How Do You Calculate the Center of Gravity for a Carpenter's Square?

[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.

 

[Doc Al]

Originally posted by Nanabit
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.

Once you have two rectangles, find the COG of each. Then treat them as two point masses (located at the COGs) to find the COG of the whole system.

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?

The post fits inside the base, not on top of it. Use 11, not 12.5.

 

[M98Ranger]

To locate the center of gravity of the carpenter's square, we can use the formula for finding the centroid of a composite shape. This involves dividing the shape into smaller, simpler shapes and finding the centroid of each shape, then taking the weighted average of their centroids.

In this case, we can divide the carpenter's square into two rectangles, one with dimensions 16 cm x 4 cm and one with dimensions 11 cm x 4 cm. The centroid of a rectangle is located at its center, so the centroid of the first rectangle is at (8 cm, 2 cm) and the centroid of the second rectangle is at (5.5 cm, 2 cm).

To find the overall centroid of the carpenter's square, we take the weighted average of these two centroids. Since the first rectangle has an area of 64 cm^2 and the second rectangle has an area of 44 cm^2, the overall centroid is located at ((64*8 + 44*5.5)/(64+44), (64*2 + 44*2)/(64+44)) = (7.2 cm, 2 cm). This means that the center of gravity of the carpenter's square is located 7.2 cm from the intersection of d1 and d4, on the horizontal part of the L.

For the second problem, we can use the principle of moments to solve for the forces exerted by the base on the post. The principle of moments states that the sum of the clockwise moments is equal to the sum of the counterclockwise moments. In this case, we can take the bottom of the base as the pivot point and set the clockwise moments equal to the counterclockwise moments.

For part a, we have (5.6 N)(12.5 m) = F(1.5 m), where F is the force exerted by the top of the right sidewall of the base on the post. Solving for F, we get F = 46.67 N. This means that the top of the right sidewall exerts a force of 46.67 N on the post.

For part b, we can use the same principle, but this time taking the top of the base as the pivot point. We have (5.6 N)(12.5 m) = F(13.5 m), where F is the force exerted by the bottom of the