D’alembert’s principle states that the sum of the differences between the forces acting on a mass particle and the rate of change of momentum of the system itself along any virtual displacement is zero.

$\sum_i (F_i - m_i a_i ) \delta r_i=0\space F_i= $Net force acting on $i^{th}$ particle

$m_i = $Mass of $i^{th}$ particle

$a_i= $Acceleration of $i^{th}$ particle

$r_i = $ Virtual displacement of $i^{th}$ particle

Note:-In many textbooks, it is given as $\sum F-ma=0,$ that is a special case (refer Wikipedia).

Note: Please ask your profs regarding what to write in the exams.

Example 1:-

Consider 2 teams playing tug-of-war. A box of mass ‘m’ is attached to a rope at two opposite places. Team A pulls the box with force $F_A$ and Team B pulls it with force $F_B.$ The force of Team A is more, so the box is accelerated towards Team A.

By D’alembert’s principle,

$ [F_A+F_B ]-ma=0$

Example 2:-

A box of mass ‘m’ is kept on an inclined surface with co-efficient of friction $‘µ’$ and is accelerated downwards.

By D’alembert’s principle (along x-axis)

$[-f+mg \sin\theta ]-ma=0 \\ i.e. \space -μ(mg \cos\theta )+ mg \sin\theta -ma=0$