Find solutions to your problems with the expert advice available on IDNLearn.com. Our experts are available to provide accurate, comprehensive answers to help you make informed decisions about any topic or issue you encounter.
Sagot :
To solve the problem of finding the upward thrust on the large piston in a hydraulic press system when a 20-pound effort is applied to the small piston, we can follow these steps:
### Step 1: Understand Given Information
- Radius of the small piston, [tex]\( r_s \)[/tex]: 2 inches.
- Radius of the large piston, [tex]\( r_l \)[/tex]: 16 inches.
- Effort applied to the small piston, [tex]\( F_s \)[/tex]: 20 pounds.
### Step 2: Calculate the Area of the Small Piston
The area of a circle is given by the formula [tex]\( A = \pi r^2 \)[/tex], where [tex]\( r \)[/tex] is the radius of the circle.
For the small piston:
[tex]\[ A_s = \pi r_s^2 = \pi \times (2 \text{ inches})^2 = 4\pi \text{ square inches} \][/tex]
### Step 3: Calculate the Area of the Large Piston
Again, using the area formula for a circle:
For the large piston:
[tex]\[ A_l = \pi r_l^2 = \pi \times (16 \text{ inches})^2 = 256\pi \text{ square inches} \][/tex]
### Step 4: Apply Pascal's Principle
Pascal's principle states that pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container. Mathematically, it can be expressed as:
[tex]\[ \frac{F_s}{A_s} = \frac{F_l}{A_l} \][/tex]
where:
- [tex]\( F_s \)[/tex] is the force applied to the small piston.
- [tex]\( A_s \)[/tex] is the area of the small piston.
- [tex]\( F_l \)[/tex] is the upward thrust on the large piston.
- [tex]\( A_l \)[/tex] is the area of the large piston.
### Step 5: Solve for the Upward Thrust on the Large Piston
Rearrange the formula to solve for [tex]\( F_l \)[/tex]:
[tex]\[ F_l = F_s \times \frac{A_l}{A_s} \][/tex]
Plug in the known values:
[tex]\[ F_l = 20 \text{ pounds} \times \frac{256\pi \text{ square inches}}{4\pi \text{ square inches}} \][/tex]
Notice that [tex]\( \pi \)[/tex] cancels out in the numerator and denominator:
[tex]\[ F_l = 20 \times \frac{256}{4} \][/tex]
[tex]\[ F_l = 20 \times 64 \][/tex]
[tex]\[ F_l = 1280 \text{ pounds} \][/tex]
### Conclusion
The upward thrust on the large piston is [tex]\( 1280 \)[/tex] pounds when a 20-pound effort is applied to the small piston.
### Step 1: Understand Given Information
- Radius of the small piston, [tex]\( r_s \)[/tex]: 2 inches.
- Radius of the large piston, [tex]\( r_l \)[/tex]: 16 inches.
- Effort applied to the small piston, [tex]\( F_s \)[/tex]: 20 pounds.
### Step 2: Calculate the Area of the Small Piston
The area of a circle is given by the formula [tex]\( A = \pi r^2 \)[/tex], where [tex]\( r \)[/tex] is the radius of the circle.
For the small piston:
[tex]\[ A_s = \pi r_s^2 = \pi \times (2 \text{ inches})^2 = 4\pi \text{ square inches} \][/tex]
### Step 3: Calculate the Area of the Large Piston
Again, using the area formula for a circle:
For the large piston:
[tex]\[ A_l = \pi r_l^2 = \pi \times (16 \text{ inches})^2 = 256\pi \text{ square inches} \][/tex]
### Step 4: Apply Pascal's Principle
Pascal's principle states that pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container. Mathematically, it can be expressed as:
[tex]\[ \frac{F_s}{A_s} = \frac{F_l}{A_l} \][/tex]
where:
- [tex]\( F_s \)[/tex] is the force applied to the small piston.
- [tex]\( A_s \)[/tex] is the area of the small piston.
- [tex]\( F_l \)[/tex] is the upward thrust on the large piston.
- [tex]\( A_l \)[/tex] is the area of the large piston.
### Step 5: Solve for the Upward Thrust on the Large Piston
Rearrange the formula to solve for [tex]\( F_l \)[/tex]:
[tex]\[ F_l = F_s \times \frac{A_l}{A_s} \][/tex]
Plug in the known values:
[tex]\[ F_l = 20 \text{ pounds} \times \frac{256\pi \text{ square inches}}{4\pi \text{ square inches}} \][/tex]
Notice that [tex]\( \pi \)[/tex] cancels out in the numerator and denominator:
[tex]\[ F_l = 20 \times \frac{256}{4} \][/tex]
[tex]\[ F_l = 20 \times 64 \][/tex]
[tex]\[ F_l = 1280 \text{ pounds} \][/tex]
### Conclusion
The upward thrust on the large piston is [tex]\( 1280 \)[/tex] pounds when a 20-pound effort is applied to the small piston.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com is committed to providing accurate answers. Thanks for stopping by, and see you next time for more solutions.
