A homogeneous rod of mass 3 kg is pushed along the smooth horizontal surface by a horizontal suface by a horizontal force F equal to 40 N. The angle `theta` for which rod hasd pure translation motion is `(g=10 m//s^(2))`
45 ∘
37 ∘
53 ∘
60 ∘

Options

A. `45^(@)`
B. `37^(@)`
C. `53^(@)`
D. `60^(@)`

Answer

To solve the problem of determining the angle `theta` for which the rod has pure translational motion, we can follow these steps:

### Step 1: Understand the Forces Acting on the Rod
We have a homogeneous rod of mass \( m = 3 \, \text{kg} \) being pushed by a horizontal force \( F = 40 \, \text{N} \). The weight of the rod acts downwards and is given by \( mg \), where \( g = 10 \, \text{m/s}^2 \).

### Step 2: Calculate the Weight of the Rod
The weight \( W \) of the rod can be calculated as:
\[ W = mg = 3 \, \text{kg} \times 10 \, \text{m/s}^2 = 30 \, \text{N} \]

### Step 3: Draw the Free Body Diagram
In the free body diagram, we have:
- The force \( F \) acting horizontally.
- The normal force \( N \) acting vertically upwards.
- The weight \( W \) acting vertically downwards.

### Step 4: Resolve Forces
When the rod is inclined at an angle \( \theta \):
- The horizontal component of the force \( F \) is \( F \cos \theta \).
- The vertical component of the force \( F \) is \( F \sin \theta \).

### Step 5: Apply Equilibrium Conditions
For pure translational motion, the net torque about the center of mass must be zero. Thus, we can set up the following equations based on the equilibrium of forces and torques.

1. **Vertical Forces**:
\[ N = mg \]

2. **Horizontal Forces**:
The horizontal component of the force must balance the torque due to the weight:
\[ F \sin \theta \cdot \frac{L}{2} = N \cos \theta \cdot \frac{L}{2} \]

### Step 6: Substitute for Normal Force
Substituting \( N = mg \) into the torque equation gives:
\[ F \sin \theta = mg \cos \theta \]

### Step 7: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{F}{mg} = \frac{\cos \theta}{\sin \theta} = \cot \theta \]
Thus, we can write:
\[ \tan \theta = \frac{mg}{F} \]

### Step 8: Substitute Known Values
Substituting the known values:
\[ \tan \theta = \frac{30 \, \text{N}}{40 \, \text{N}} = \frac{3}{4} \]

### Step 9: Find the Angle
To find the angle \( \theta \), we can use the inverse tangent function:
\[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \]
Using a calculator or trigonometric tables, we find:
\[ \theta \approx 37^\circ \]

### Conclusion
Thus, the angle \( \theta \) for which the rod has pure translational motion is:
\[ \boxed{37^\circ} \]