Answer :
Certainly! Let's solve the problem step by step:
1. Understanding the given information:
- The weight (force) to be lifted using the hydraulic lift is [tex]\(2500 \, \text{N}\)[/tex].
- The cross-sectional area of the small piston is [tex]\(\frac{1}{10}\)[/tex] of the cross-sectional area of the big piston.
2. Hydraulic lift principle:
- A hydraulic system works on the principle of Pascal's law which states that pressure applied at any point in a confined fluid is transmitted equally in all directions throughout the fluid.
- Therefore, the pressure exerted by the big piston and the small piston in the hydraulic lift must be the same.
3. Relationship between force and area:
- Let [tex]\(F_{\text{big}}\)[/tex] be the force applied by the big piston, and [tex]\(A_{\text{big}}\)[/tex] be the area of the big piston.
- Similarly, let [tex]\(F_{\text{small}}\)[/tex] be the force applied by the small piston, and [tex]\(A_{\text{small}}\)[/tex] be the area of the small piston.
4. Setting up the equations based on the principle of equal pressure:
- According to Pascal's law:
[tex]\[ \text{Pressure}_{\text{big}} = \text{Pressure}_{\text{small}} \][/tex]
- Thus,
[tex]\[ \frac{F_{\text{big}}}{A_{\text{big}}} = \frac{F_{\text{small}}}{A_{\text{small}}} \][/tex]
5. Expressing [tex]\(A_{\text{small}}\)[/tex] in terms of [tex]\(A_{\text{big}}\)[/tex]:
- We are given that [tex]\(A_{\text{small}} = \frac{1}{10} A_{\text{big}}\)[/tex].
6. Substituting [tex]\(A_{\text{small}}\)[/tex] in the pressure equation:
- Substituting [tex]\(A_{\text{small}} = \frac{1}{10} A_{\text{big}}\)[/tex] into the pressure equation, we get:
[tex]\[ \frac{F_{\text{big}}}{A_{\text{big}}} = \frac{F_{\text{small}}}{\frac{1}{10} A_{\text{big}}} \][/tex]
- Simplifying, we find:
[tex]\[ \frac{F_{\text{big}}}{A_{\text{big}}} = \frac{10 F_{\text{small}}}{A_{\text{big}}} \][/tex]
7. Simplifying the equation to find [tex]\(F_{\text{small}}\)[/tex]:
- Eliminating [tex]\(A_{\text{big}}\)[/tex] from both sides,
[tex]\[ F_{\text{big}} = 10 F_{\text{small}} \][/tex]
- Solving for [tex]\(F_{\text{small}}\)[/tex],
[tex]\[ F_{\text{small}} = \frac{F_{\text{big}}}{10} \][/tex]
8. Substituting the known value of [tex]\(F_{\text{big}}\)[/tex]:
- We know [tex]\(F_{\text{big}} = 2500 \, \text{N}\)[/tex], so:
[tex]\[ F_{\text{small}} = \frac{2500 \, \text{N}}{10} = 250 \, \text{N} \][/tex]
Therefore, the force required to be applied on the small piston is [tex]\(250 \, \text{N}\)[/tex].
1. Understanding the given information:
- The weight (force) to be lifted using the hydraulic lift is [tex]\(2500 \, \text{N}\)[/tex].
- The cross-sectional area of the small piston is [tex]\(\frac{1}{10}\)[/tex] of the cross-sectional area of the big piston.
2. Hydraulic lift principle:
- A hydraulic system works on the principle of Pascal's law which states that pressure applied at any point in a confined fluid is transmitted equally in all directions throughout the fluid.
- Therefore, the pressure exerted by the big piston and the small piston in the hydraulic lift must be the same.
3. Relationship between force and area:
- Let [tex]\(F_{\text{big}}\)[/tex] be the force applied by the big piston, and [tex]\(A_{\text{big}}\)[/tex] be the area of the big piston.
- Similarly, let [tex]\(F_{\text{small}}\)[/tex] be the force applied by the small piston, and [tex]\(A_{\text{small}}\)[/tex] be the area of the small piston.
4. Setting up the equations based on the principle of equal pressure:
- According to Pascal's law:
[tex]\[ \text{Pressure}_{\text{big}} = \text{Pressure}_{\text{small}} \][/tex]
- Thus,
[tex]\[ \frac{F_{\text{big}}}{A_{\text{big}}} = \frac{F_{\text{small}}}{A_{\text{small}}} \][/tex]
5. Expressing [tex]\(A_{\text{small}}\)[/tex] in terms of [tex]\(A_{\text{big}}\)[/tex]:
- We are given that [tex]\(A_{\text{small}} = \frac{1}{10} A_{\text{big}}\)[/tex].
6. Substituting [tex]\(A_{\text{small}}\)[/tex] in the pressure equation:
- Substituting [tex]\(A_{\text{small}} = \frac{1}{10} A_{\text{big}}\)[/tex] into the pressure equation, we get:
[tex]\[ \frac{F_{\text{big}}}{A_{\text{big}}} = \frac{F_{\text{small}}}{\frac{1}{10} A_{\text{big}}} \][/tex]
- Simplifying, we find:
[tex]\[ \frac{F_{\text{big}}}{A_{\text{big}}} = \frac{10 F_{\text{small}}}{A_{\text{big}}} \][/tex]
7. Simplifying the equation to find [tex]\(F_{\text{small}}\)[/tex]:
- Eliminating [tex]\(A_{\text{big}}\)[/tex] from both sides,
[tex]\[ F_{\text{big}} = 10 F_{\text{small}} \][/tex]
- Solving for [tex]\(F_{\text{small}}\)[/tex],
[tex]\[ F_{\text{small}} = \frac{F_{\text{big}}}{10} \][/tex]
8. Substituting the known value of [tex]\(F_{\text{big}}\)[/tex]:
- We know [tex]\(F_{\text{big}} = 2500 \, \text{N}\)[/tex], so:
[tex]\[ F_{\text{small}} = \frac{2500 \, \text{N}}{10} = 250 \, \text{N} \][/tex]
Therefore, the force required to be applied on the small piston is [tex]\(250 \, \text{N}\)[/tex].
