This **inclined plane calculator** is used to analyze the **forces** and **motion** of objects on **sloped surfaces** using **F = W sin(θ) + μ W * cos(θ)** formula.

For example, a box of **mass (m = 10 kg)** is placed on an inclined plane at an **angle (θ = 30°)**. Using the inclined plane calculator, one can input the mass and angle to find:

Force perpendicular to the incline: F_perpendicular = mgcos(θ) = 109.81 * cos(30°) = 84.87 N

Force parallel to the incline: F_parallel = mgsin(θ) = 109.81 * sin(30°) = 49.05 N

## Inclined Plane Calculator

Mass (kg) | Angle (°) | Friction Coefficient | Force Required (N) | Acceleration (m/s²) |
---|---|---|---|---|

10 | 15 | 0.05 | 43.2 | 2.01 |

25 | 30 | 0.2 | 164.9 | 3.42 |

50 | 45 | 0.1 | 417.3 | 5.89 |

100 | 20 | 0.15 | 451.6 | 2.57 |

75 | 35 | 0.25 | 534.8 | 3.76 |

15 | 10 | 0.1 | 25.5 | 1.17 |

40 | 25 | 0.3 | 214.5 | 4.12 |

60 | 40 | 0.2 | 471.5 | 6.25 |

80 | 50 | 0.15 | 682.2 | 4.84 |

120 | 15 | 0.1 | 295.4 | 3.09 |

90 | 30 | 0.25 | 432.7 | 5.01 |

30 | 45 | 0.2 | 245.0 | 4.10 |

55 | 20 | 0.05 | 207.6 | 2.89 |

70 | 35 | 0.3 | 556.3 | 4.78 |

110 | 10 | 0.2 | 194.5 | 2.45 |

125 | 25 | 0.15 | 676.7 | 5.67 |

## Inclined Plane Formula

**F = W * sin(θ)**

This formula assumes ideal conditions with no **friction**. In real-world scenarios, we must consider friction, leading to a more comprehensive formula:

**F = W ***sin(θ) + μ* W * cos(θ)

Where:

μis thecoefficient of frictionW * cos(θ)represents thenormal force

A **100 kg crate** on a **30° ramp** with a friction coefficient of **0.2**.

Weight (W)= 100 kg9.8 m/s² = *980 N

F = 980sin(30°) + 0.2980 * cos(30°)

F = 490 + 169.7 = 659.7 N

### Force Parallel to the Incline (F_parallel):

`F_parallel = mg sin(theta)`

Where:

- m = mass (kg)
- g = gravitational acceleration (9.81 m/s²)
- theta = angle of the incline (degrees)

### Normal Force (F_normal):

`F_normal = mg cos(theta)`

### Frictional Force (F_friction, if there’s friction):

`F_friction = μ F_normal = μ mg cos(theta)`

Where:

- μ = coefficient of friction

### Net Force (if friction is present):

`F_net = F_parallel - F_friction`

## Formula for Inclined Plane with Pulley

**F = (W ***sin(θ) + μ* W * cos(θ)) / n

Where

nis thenumber of supporting rope strandsin the pulley system.

With a **double pulley system** (n=2): **F = 659.7 / 2 = 329.85 N**

When a **pulley system** is incorporated with an inclined plane, it can significantly reduce the **force required** to move an object. The basic formula remains similar, but the **mechanical advantage** of the pulley system is factored in:

## How to Calculate an Inclined Plane

**Identify given information**: Mass, angle, coefficient of friction, etc.**Convert units**if necessary (e.g., kg to N for weight)**Draw a free body diagram**showing all forces**Apply the appropriate formula**based on the scenario**Solve for the unknown variable**

**Problem**: Calculate the **acceleration** of a **5 kg block** sliding down a **25° incline** with a friction coefficient of **0.1**.

**Step 1**: Identify given information

**Mass (m)**= 5 kg**Angle (θ)**= 25°**Coefficient of friction (μ)**= 0.1**Acceleration due to gravity (g)**= 9.8 m/s²

**Step 2**: No unit conversion needed

**Step 3**: Draw a **free body diagram** (forces: weight, normal force, friction)

**Step 4**: Apply **Newton’s Second Law** along the incline **F = ma****mg sin(θ) – μ mg * cos(θ) = ma**

**Step 5**: Solve for **acceleration (a)****a = g (sin(θ) – μ cos(θ))**

**a = 9.8**

*(sin(25°) – 0.1*cos(25°))**a = 9.8**

*(0.4226 – 0.1*0.9063)**a = 3.24 m/s²**

The **block accelerates** down the incline at **3.24 m/s²**.

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