Calculation of Applied Force P to Keep a Block Moving Up an Inclined Plane

A common scenario in statics involves determining the applied force required to keep a block moving up an inclined plane. Here is a detailed step-by-step guide to solve such a problem, including the necessary calculations and concepts.

Problem Statement

A 600 N block rests on a 30° plane. Find the value of the horizontal force P applied to keep the block moving up the plane if the coefficient of friction is 0.2.

Step-by-Step Solution

To solve this problem, follow these five steps:

Step 1: Identify Forces Acting on the Block

Weight W: Acts vertically downward with a magnitude of 600 N. Normal Force N: Acts perpendicular to the incline. Frictional Force F_f: Acts to oppose motion, given by F_f μN. Here, μ 0.2. Applied Force P: Applied horizontally to move the block up the incline.

Step 2: Resolve the Weight into Components

The weight W can be resolved into two components along and perpendicular to the incline:

Parallel to the Incline (W_{parallel}): Given by W_{parallel} W sinθ. Perpendicular to the Incline (W_{perpendicular}): Given by W_{perpendicular} W cosθ.

For a 30° incline, these components are:

W_{parallel} 600 sin(30°) 600 × 0.5 300 N

W_{perpendicular} 600 cos(30°) ≈ 600 × 0.866 519.62 N

Step 3: Calculate the Normal Force

The normal force N is approximately equal to the perpendicular component of the weight:

N W_{perpendicular} 519.62 N

Step 4: Calculate the Frictional Force

The frictional force F_f is calculated as:

F_f μN 0.2 × 519.62 ≈ 103.92 N

Step 5: Set Up the Equation of Motion

To keep the block moving up the plane, the net force along the incline must be zero. The forces along the incline include the component of the weight, the frictional force, and the applied force P. The equation is:

P cos(30°) - W_{parallel} - F_f 0

Substituting the values:

P cos(30°) - 300 - 103.92 0

Step 6: Solve for P

Expressing P in terms of known quantities:

P cos(30°) 300 103.92 403.92 N

Since cos(30°) √3 / 2:

P × (√3 / 2) 403.92

P (403.92 × 2) / √3 ≈ 807.84 / 1.732 ≈ 466.00 N

Conclusion

The value of the force P applied horizontally to keep the block moving up the plane is approximately 466 N.