Determine the horizontal force F to be applied to the block weighing 1500N to hold it in position on a smooth inclined plane PQ which moves an angle of 30 degree with the horizontal.

To find the horizontal force \( F \) required to hold a block weighing 1500 N on a smooth inclined plane that is inclined at an angle of 30 degrees, we need to analyze the forces acting on the block. The block is subject to gravitational force, and since the plane is smooth, there is no friction to consider. Let's break this down step by step.

Understanding the Forces at Play

The weight of the block acts vertically downward and can be resolved into two components relative to the inclined plane:

• The component acting perpendicular to the incline, which we will call \( W_{\perp} \).
• The component acting parallel to the incline, which we will call \( W_{\parallel} \).

Given that the weight \( W \) of the block is 1500 N, we can calculate these components using trigonometric functions:

Calculating the Components of Weight

Using the angle of the incline (\( \theta = 30^\circ \)), we can find the components:

• The perpendicular component \( W_{\perp} = W \cdot \cos(\theta) = 1500 \cdot \cos(30^\circ) \).
• The parallel component \( W_{\parallel} = W \cdot \sin(\theta) = 1500 \cdot \sin(30^\circ) \).

Now, substituting the values:

• Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \), we have:
• \( W_{\perp} = 1500 \cdot 0.866 \approx 1299 \, \text{N} \).
• And \( \sin(30^\circ) = \frac{1}{2} = 0.5 \), thus:
• \( W_{\parallel} = 1500 \cdot 0.5 = 750 \, \text{N} \).

Determining the Required Horizontal Force

To keep the block stationary on the incline, the horizontal force \( F \) must counteract the parallel component of the weight \( W_{\parallel} \). However, we also need to consider the angle at which this force is applied. The horizontal force can be resolved into two components as well:

• A component acting parallel to the incline, which is \( F \cdot \cos(30^\circ) \).
• A component acting perpendicular to the incline, which is \( F \cdot \sin(30^\circ) \).

For the block to remain stationary, the component of the horizontal force acting parallel to the incline must equal the parallel component of the weight:

Setting Up the Equation

Thus, we can set up the equation:

\( F \cdot \cos(30^\circ) = W_{\parallel} \)

Substituting the known values:

\( F \cdot 0.866 = 750 \)

Now, solving for \( F \):

\( F = \frac{750}{0.866} \approx 866 \, \text{N} \).

Final Thoughts

The horizontal force \( F \) that needs to be applied to hold the block in position on the inclined plane is approximately 866 N. This analysis illustrates how forces can be resolved and balanced in a system, which is a fundamental concept in physics and engineering.