The radius of one arm of a hydraulic lift is three times the radius of the other arm. What force should be applied on the narrow arm so as to lift 50 kg at the wider arm?

Options

A. 60 N
B. 54.4 N
C. 26.7 N
D. 30 N

Answer

To solve the problem of determining the force that should be applied on the narrow arm of a hydraulic lift to lift a 50 kg weight on the wider arm, we can follow these steps:

### Step 1: Understand the relationship between the radii
Given that the radius of the wider arm (R2) is three times the radius of the narrow arm (R1), we can express this relationship mathematically:
\[ R2 = 3R1 \]

### Step 2: Calculate the force due to the weight
The weight (F2) of the object being lifted can be calculated using the formula:
\[ F2 = m \cdot g \]
where:
- \( m = 50 \, \text{kg} \) (mass of the object)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating F2:
\[ F2 = 50 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 490 \, \text{N} \]

### Step 3: Use the hydraulic lift principle
According to the principle of hydraulic lifts, the pressure is the same on both sides:
\[ \frac{F1}{A1} = \frac{F2}{A2} \]
where:
- \( F1 \) is the force applied on the narrow arm,
- \( A1 \) is the area of the narrow arm,
- \( A2 \) is the area of the wider arm.

### Step 4: Calculate the areas in terms of the radii
The areas can be expressed in terms of the radii:
\[ A1 = \pi R1^2 \]
\[ A2 = \pi R2^2 = \pi (3R1)^2 = 9\pi R1^2 \]

### Step 5: Substitute the areas into the pressure equation
Substituting the areas into the pressure equation gives:
\[ \frac{F1}{\pi R1^2} = \frac{490 \, \text{N}}{9\pi R1^2} \]

### Step 6: Simplify the equation
The \(\pi R1^2\) terms cancel out:
\[ F1 = \frac{490 \, \text{N}}{9} \]

### Step 7: Calculate F1
Now, calculating \( F1 \):
\[ F1 = \frac{490}{9} \approx 54.44 \, \text{N} \]

### Final Answer
The force that should be applied on the narrow arm is approximately:
\[ F1 \approx 54.44 \, \text{N} \]