Question

Part 1: A square plate is produced by welding together four smaller square plates, each of side a. The weight of each of the four plates is shown in the figure [ (0,2a)- 70N, (2a,2a)- 20N, (0,0)- 90N, (2a,0)- 60N ] Find the x-coordinate of the center of gravity (as a multiple of a). Answer in units of a.

Part 2: Find the y-coordinate of the center of gravity (as a multiple of a). Answer in units of a

Answer 1

To find the x-coordinate and y-coordinate of the center of gravity of the combined square plate formed by the four smaller plates, we will first identify the coordinates and weights of each smaller plate.

The smaller plates are located at the following coordinates with their respective weights:
1. Plate at \((0, 2a)\) with a weight of 70 N
2. Plate at \((2a, 2a)\) with a weight of 20 N
3. Plate at \((0, 0)\) with a weight of 90 N
4. Plate at \((2a, 0)\) with a weight of 60 N

### Part 1: Finding the x-coordinate of the center of gravity

The formula for the x-coordinate of the center of gravity (CG) is given by:

\[
\bar{x} = \frac{\sum (x_i \cdot W_i)}{\sum W_i}
\]

where \(x_i\) is the x-coordinate of each plate and \(W_i\) is the weight of each plate.

#### Weights and Coordinates
Now let's calculate \(\sum (x_i \cdot W_i)\) and \(\sum W_i\):

- For the plate at \((0, 2a)\):
- \(x_1 = 0\), \(W_1 = 70\,N\)
- Contribution: \(0 \cdot 70 = 0\)

- For the plate at \((2a, 2a)\):
- \(x_2 = 2a\), \(W_2 = 20\,N\)
- Contribution: \(2a \cdot 20 = 40a\)

- For the plate at \((0, 0)\):
- \(x_3 = 0\), \(W_3 = 90\,N\)
- Contribution: \(0 \cdot 90 = 0\)

- For the plate at \((2a, 0)\):
- \(x_4 = 2a\), \(W_4 = 60\,N\)
- Contribution: \(2a \cdot 60 = 120a\)

Now, summing these contributions:

\[
\sum (x_i \cdot W_i) = 0 + 40a + 0 + 120a = 160a
\]

Next, summing the weights:

\[
\sum W_i = 70 + 20 + 90 + 60 = 240\,N
\]

Now we can find \(\bar{x}\):

\[
\bar{x} = \frac{160a}{240} = \frac{2}{3}a
\]

### Part 2: Finding the y-coordinate of the center of gravity

Similarly, the formula for the y-coordinate of the center of gravity is:

\[
\bar{y} = \frac{\sum (y_i \cdot W_i)}{\sum W_i}
\]

#### Weights and Coordinates
Now let's calculate \(\sum (y_i \cdot W_i)\):

- For the plate at \((0, 2a)\):
- \(y_1 = 2a\), \(W_1 = 70\,N\)
- Contribution: \(2a \cdot 70 = 140a\)

- For the plate at \((2a, 2a)\):
- \(y_2 = 2a\), \(W_2 = 20\,N\)
- Contribution: \(2a \cdot 20 = 40a\)

- For the plate at \((0, 0)\):
- \(y_3 = 0\), \(W_3 = 90\,N\)
- Contribution: \(0 \cdot 90 = 0\)

- For the plate at \((2a, 0)\):
- \(y_4 = 0\), \(W_4 = 60\,N\)
- Contribution: \(0 \cdot 60 = 0\)

Now, summing these contributions:

\[
\sum (y_i \cdot W_i) = 140a + 40a + 0 + 0 = 180a
\]

We already calculated the total weight \(\sum W_i = 240\,N\). Now we can find \(\bar{y}\):

\[
\bar{y} = \frac{180a}{240} = \frac{3}{4}a
\]

### Final Answers
- The x-coordinate of the center of gravity is \(\frac{2}{3}a\).
- The y-coordinate of the center of gravity is \(\frac{3}{4}a\).

Thus, in units of \(a\):
- **Part 1**: \(\bar{x} = \frac{2}{3}\)
- **Part 2**: \(\bar{y} = \frac{3}{4}\)