# Locate its center of gravity

Nanabit
A carpenter's square has the shape of an L, as in Figure P12.7 (d1 = 16.0 cm, d2 = 4.00 cm, d3 = 4.00 cm, d4 = 11.0 cm). Locate its center of gravity. (Hint: Take (x,y) = (0,0) at the intersection of d1and d4) (I tried to attach the picture but I couldn't get it to work, so just know the width of the whole thing is 4 cm, the length of the vertical part of the L is 16 cm, and the length of the horizontal part of the L is 11 cm.)

I know this has to do with taking the area of the t-square by separating it into 2 rectangles, but I'm not sure what to do from there.

A vertical post with a square cross section is 11.0 m tall. Its bottom end is encased in a base 1.50 m tall, which is precisely square but slightly loose. A force 5.60 N to the right acts on the top of the post. The base maintains the post in equilibrium.
Find the force which the top of the right sidewall of the base exerts on the post. Find the force which the bottom of the left sidewall of the base exerts on the post.

This one I think I am a centimeter away from and don't know why I'm getting it wrong. I know it's the sum of the torques = zero. So I set the centerpoint for part a at the bottom of the base. Then I had (5.6N)(12.5m)=F(1.5m). I did the same thing for part b but set the centerpoint at the top of the base. Am I missing something?

thanks.

#### Attachments

• p12-07alt.gif
5.3 KB · Views: 542
Last edited: