10. The radii of the small piston and the large piston of a hydraulic press are 2 in and 16 in, respectively. Find the upward thrust on the large piston when a 20 lb effort is applied to the small piston.



Answer :

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.