Answer :
To solve the problem of finding the maximum extension and retraction force available from an air cylinder, we need to follow these steps:
1. Calculate the Bore Area:
- The bore of the cylinder is like the circular area where the piston moves. The formula for the area of a circle is [tex]\( \text{Area} = \pi \times (\text{radius})^2 \)[/tex].
- The bore diameter is given as 2 inches, so the radius is [tex]\( \frac{2}{2} = 1 \)[/tex] inch.
- Therefore, the bore area is [tex]\( \pi \times (1)^2 = 3.1416 \)[/tex] square inches (considering [tex]\(\pi \approx 3.1416\)[/tex]).
2. Calculate the Rod Area:
- The rod is also round but smaller, with a diameter of 1 inch. Therefore, its radius is [tex]\( \frac{1}{2} = 0.5 \)[/tex] inches.
- Using the same area formula, the rod area is [tex]\( \pi \times (0.5)^2 = 0.7854 \)[/tex] square inches.
3. Calculate the Extension Force:
- The extension force is simply the force exerted on the entire bore area.
- Given that the air pressure available is 100 psig (pounds per square inch gauge), the extension force is calculated as:
[tex]\[
\text{Extension Force} = 100 \, \text{psig} \times 3.1416 \, \text{square inches} = 314.16 \, \text{pounds}
\][/tex]
4. Calculate the Retraction Force:
- Retraction involves the movement of the rod as well, so we need to subtract the rod area from the bore area for the effective area.
- Therefore, the effective area for retraction is [tex]\( 3.1416 - 0.7854 = 2.3562 \)[/tex] square inches.
- Thus, the retraction force is:
[tex]\[
\text{Retraction Force} = 100 \, \text{psig} \times 2.3562 \, \text{square inches} = 235.62 \, \text{pounds}
\][/tex]
Therefore, the maximum extension and retraction forces available from the cylinder are approximately 314 pounds and 236 pounds, respectively.
1. Calculate the Bore Area:
- The bore of the cylinder is like the circular area where the piston moves. The formula for the area of a circle is [tex]\( \text{Area} = \pi \times (\text{radius})^2 \)[/tex].
- The bore diameter is given as 2 inches, so the radius is [tex]\( \frac{2}{2} = 1 \)[/tex] inch.
- Therefore, the bore area is [tex]\( \pi \times (1)^2 = 3.1416 \)[/tex] square inches (considering [tex]\(\pi \approx 3.1416\)[/tex]).
2. Calculate the Rod Area:
- The rod is also round but smaller, with a diameter of 1 inch. Therefore, its radius is [tex]\( \frac{1}{2} = 0.5 \)[/tex] inches.
- Using the same area formula, the rod area is [tex]\( \pi \times (0.5)^2 = 0.7854 \)[/tex] square inches.
3. Calculate the Extension Force:
- The extension force is simply the force exerted on the entire bore area.
- Given that the air pressure available is 100 psig (pounds per square inch gauge), the extension force is calculated as:
[tex]\[
\text{Extension Force} = 100 \, \text{psig} \times 3.1416 \, \text{square inches} = 314.16 \, \text{pounds}
\][/tex]
4. Calculate the Retraction Force:
- Retraction involves the movement of the rod as well, so we need to subtract the rod area from the bore area for the effective area.
- Therefore, the effective area for retraction is [tex]\( 3.1416 - 0.7854 = 2.3562 \)[/tex] square inches.
- Thus, the retraction force is:
[tex]\[
\text{Retraction Force} = 100 \, \text{psig} \times 2.3562 \, \text{square inches} = 235.62 \, \text{pounds}
\][/tex]
Therefore, the maximum extension and retraction forces available from the cylinder are approximately 314 pounds and 236 pounds, respectively.
