these values)
• May 1st 2009, 08:01 AM
jmoran
OMG !!!

I actually understand.....

Thank you thank you thank you!!!

Jennifer
• January 31st 2013, 02:54 PM
Wulfman
Re: Tension & Magnitude
?
• January 31st 2013, 11:52 PM
topsquark
Re: Tension & Magnitude
Quote:

Originally Posted by Wulfman
?

At angle $\theta$, tension T, in pounds, the horizontal force on the ball is [tex]Tsin(\theta)[/itex] and that must be u. The vertical force is $T cos(\theta)$ and that must be the balls weight, 1 lb. Solve the equations $T sin(\theta)= u$ and $T cos(\theta)= 1$ for T and $\theta$.
Notice that if you square each equation and add you get $T^2 sin^2(\theta)+ T^2 cos^2(\theta)= T^2(sin^2(\theta)+ cos^2(\theta))= T^2= u^2+ 1$. If you divide the first equation by the second, you get $\frac{T sin(\theta)}{T cos(\theta)}= tan(\theta)= frac{u}{1}= u$.