1. ## center of gravity/mass

1. The problem statement, all variables and given/known data
Center of gravity for a shape T both arm are 60mm long and 6mm wide.

2. Relevant equations

dont know the equation but seeing as there is no wieght involved am guessing the formula isnt that hard. a discription of the formula and how to use it would help alot

3. The attempt at a solution

if both arms are 60mm and 6mm wide shouldent the center og gravity be where the to arms connect?

2. You need to supply a picture. It depends on the shape of this arm.

3. Originally Posted by JBswe
1. The problem statement, all variables and given/known data
Center of gravity for a shape T both arm are 60mm long and 6mm wide.
For a two dimensional object with uniform density the center of mass for in the x-direction is given by

$\displaystyle \frac{\int_{}^{}xf(x)dx}{\int_{}^{}f(x)dx}$,

where f(x) is the total height or extent in y direction at point x. The y coordinate is found by changing to y and f(y) and taking the integral along the y-axis.
For a standing T the CoG x-coordinate lies midway(x-wise) of the top bar due to symmetry. Let y=0 be at the bottom point of the T, then the CoG y-coordinate of your T is found through

$\displaystyle \frac{\int_{0}^{60}6ydy+\int_{60}^{66}60ydy}{\int_ {0}^{60}6dy+\int_{60}^{66}60dy}=\frac{6\cdot 60^2/2+60\cdot (66^2/2-60^2/2)}{360+360}=46.5$

Thus, for your standing T the center of gravity is located at x=0,y=46.5mm, when the bottom tip of the T is put to (0,0)...

This can probably also be done without integrals by exploiting the geometry of the T...

4. Originally Posted by F.A.P
This can probably also be done without integrals by exploiting the geometry of the T...
Yes. Since we are assuming the material is uniform we can find the CM coordinates of the cross piece and the vertical piece, then find the midpoint of the line segment connecting these two points.

-Dan