Determine the magnitude of force P to be applied to a block of weight 1500N to hold it in the position on
a smooth inclined plane AB which makes an angle 30 with the horizontal.

To find the magnitude of the force P required to hold a block of weight 1500 N on a smooth inclined plane at an angle of 30 degrees, we need to analyze the forces acting on the block. Since the plane is smooth, we can ignore friction in our calculations.

Understanding the Forces at Play

The block experiences two main forces: its weight (acting vertically downward) and the normal force (acting perpendicular to the surface of the incline). The weight of the block can be resolved into two components: one parallel to the incline and one perpendicular to it.

Weight Components

Given that the weight of the block (W) is 1500 N, we can calculate the components of this weight:

• The component of weight acting parallel to the incline (W_parallel) can be calculated using the sine function:
• The component of weight acting perpendicular to the incline (W_{perpendicular}) can be calculated using the cosine function:

Using the angle θ = 30 degrees:

• W_parallel = W * sin(θ) = 1500 N * sin(30°) = 1500 N * 0.5 = 750 N
• W_{perpendicular} = W * cos(θ) = 1500 N * cos(30°) = 1500 N * (√3/2) ≈ 1299.04 N

Balancing Forces

To keep the block stationary on the incline, the applied force P must counteract the component of the weight acting parallel to the incline. Since there is no friction, the force P must equal W_parallel.

Calculating the Required Force

Thus, we have:

P = W_parallel = 750 N

Final Thoughts

In conclusion, the magnitude of the force P that needs to be applied to hold the block in position on the inclined plane is 750 N. This force effectively balances the gravitational pull acting down the slope, ensuring that the block remains stationary.