Question

A pair of bullocks exerts 120 kg pull at 30$^\circ$ to the vertical. The speed of ploughing is 1.35 kmph and V-shape furrow having 25 cm width and 10 cm depth is formed. Calculate the horse power developed by the bullocks.

• A. 0.3
• B. 0.6
• C. 1.0
• D. 1.2

Hint:

Be careful with the angle definition:
- If angle is with the vertical: use \( \sin(\theta) \) for the horizontal component.
- If angle is with the horizontal: use \( \cos(\theta) \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Power is defined as the rate of doing work.
In animal-drawn implements, the power developed depends on the force exerted in the direction of motion (referred to as the draft) and the forward speed of the animals.
When a pull is applied at an angle to the line of travel, only the horizontal component of this force (the draft) contributes to the work done.
Key Formula or Approach:
1. Draft (\( D \)), which is the horizontal component of the pull (\( P \)):
\[ D = P \sin(\theta) \]
where \( \theta \) is the angle made by the pull with the vertical line.
2. Speed of operation (\( v \)) converted to m/s:
\[ v = \frac{\text{Speed in km/h} \times 1000}{3600} \]
3. Power developed in kg-m/s:
\[ \text{Power} = D \times v \]
4. Horsepower (HP) developed:
\[ \text{HP} = \frac{\text{Power in kg-m/s}}{75} \]
Note: \( 1 \text{ metric HP} = 75 \text{ kg-m/s} \).

Step 2: Detailed Explanation:
Let us perform the calculations step-by-step:
1. Identify the given values:
- Pull, \( P = 120 \text{ kg} \).
- Angle with the vertical, \( \theta = 30^\circ \).
- Speed of ploughing, \( v = 1.35 \text{ km/h} \).
2. Calculate the horizontal draft force:
\[ D = 120 \times \sin(30^\circ) \]
Since \( \sin(30^\circ) = 0.5 \):
\[ D = 120 \times 0.5 = 60 \text{ kg} \]
3. Convert the speed to meters per second:
\[ v = \frac{1.35 \times 1000}{3600} = 0.375 \text{ m/s} \]
4. Calculate the work done per second (Power):
\[ \text{Power} = D \times v = 60 \text{ kg} \times 0.375 \text{ m/s} = 22.5 \text{ kg-m/s} \]
5. Convert power into horsepower:
\[ \text{HP} = \frac{22.5}{75} = 0.3 \text{ HP} \]
The horse power developed by the pair of bullocks is \( 0.3 \text{ HP} \).

Step 3: Final Answer:
The horsepower developed by the bullocks is 0.3.