A body of weight 20N is on a horizontal surface, minimum force applied to pull it when applied force makes an angle 60° with horizontal (angle of friction a = 30°) is

Options

A. 20 N
B. `20sqrt(3)N`
C. `20/sqrt(3)`N
D. 20 N

Answer

To solve the problem, we need to determine the minimum force required to pull a body of weight 20 N when the applied force makes an angle of 60° with the horizontal, given that the angle of friction (α) is 30°.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Weight of the body (W) = 20 N
- Angle of applied force (θ) = 60°
- Angle of friction (α) = 30°

2. **Calculate the Coefficient of Static Friction (μ)**:
- The coefficient of static friction can be calculated using the angle of friction:
\[ \mu = \tan(\alpha) = \tan(30°) = \frac{1}{\sqrt{3}} \]

3. **Resolve the Applied Force (F)**:
- Let the applied force be \( F \).
- The horizontal component of the applied force is:
\[ F_{\text{horizontal}} = F \cos(60°) = \frac{F}{2} \]
- The vertical component of the applied force is:
\[ F_{\text{vertical}} = F \sin(60°) = F \frac{\sqrt{3}}{2} \]

4. **Calculate the Normal Force (N)**:
- The normal force can be calculated as:
\[ N = W - F_{\text{vertical}} = 20 N - F \frac{\sqrt{3}}{2} \]

5. **Use the Condition for Motion**:
- For the body to just start moving, the horizontal component of the applied force must equal the maximum static friction:
\[ F_{\text{horizontal}} = \mu N \]
- Substituting the values we have:
\[ \frac{F}{2} = \mu \left(20 N - F \frac{\sqrt{3}}{2}\right) \]
- Substitute \( \mu = \frac{1}{\sqrt{3}} \):
\[ \frac{F}{2} = \frac{1}{\sqrt{3}} \left(20 N - F \frac{\sqrt{3}}{2}\right) \]

6. **Solve for F**:
- Multiply both sides by \( \sqrt{3} \):
\[ \frac{F \sqrt{3}}{2} = 20 N - F \frac{\sqrt{3}}{2} \]
- Rearranging gives:
\[ \frac{F \sqrt{3}}{2} + F \frac{\sqrt{3}}{2} = 20 N \]
\[ F \sqrt{3} = 20 N \]
\[ F = \frac{20 N}{\sqrt{3}} \approx 11.55 N \]

### Final Answer:
The minimum force required to pull the body is:
\[ F = \frac{20}{\sqrt{3}} \text{ N} \]