Question

6. 1273 5.7 NOT TO SCALE. All coordinates in inches. The dimension given for the hole is its diameter Assume uniform thickness and density Here is your shape drawn to scale: This problem DOES need to be written out for your portfolo EXAMPLE NOTES IMAGES DISCUSS select from tagitems where type-schw41' AND link-'1273 UNITS STATS HELP Part Description Answer Chk History What is the area of the object shown above? (include units with answer) Format Check Missing 10.56 pts. 106% 296 try penalty A. # tries: 2 Show Details Units Hints: 12 What is the X coordinate of the center of mass the object shownForma above? (include units with answer) 16.5 pts. 110% Check2%try penalty B. tries: 0 Show Details What is the Y coordinate of the center of mass the object shown above? (include units with answer) Format 163 pts, 110% Check 2%try penalty # tries: 0 Show Details

Hint for A:

Hint #1: The dimension given for the hole is its DIAMETER

Hints for B:

Hint #1:	Do not assume that the hole is centered in the right part of the shape.
Hint #2:	Be sure to measure all distances from the given coordinate system
Hint #3:	Be sure to subtract the value for hole

Hint for C:

Hint #1: Be sure to measure all distances from the given coordinate system.

Answer 1

Area of the object A = A_1 + A_2 + A_3 - A_4 A = (24.2 times 6) + (10.5 times 19.4) + (1/2 times 19.4 times 4.3) - (pi/4 times 5.7^2) A = 145.2 + 203.7 + 41.71 - 25.518 A = 365.092 in^2 x coordinate of center of mass x = sigma mx/sigma m = sigma Ax/sigma A x = A_1 x_1 + A_2 x_2 + A_3 x_3 - A_4 x_4/A_1 + A_2 + A_3 - A_4 x = (145.2 times 24.2/2) + [203.7 times (4.8 + 19.4/2)] + [41.71 times (4.8 + 2/3 times 19.4)] - [25.518 times 16.6]/365.092 x = 13.768 inch y coordinate of center of mass y = sigma m y/sigma m = sigma Ay/sigma A y = A_1 y_1 + A_2 y_2 + A_3 y_3 - A_4 y_4/A_1 + A_2 + A_3 - A_4 y (145.2 times 6/2) + [203.7 times (6 + 10.5/2)] + [41.71 times (16.5 + 4.3/3)] - [25.518