Centre of Mass of carpenter's L square -

Click For Summary

Homework Help Overview

The problem involves finding the center of mass of an L-shaped carpenter's square, assuming uniform material. The original poster attempts to divide the shape into two rectangles to facilitate the calculation of the center of mass coordinates.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the separation of the L-shape into two rectangles and the calculation of their respective centers of mass. Questions arise regarding the correctness of the y-coordinate calculations and the placement of the coordinate system.

Discussion Status

Some participants provide feedback on the calculations, indicating that while the x-coordinate appears correct, there are concerns about the y-coordinate. Suggestions are made to reconsider the centroid positions based on the coordinate system used. There is an ongoing exploration of different methods to verify the results.

Contextual Notes

Participants note the importance of considering the height of the rectangles and the overall dimensions of the L-shape. There is mention of using area as mass due to the uniform material assumption, and the discussion includes references to external resources for clarification.

wizzle
Messages
26
Reaction score
0

Homework Statement



A carpenter's square has the shape of an
"L," as shown at the right. Find the
coordinates of its center of mass, assuming
it to be made of uniform, material. (Hint:
divide the L-shape into two rectangles.)

Homework Equations



MaXa+MbXb/(Ma + Mb)


The Attempt at a Solution



I separated the L shape into two rectangles, with the upper part of the L being a rectangle of 14 cm x 4 cm, and the lower part of the L is a rectangle measuring 4 cm by 12 cm. I found the centre of mass of the upper rectangle to be (2,9) and of the lower rectangle to be (6,2). For the mass, since it's of uniform material, I used the area as its mass (not sure if that's right). The technique I tried was to first find the x-coordinate of the object's centre of mass:

(2x56)+(6x48) / (56+48) = 3.846 cm

For the y coordinate I used the y values of each rectangle:

(9 x 56)+(2x48) / (56+48) = 5.77 cm

So the centre of mass I got was (3.8 cm, 5.8 cm). Were the steps I used sensible?

Thanks a lot!
-Lauren
 

Attachments

  • COM question.jpg
    COM question.jpg
    11 KB · Views: 4,710
Physics news on Phys.org
The x is correct, the y is wrong. You are calculating it with respect to the bottom of the L. Is the y-coordinate of the centroid of the upper rectrangle then equal to 9?

Note: you can check your answer by considering two different rectangles. One large one that covers the entire L, substracted by a smaller one that is enclosed within the legs.
 
Last edited:
Ahh...thank you. Since I separated the two at y=4 cm, then the upper rectangle is 14 cm high. Should I have put

(7x56)+(2x48) / (56+48) = 4.69 cm for the y coordinate? Thanks for your help!
 
The upper is indeed 14 cm high, and if considered alone, its centroid would be at y = 7 cm. However, since there is a horizontal leg element as well, and you place the coordinate system in the lower left corner of the L, the centroid of the upper retrangle moves up by 4 cm, which is the height of the lower leg. See this webpage for some illustrations.
 
Hi! Thanks a lot for attaching that link, it really helped. I think I've got it this time, since I'm using the right coordinates -
(11x56)+(2x48) / (56+48) = 6.85 cm
I appreciate your time srvs!
 
Pleasure. To double-check you could different rectangles as well. Haven't got a picture but if you consider a rectangle that spans from the most bottom-left corner of your L, and across the entire L then its dimensions are 18 by 12. The rectangle between the legs of the L is 14 by 12, so you could also say that with respect to the bottom of the larger rectangle, y = [ (14+4)/2 * (14+4 * 12) - (4 + 14/2) * 14 * (12 - 4) ] / [ (14+4 * 12) - 14 * (12 - 4) ] = 6.85 cm. So you consider instead the surface area and y coordinate of the large rectangle and substract from that the surface area and y coordinate of the smaller one.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
13
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
13K
Replies
2
Views
3K
Replies
2
Views
4K
Replies
5
Views
8K