MHB Finding the Center of Mass of a Plywood Sheet: A Scientific Approach

AI Thread Summary
The discussion focuses on calculating the center of mass of a plywood sheet with a removed quadrant, emphasizing the uniform density of the material. The coordinates of the center of mass were initially miscalculated, but through integration of the height function and moments about the axes, the correct values were determined. The area of the plywood was calculated to be 24, leading to moments of 80ρ and 112ρ for the x and y directions, respectively. After correcting the moment calculations, the final coordinates of the center of mass were established as (14/3, 5/3). The thread highlights the importance of accurately applying the moments to find the correct center of mass.
cbarker1
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Dear Everybody,

Shown below is a 4.00-ft by 8.00-ft sheet of plywood with the upper left quadrant removed. Assume the plywood is uniform and determine the x and y coordinates of the center of gravity. Hint: The Earth's gravitational field is also uniform for the entire sheet of plywood.

9-p-035.gif


I know the center of mass of the smaller rectangle is at (2,1) and the larger rectangle is at (6,2). I know the density is twice the area of the small rectangle of wood. I need some help with the total center of mass of the object above.
 
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Let $\rho$ be the uniform mass density of the sheet and $A$ be the area, and so we know:

$$\rho=\frac{m}{A}\implies m=\rho A$$

Now, the find the area, we can define the height $h$ (along the $y$-axis) of the sheet using a piecewise defined function:

$$h(x)=\begin{cases}2, & 0\le x<4 \\[3pt] 4, & 4\le x\le8 \\ \end{cases}$$

Hence:

$$A=\int_0^8 h(x)\,dx=\int_0^4 2\,dx+\int_4^8 4\,dx=2(4-0)+4(8-4)=24$$

And so:

$$m=24\rho$$

Now we need our moments:

$$M_x=\rho\int_0^8 h^2(x)\,dx$$

$$M_y=\rho\int_0^8 xh(x)\,dx$$

What do you find for the moments?
 
For the moments in x direction: 80$\rho$ and for y direction 112$\rho$.
 
Cbarker1 said:
For the moments in x direction: 80$\rho$ and for y direction 112$\rho$.

Yes, I get the same. So now, the center of mass is:

$$\left(\overline{x},\overline{y}\right)=\left(\frac{M_y}{m},\frac{M_x}{m}\right)=\,?$$
 
(10/3,14/3)
 
Cbarker1 said:
(10/3,14/3)

You've got your coordinates reversed...you see $M_x$ is the moment about the $x$-axis and so it is used to find the $y$-coordinate, and likewise $M_y$ is the moment about the $y$-axis, and is used to find the $x$-coordinate.

I made an error in the moment about $x$...it should be:

$$M_x=\frac{\rho}{2}\int_0^8 h^2(x)\,dx$$

This results in the center of mass of:

$$\left(\overline{x},\overline{y}\right)=\left(\frac{14}{3},\frac{5}{3}\right)$$

View attachment 7442
 

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