# Torque confusion

1. May 16, 2007

### yetar

1. The problem statement, all variables and given/known data
On my book, the torque is defined as:
t = rxF
F is the a[llied force on a particle, r is the distance of the particle from the origin.
However, later on, the book tries to find the center of mass of a 3D particle system and says the torque of a particle is m_i*g*(x_i, y_i, z_i) when gravity is present.
However gravity is G=g*(0, -1, 0), so I dont see how the torque of the particle in gravity relates to rxF, because the y coordinate of the particle torque should be zero.

Thank you.

2. Relevant equations

3. The attempt at a solution

2. May 16, 2007

### variation

Hello,

It is not clear for me where is emphasis of your question.
But i just know that:
If a particle with mass $$m_i$$ at $$(x_i,y_i,z_i)$$ in the gravitational field of the question, then the torque respective to the origin is
$$m_i(x_i,y_i,z_i)\times g(0,-1,0) = m_ig(z_i,0,-x_i)$$
In the last step, I expact that you have known the outer product (cross product) of vectors.

Regards

3. May 16, 2007

### yetar

Ok, the purpose of what I read was to find the center of mass of the particle system.
This point needs to be the point that when some wedge is holding it, the particle system is in equilibrium. So the sum of the net Force is zero and also the sum of the net torque is zero.
From the sum of the net torque we get the center of mass.
But the problem is that the y coordinate of rxF for every force on every particle is zero.
And somehow from the equality of the sum of net torque equals to zero we get the center of mass?
I know what is a cross product and how to calculate it if that is what you ask.

4. May 16, 2007

### variation

Do you mean that how to use the net torque vanishes and find the position of center of mass ? If yes, then ...

For a rigid system composed of particles which have mass $$m_i$$ and position $$\mathbf{r}_i$$ respectively, a pivot is fixed on the center of mass and the net torque of the total system is zero (respective to the pivot or center of mass). Of course, the position of the center of mass is nuknown and we can assume it as $$\mathbf{x}$$. Therefore,
$$\mathbf{N}=\sum_i(\mathbf{r}_i-\mathbf{x})\times m_i\mathbf{g}=0\quad\Rightarrow\quad\left(\sum_im_i\mathbf{r}_i\times\mathbf{g}\right)-\mathbf{x}\times\left(\sum_im_i\right)\mathbf{g}=0$$
$$\Rightarrow\quad\left(\sum_im_i\mathbf{r}_i\right)\times\mathbf{g}=\left(\sum_im_i\right)\mathbf{x}\times\mathbf{g}$$
,where the uniform gravitational field is used. Because the arbitrariness of the $$\mathbf{g}$$ and one can conclude that $$\mathbf{x}=\frac{\sum_im_i\mathbf{r}_i}{\sum_im_i}$$.