Engineer's Notebook §1.1

Free-body diagrams

Cut one body free, replace everything touching it with the force it exerts, and the forces left must balance. For a block resting on a slope at angle θ:

N = mg cos θ,F = mg sin θ
m
mass
kg
g
gravitational acceleration
m/s²
θ
slope angle
°
W
weight, mg
N
N
normal force
N
F
friction force
N
μ
coefficient of friction
—
P
applied force
N
Left, a block resting on a slope. Right, the same block drawn on its own with the three forces on it: weight straight down, the normal force perpendicular to the slope and friction up the slope. θ = 20° FREE BODY W N
Fig. 1.1-1Arrow lengths are to scale with the weight.
W = mg98.1 N
N = mg cos θ92.2 N
μN, friction available46.1 N
F = mg sin θ33.6 N

Worked example

Uses the values in Fig. 1.1-1. Change them and the working updates.

Given
  • mmass of the block10 kg
  • θslope angle20°
  • μcoefficient of friction0.5
Find

The normal force N and the friction force F that holds the block, and whether friction can supply it.

Solution
W = mg = 10 kg × 9.81 m/s² = 98.1 N N = W cos θ = 98.1 N × cos 20° = 92.2 N F = W sin θ = 98.1 N × sin 20° = 33.6 N μN = 0.5 × 92.2 N = 46.1 N
Answer
F = 33.6 N

Friction can supply up to 46.1 N, so the block holds.

Related cases

  • Level ground, horizontal pull P

    N = mgF = P ≤ μmg

  • Mass hanging on a cord

    T = mg

  • Pushed up a slope at steady speed

    P = mg(sin θ + μ cos θ)

Assumptions
  • The block is treated as a particle: every force acts through one point, so tipping is not checked.
  • Friction follows the Coulomb model. It supplies whatever force is needed, up to μN, and beyond that the block slides.
  • One μ is used for both starting and continued sliding. Real sliding friction is usually a little lower than static friction.
  • g = 9.81 m/s², and the slope is rigid and fixed.

Derivation

Take axes along the slope and perpendicular to it. The weight mg then has a component mg sin θ down the slope and mg cos θ into it.

  1. Perpendicular to the slope: N − mg cos θ = 0, so N = mg cos θ.
  2. Along the slope: F − mg sin θ = 0, so F = mg sin θ.
  3. The block holds while F ≤ μN, that is mg sin θ ≤ μmg cos θ.

The mass cancels, so the block starts to slide when tan θ = μ, whatever it weighs. That angle is the angle of repose.