# Center of Gravity Question

1. Apr 9, 2012

### hardygirl989

1. The problem statement, all variables and given/known data

A carpenter's square has the shape of an L, where d1 = 19.0 cm, d2 = 2.00 cm, d3 = 2.00 cm, d4 = 9.0 cm. Locate its center of gravity. (Take (x, y) = (0, 0) at the intersection of d1 and d4.)

Picture ---> http://www.webassign.net/pse/p12-07alt.gif

Answer = (____ , ____) cm

2. Relevant equations
Xcg=(A1X1+A2X2)/(A1+A2)
Ycg=(A1Y1+A2Y2)/(A1+A2)

3. The attempt at a solution

x1=1cm
x2=3.50cm
y1=9.5cm
y2=1cm

Xcg=(38.8*1+14*3.5)/(38.8+14)=1.66cm
Ycg=(38.8*9.5+14*1)/(38.8+14)=7.47cm

This seems to be the incorrect answer, and I am not sure why...Can anyone help? Thanks.

2. Apr 10, 2012

### Filip Larsen

It seems you have correctly divided the L-shape into two rectangles, one covering d1 and d2 and one covering d4 - d2 and d3. If that is so, then your numbers 38.8 and x2 = 3.5 are wrong.

When you get stuck like this it is often a good idea to track backwards through your work and check all results once more. :tongue:

3. Apr 10, 2012

### hardygirl989

I recalculated the area to get 38 cm^2 instead of 38.8cm^2, but I am still confused on what the x value should be...why is it not 3.5, which is half of 7, be the correct value. 7 is the length of the second rectangle for (d4-d2) and d3? What should the value of X be then?

4. Apr 10, 2012

### Filip Larsen

The x and y coordinates should represent the center of each rectangle as measured from the origin. Since the first rectangle have the origin (0,0) on its lower left corner the center its center is easily found (your x1 and y1). The other rectangle however is offset a bit away from the origin such that the coordinates of its lower left corner is (d2, 0). The number you found (3.5 cm, 1 cm) is "only" the distance from the lower left of this rectangle to its center, so you are missing to include the offset in the coordinates.

5. Apr 10, 2012

### hardygirl989

thank you! I got the right answer now. :)

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