Radius of one arm of hyddraulic lift is foud times off radius of other arm. What force should be applied on narrow arm to lift 100 kg?

Options

A. 26.5 N
B. 62.5 N
C. 6.25 N
D. 8.3 N

Explanation

To solve the problem, we will use the principle of hydraulic lifts, which states that the pressure applied on one side of the lift is equal to the pressure on the other side.

### Step-by-Step Solution:

1. **Identify the Radii**:
- Let the radius of the narrow arm (r1) be \( r \).
- Then, the radius of the wider arm (r2) is \( 4r \) (since it is four times the radius of the narrow arm).

2. **Calculate the Areas**:
- The area of the narrow arm (A1) is given by:
\[ A_1 = \pi r_1^2 = \pi r^2 \]
- The area of the wider arm (A2) is given by:
\[ A_2 = \pi r_2^2 = \pi (4r)^2 = \pi \cdot 16r^2 \]

3. **Set Up the Pressure Equation**:
- According to the principle of hydraulics, the pressure on both sides must be equal:
\[ \frac{F}{A_1} = \frac{F_{lift}}{A_2} \]
- Here, \( F \) is the force applied on the narrow arm, and \( F_{lift} \) is the weight being lifted (which is 100 kg).

4. **Convert Weight to Force**:
- The weight \( F_{lift} \) can be calculated as:
\[ F_{lift} = m \cdot g = 100 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 980 \, \text{N} \]

5. **Substitute Areas into the Pressure Equation**:
- Substitute \( A_1 \) and \( A_2 \) into the pressure equation:
\[ \frac{F}{\pi r^2} = \frac{980}{\pi \cdot 16r^2} \]

6. **Cancel Common Terms**:
- The \( \pi \) and \( r^2 \) terms cancel out:
\[ F = \frac{980}{16} \]

7. **Calculate the Force**:
- Now, calculate \( F \):
\[ F = \frac{980}{16} = 61.25 \, \text{N} \]

### Final Result:
The force that should be applied on the narrow arm to lift 100 kg is approximately **61.25 N**.