# Equilibrium question.

1. Nov 23, 2007

### kjthesandman

I was having trouble calculating the equilibrium of this object. I thought I had to solve it using the scalar cross product...but i'm not certain I am on the right track. Can some one help me? This is the question.

A carpenter's square has the shape of an L, where d1 = 19.0 cm, d2 = 6.00 cm, d3 = 6.00 cm, d4 = 11.0 cm. Locate its center of gravity. (Hint: Take (x, y) = (0, 0) at the intersection of d1 and d4)

2. Nov 23, 2007

### azatkgz

Can you please draw this carpenter.I'll try to help you.

Like this?

Last edited: Nov 24, 2007
3. Nov 24, 2007

### kjthesandman

Sort of, actually d3 is connected to d1, and d1 is the longer side of the carpenter square. The picture looks more like this...

Last edited: Nov 24, 2007
4. Nov 24, 2007

### PhanthomJay

Rather than using cross products, break up the shape into 2 rectangles, and sum moment areas about each axis separately , where the moment of an area is the area times the perpendicular distance from its centroid to the axis, and where
(A1)x1 + A2(x2) = (A1 + A2)(X_), etc., (which you are familiar with?).

5. Nov 24, 2007

### kjthesandman

Well I was using the equation ((m1)(x1)+(m2)(x2))/(m1+m2)= but i keep getting the wrong answer...using this method how do i find y1, y2, and x1, x2. I know it seems like a real fundamental question but I am still having trouble.

6. Nov 24, 2007

### PhanthomJay

x1, x2, y1, and y2, are the perpendicular distances from the centroids of each area to the origin. For example, one of the rectangles is 19 by 6. For that particular rectangle, since its centroid is at its geometric center, then x1 is 3 and y1 is 9.5. Now find the centroid of the second rectangle, and the corresponding x2 and y2 distances. Be sure to use A1 and A2 for what you are calling m1 and m2.

7. Nov 24, 2007

### azatkgz

Centre of mass is between two centres of mass,so in the red line.Where the red lines coincide,there is the centre of mass of the carpenter.