Center of Gravity of a Carpenter's Square

[hardygirl989]

Homework Statement

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

Homework Equations

Xcg=(A1X1+A2X2)/(A1+A2)
Ycg=(A1Y1+A2Y2)/(A1+A2)

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.

 

[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.

 

[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?

 

[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.

 

[hardygirl989]

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