Answer :
Certainly! Let's solve this step by step:
### Given Data:
- Force on the first piston ([tex]\(F_1\)[/tex]) = 30 N
- Area of the first piston ([tex]\(A_1\)[/tex]) = 100 mm²
- Force on the second piston ([tex]\(F_2\)[/tex]) = 90 N
### Objective:
We need to calculate the area of the second piston ([tex]\(A_2\)[/tex]).
### Step-by-Step Solution:
1. Convert the area of the first piston to square meters:
Since [tex]\(1 \, \text{mm}^2 = 10^{-6} \, \text{m}^2\)[/tex]:
[tex]\[ A_1 = 100 \, \text{mm}^2 = 100 \times 10^{-6} \, \text{m}^2 = 1 \times 10^{-4} \, \text{m}^2 \][/tex]
2. Use the hydraulic system principle:
The hydraulic system operates on the principle of Pascal's Law, which states that the pressure in a closed system is constant. Mathematically, this is represented as:
[tex]\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \][/tex]
Here, [tex]\(P_1 = P_2\)[/tex] where [tex]\(P = \frac{F}{A}\)[/tex].
3. Rearrange the formula to solve for [tex]\(A_2\)[/tex]:
[tex]\[ A_2 = \frac{F_2 \times A_1}{F_1} \][/tex]
4. Substitute the given values into the equation:
[tex]\[ A_2 = \frac{90 \, \text{N} \times 1 \times 10^{-4} \, \text{m}^2}{30 \, \text{N}} \][/tex]
5. Perform the calculation:
[tex]\[ A_2 = \frac{90 \times 1 \times 10^{-4}}{30} \][/tex]
[tex]\[ A_2 = \frac{90}{30} \times 10^{-4} \][/tex]
[tex]\[ A_2 = 3 \times 10^{-4} \, \text{m}^2 \][/tex]
6. Convert the area of the second piston back to square millimeters:
Since [tex]\(1 \, \text{m}^2 = 10^6 \, \text{mm}^2\)[/tex]:
[tex]\[ A_2 = 3 \times 10^{-4} \, \text{m}^2 \times 10^6 \, \text{mm}^2/\text{m}^2 = 300 \, \text{mm}^2 \][/tex]
### Conclusion:
The area of the output piston [tex]\(A_2\)[/tex] is [tex]\(300 \, \text{mm}^2\)[/tex].
### Given Data:
- Force on the first piston ([tex]\(F_1\)[/tex]) = 30 N
- Area of the first piston ([tex]\(A_1\)[/tex]) = 100 mm²
- Force on the second piston ([tex]\(F_2\)[/tex]) = 90 N
### Objective:
We need to calculate the area of the second piston ([tex]\(A_2\)[/tex]).
### Step-by-Step Solution:
1. Convert the area of the first piston to square meters:
Since [tex]\(1 \, \text{mm}^2 = 10^{-6} \, \text{m}^2\)[/tex]:
[tex]\[ A_1 = 100 \, \text{mm}^2 = 100 \times 10^{-6} \, \text{m}^2 = 1 \times 10^{-4} \, \text{m}^2 \][/tex]
2. Use the hydraulic system principle:
The hydraulic system operates on the principle of Pascal's Law, which states that the pressure in a closed system is constant. Mathematically, this is represented as:
[tex]\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \][/tex]
Here, [tex]\(P_1 = P_2\)[/tex] where [tex]\(P = \frac{F}{A}\)[/tex].
3. Rearrange the formula to solve for [tex]\(A_2\)[/tex]:
[tex]\[ A_2 = \frac{F_2 \times A_1}{F_1} \][/tex]
4. Substitute the given values into the equation:
[tex]\[ A_2 = \frac{90 \, \text{N} \times 1 \times 10^{-4} \, \text{m}^2}{30 \, \text{N}} \][/tex]
5. Perform the calculation:
[tex]\[ A_2 = \frac{90 \times 1 \times 10^{-4}}{30} \][/tex]
[tex]\[ A_2 = \frac{90}{30} \times 10^{-4} \][/tex]
[tex]\[ A_2 = 3 \times 10^{-4} \, \text{m}^2 \][/tex]
6. Convert the area of the second piston back to square millimeters:
Since [tex]\(1 \, \text{m}^2 = 10^6 \, \text{mm}^2\)[/tex]:
[tex]\[ A_2 = 3 \times 10^{-4} \, \text{m}^2 \times 10^6 \, \text{mm}^2/\text{m}^2 = 300 \, \text{mm}^2 \][/tex]
### Conclusion:
The area of the output piston [tex]\(A_2\)[/tex] is [tex]\(300 \, \text{mm}^2\)[/tex].
