Question

If 80 kg MB plough is pulled by a horizontal force of 46 kg, find the angle of friction between plough body and horizontal surface.

• A. 30°
• B. 40°
• C. 45°
• D. 60°

Hint:

In exam questions, if your calculated answer is very close to one of the options involving standard trigonometric angles (0°, 30°, 45°, 60°, 90°), it's highly likely that the standard angle is the correct answer. The given numbers are often chosen to lead to these specific results.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem relates the concepts of friction, normal force, and the angle of friction. When an object is pulled horizontally, the pulling force must overcome the force of kinetic friction to move it. The angle of friction (\(\phi\)) is related to the coefficient of friction (\(\mu\)).

Step 2: Key Formula or Approach:
1. The force of friction (\(F_f\)) is given by \( F_f = \mu \cdot N \), where \(\mu\) is the coefficient of friction and N is the normal force.
2. When an object is on a horizontal surface, the normal force (N) is equal to its weight (W), i.e., \(N = W = m \cdot g\).
3. The horizontal pulling force (P) required to move the object is equal to the force of friction, so \(P = F_f\).
4. The relationship between the coefficient of friction (\(\mu\)) and the angle of friction (\(\phi\)) is given by \( \mu = \tan(\phi) \).

Step 3: Detailed Explanation:
The problem uses 'kg' for both mass and force. This is common in agricultural engineering contexts where 'kg-force' (kgf) is used. We can assume the units are consistent.
Given:

• Mass of the plough (m) = 80 kg. This gives the normal force, N = 80 kgf.
• Horizontal pulling force (P) = 46 kg. This force is equal to the friction force, \(F_f = 46\) kgf. First, we calculate the coefficient of friction (\(\mu\)).
From the friction formula, \( F_f = \mu \cdot N \).
\[ \mu = \frac{F_f}{N} \] \[ \mu = \frac{46 \text{ kgf}}{80 \text{ kgf}} = 0.575 \] Next, we find the angle of friction (\(\phi\)) using the relationship \( \mu = \tan(\phi) \).
\[ \phi = \arctan(\mu) \] \[ \phi = \arctan(0.575) \] Calculating the value:
\[ \phi \approx 9° \] This value is very close to 30°. Let's re-check the calculation or consider if there's a simpler intended ratio. Often in exams, the numbers are chosen to give a standard angle. Let's see what the tangent of the given options are: (A) tan(30°) = \(1/\sqrt{3} \approx 0.577\) (B) tan(40°) \(\approx 0.839\) (C) tan(45°) = 1 (D) tan(60°) = \(\sqrt{3} \approx 1.732\) Our calculated coefficient of friction, \(\mu = 0.575\), is extremely close to \(\tan(30°) \approx 0.577\). The small difference is likely due to rounding in the problem's given values. Therefore, 30° is the intended answer.

Step 4: Final Answer:
The angle of friction is approximately 30°.