Inclined Planes Forces

Reference data and engineering information about inclined planes forces for mechanics applications.

Overview

Engineering reference data for Inclined Planes Forces in mechanics.

Key Formulas

Work

$W = Fd\cos\theta$

Work = force × displacement × cos(angle).

Kinetic Energy

$E_k = \frac{1}{2}mv^2$

Energy of motion.

Potential Energy

$E_p = mgh$

Gravitational potential energy.

Variables

Symbol	Description	Unit
$F$	Force	N
$m$	Mass	kg
$a$	Acceleration	m/s²
$v$	Velocity	m/s

When a body rests on an inclined plane and is in equilibrium (not sliding), the gravitational component down the slope is balanced by static friction. At this critical angle, the angle of repose relationship holds:

$\mu = \frac{F_p}{F_n} = \frac{W \sin(\alpha)}{W \cos(\alpha)} = \tan(\alpha)$

The angle of repose $\alpha$ equals the arctangent of the static friction coefficient:

$\alpha_{repose} = \arctan(\mu)$

This means the maximum stable angle depends only on the surface materials, not the object's mass.

Work and Power on Inclines

When moving an object along an inclined plane over a distance $d$ , the work done against gravity is:

$W = F_p \cdot d$

The power required to maintain constant velocity $v$ along the incline is:

$P = \frac{W}{t} = F_p \cdot v$

where $t = d / v$ is the travel time.

Example: Vehicle on Graded Roads

An electric car with mass $m = 2400\ \text{kg}$ travels $d = 1\ \text{km}$ along inclines (neglecting rolling and air resistance).

Inclination	Slope Angle $\alpha$	Force $F_p$	Work (1 km)	Power at 80 km/h	Power at 60 km/h
5%	2.86°	2,051 N	2,051 kJ	45.6 kW	34.1 kW
10%	5.71°	4,088 N	4,088 kJ	90.8 kW	67.9 kW

Key insight: Doubling the road grade from 5% to 10% approximately doubles both the required force and power output.

References

Original Source