# Rotational Force Problem

[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 alot 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
Homework Helper
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?

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.

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 alot of trouble on

ohhhhhhhh I think I got it

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

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

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

$$F_x = ma_r$$

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

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