In one line
D'Alembert's principle says that adding a fictitious inertia force \( -ma \) to the real forces turns a dynamics problem into an equilibrium problem: \( \Sigma F - ma = 0 \).
Explained simply
When a bus brakes suddenly, you feel thrown forward, as if some force pushed you. No real object pushed you; it is your body resisting the change in motion. D'Alembert gave this feeling a name, the inertia force, and drew it as a force opposite to the acceleration.
Technically, rewriting \( \Sigma F = ma \) as \( \Sigma F + (-ma) = 0 \), the term \( -ma \) is the inertia force. With it included, the body is said to be in dynamic equilibrium, and all statics methods (\( \Sigma F_x = 0 \), \( \Sigma F_y = 0 \)) apply.
How it works
- Assume the direction of acceleration a.
- Draw the FBD with all real forces.
- Add the inertia force ma opposite to a, usually as a dashed arrow.
- Write equations of dynamic equilibrium: \( \Sigma F_x = 0 \), \( \Sigma F_y = 0 \) including the inertia force.
- Solve; a negative a means the assumed direction was wrong.
Key formula
Inertia force \( F_i = -m\vec{a} \) (N); dynamic equilibrium \( \Sigma \vec{F} + F_i = 0 \).
Quick example
A lift of mass 500 kg accelerates upward at 2 m/s\(^2\). Real forces: tension T up, weight 4905 N down; inertia force \( 500 \times 2 = 1000 \) N down. Dynamic equilibrium: \( T - 4905 - 1000 = 0 \), so T = 5905 N. When it accelerates downward at 2 m/s\(^2\), the inertia force is upward and T = 3905 N.
Common mistakes
- Drawing the inertia force in the same direction as the acceleration.
- Including the inertia force and then also writing \( \Sigma F = ma \) (double counting).
- Calling the inertia force a real force exerted by some body.
How MU asks it
A 3 to 5-mark theory part: 'State and explain D'Alembert's principle' with a lift or block example. Numericals (6 to 8 marks) say 'using D'Alembert's principle, find the acceleration and tension'. Show the dashed inertia force on the FBD, otherwise the method is not recognised.