Why Is the Tension Calculated Using Cosine in a Conical Pendulum Problem?

[Sheneron]

[SOLVED] Rotational Force Problem

Homework Statement

Consider a conical pendulum with a 89.0 kg bob on a 10.0 m wire making an angle of = 5.00° with the vertical (Fig. P6.9).

http://img89.imageshack.us/my.php?image=p613qy9.gif

The Attempt at a Solution

I am having a lot of trouble figuring this out. I have tried to set up force diagrams but I am getting confused. I know that the force is the y direction is just mg but to make it this way it would have to be
Tsin(\theta) - mgsin(\theta) = 0 and I don't understand why it would be mg multiplied by the sine of the angle.
Also, for the second part I don't understand why the answer is not just mg times the sin of the angle.
Can someone help me figure out this problem? Thanks.

 

[alphysicist]

Hi Sheneron,

I'm not certain which direction you're trying to write that equation for. However, you're right that mg is completely vertical; also, the tension has both vertical and horizontal components.

For this problem, you'll also need the acceleration (magnitude and direction). What do you get?

 

[Sheneron]

For the horizontal tension the equation I get is
Tcos(\theta) - mgsin(\theta) = 0
For vertical I get
Tsin(\theta) - mgsin(\theta) = 0

I am pretty sure those have to be the answers, but I can't see where they came from. I don't understand how to set the force diagram up.

 

[Sheneron]

Now I am very confused... ma would not equal 0 would it? To find the acceleration you would first have to know the tension, which I am having a lot of trouble on

 

[Sheneron]

ohhhhhhhh I think I got it

Tcos(\theta) - mg = 0
T = \frac{mg}{cos(\theta)}

T_y = Tcos(\theta) = \frac{mg}{cos(\theta)}cos(\theta) = mg

T_x = Tsin(\theta) = \frac{mg}{cos(\theta)}sin(\theta)

F_x = ma_r

\frac{mg}{cos(\theta)}sin(\theta) = ma_r

a_r = \frac{mgtan(\theta)}{m}