A homogenous rod of mass 3kg is pushed along the horizontal surface by a horizontal force 40 N. the rod makes angle ‘theta’ with the horizontal. The angle ‘theta’ for which rod has pure translation motion. take g=10m/s². Please help in this question

To determine the angle θ at which the rod experiences pure translational motion while being pushed by a horizontal force, we need to analyze the forces acting on the rod. Pure translational motion means that the rod is not rotating; thus, the net torque about any point must be zero. Let's break this down step by step.

Understanding the Forces Involved

When a horizontal force is applied to the rod, several forces come into play:

• The applied force (F) of 40 N.
• The weight of the rod (W), which acts vertically downward and is equal to the mass of the rod multiplied by the acceleration due to gravity (W = mg).
• The normal force (N) acting perpendicular to the surface.
• The frictional force (f) that opposes the motion, which can be affected by the angle θ.

Calculating the Weight of the Rod

The weight of the rod can be calculated as follows:

W = mg = 3 \, \text{kg} \times 10 \, \text{m/s}^2 = 30 \, \text{N}

Setting Up the Equations

For the rod to have pure translational motion, the torque about the center of mass must be zero. The torque (τ) due to the applied force can be expressed as:

τ = F \cdot d \cdot \sin(θ)

Where:

• F = applied force (40 N)
• d = distance from the center of mass to the point of application of the force (which we can assume to be the length of the rod divided by 2 for simplicity).
• θ = angle of inclination of the rod with respect to the horizontal.

Condition for Pure Translation

For pure translation, the torque due to the weight of the rod must balance the torque due to the applied force. The weight acts at the center of mass, and its torque can be expressed as:

τ_w = W \cdot \frac{L}{2} \cdot \cos(θ)

Setting the torques equal gives us:

F \cdot \frac{L}{2} \cdot \sin(θ) = W \cdot \frac{L}{2} \cdot \cos(θ)

We can simplify this equation by canceling out the common terms:

F \cdot \sin(θ) = W \cdot \cos(θ)

Substituting Known Values

Now, substituting the known values of F and W:

40 \cdot \sin(θ) = 30 \cdot \cos(θ)

Finding the Angle θ

To find θ, we can rearrange the equation:

tan(θ) = \frac{30}{40} = \frac{3}{4}

Now, taking the arctangent of both sides gives:

θ = arctan\left(\frac{3}{4}\right)

Calculating the Angle

Using a calculator or trigonometric tables, we find:

θ ≈ 36.87°

Thus, the angle θ at which the rod experiences pure translational motion while being pushed by a horizontal force of 40 N is approximately 36.87 degrees. This angle ensures that the torques due to the applied force and the weight of the rod are balanced, preventing any rotational motion.