   Chapter 8, Problem 11P

Chapter
Section
Textbook Problem

Find the x- and y-coordinates of the center of gravity of a 4.00-ft by 8.00-ft uniform sheet of plywood with the upper right quadrant removed as shown in Figure P8.ll. Hint: The mass of any segment of the plywood sheet is proportional to the area of that segment. Figure P8.11

To determine
The x -and y -coordinates of the center of gravity of a 4.00ft by 8.00ft uniform sheet of plywood with the upper right quadrant removed.

Explanation

Section1:

To determine: The x -coordinate of the center of gravity of a 4.00ft by 8.00ft uniform sheet of plywood with the upper right quadrant removed.

Answer: The x -coordinate of the center of gravity of a 4.00ft by 8.00ft uniform sheet of plywood with the upper right quadrant removed is 3.33ft .

Explanation:

Given info: The area of the part A1 of the plywood is 16.0ft2 , the area of the part A2 is 8.00ft2 , and x coordinates of the center of gravity of parts of the plywood A1 , A2 are 2.00ft and 6.00ft .

Assume the remaining parts of plywood as two parts A1 with an area of 4.00ft by 4.00ft and its center of gravity is located at (2.00ft,2.00ft) . Similarly, another part of the plywood is A2 with area of 2.00ft by 4.00ft and the location of the center of gravity of this part is (6.00ft,1.00ft) . The mass of the plywood is proportion to its area; its mass is defined as m=σA where σ is proportionality constant. In this way, the masses of each part can be determined.

The formula for the x co-ordinate of the center of gravity is,

xcm=ρA1x1+ρA2x2ρA1+ρA2

• ρ is proportionality constant.
• A1 and A2 are areas of the plywood respectively.
• x1 and x2 are the x -coordinates of the center of gravity of the parts of plywood.

Substitute 16.0ft2 for A1 , 8.00ft2 for A2 , 2.00ft for x1 , and 6.00ft for x2 to find xcm .

xcm=ρ(16.0ft2)(2.00ft)+ρ(8.00ft2)(6.00ft)ρ(16.0ft2)+ρ(8.00ft2)=3.33ft

Thus, the x -coordinate of the center of gravity of a 4

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