La statique est l'étude des corps au repos ou en mouvement uniforme. L'un des cas les plus classiques et les plus importants en mécanique est celui d'un solide soumis à trois forces. C'est un pilier des programmes de physique (lycée et premier cycle universitaire).
Si vous cherchez à maîtriser ce concept pour réussir vos interrogations ou simplement comprendre comment fonctionne l'équilibre, cet article est fait pour vous. Nous allons voir la théorie indispensable, puis détailler un exercice corrigé étape par étape.
La poutre est le solide étudié. On note :
Step 1 – Identify forces
Three forces act on the rod:
Step 2 – Check concurrency
Lines of action of ( \vecW ) and ( \vecT ) intersect at point C (extend W vertically down from center, extend T from B to wall – they meet above rod). Therefore, ( \vecR_A ) must also pass through C.
Thus, the three forces are concurrent at C.
Step 3 – Draw force triangle
Compute angles:
Find ( \angle ) of ( R_A ):
From geometry:
Triangle ABC: AB = 2 m, midpoint M, AM = 1 m.
In triangle MBC: MB = 1 m.
Vertical line from M, string from B at 30° → C is above rod.
Coordinates: A(0,0), B(2,0), M(1,0).
String equation from B: y = tan(30°)(x – 2) = 0.577(x – 2).
Vertical line at x=1: y = 0.577(1 – 2) = -0.577? Negative means intersection below – impossible.
Wait – error: String is from B to wall above horizontal? If angle with horizontal is 30°, from B to wall left-up, slope positive? No, if wall is vertical left of A, string goes from B(2,0) to a point on wall (0, y_w). Slope = (y_w – 0)/(0 – 2) = -y_w/2. Given angle 30° with horizontal, slope = tan(30°) = 0.577 if measured from horizontal, but direction left-up means slope positive? Actually from B to wall, Δx = -2, Δy positive, so slope = Δy/Δx negative? No – slope magnitude = 0.577 but sign negative because Δx negative. Let’s do properly:
String direction: from B to D (0, d). Vector BD = (-2, d).
Angle with horizontal: tan(θ) = d/2? No, angle with horizontal given 30°, so |d/2| = tan 30°=0.577 → d = 1.154 m. So D(0, 1.154).
Then slope of string = (1.154 – 0)/(0 – 2) = -0.577. So line eq from B: y – 0 = -0.577(x – 2) → y = -0.577x + 1.154.
Vertical line at x=1 (midpoint): y = -0.577(1) + 1.154 = 0.577 m above rod. Good. So C = (1, 0.577).
Now line AC: from A(0,0) to C(1,0.577). Slope = 0.577 → angle = 30° above horizontal.
Thus ( \vecR_A ) makes 30° above horizontal (to the right-up). Good.
Step 4 – Force triangle angles
Force triangle:
Now in force triangle, place ( \vecW ) first vertical down. Then from its tip, draw ( \vecT ) parallel to 150° direction (i.e., 30° above negative x-axis). Then from tail of W, draw ( \vecR_A ) parallel to 30° direction (30° above positive x-axis). La statique est l'étude des corps au repos
Angles in force triangle:
So force triangle is equiangular (all 60° interior) → equilateral.
Step 5 – Magnitudes
Equilateral triangle with one side W = 100 N → all sides = 100 N.
Thus:
Step 6 – Check by analytical method (ΣFx=0, ΣFy=0)
Rx = R_A cos30° = 100 × 0.866 = 86.6 N
Ry = R_A sin30° = 50 N
Forces:
Horizontal: T cos30°? Wait T direction: 150° from +x → cos150° = -cos30° = -0.866, sin150° = 0.5.
So Tx = T cos150° = 100 × (-0.866) = -86.6 N (leftward)
Ty = T sin150° = 100 × 0.5 = 50 N (upward)
Weight W: (0, -100)
Reaction R_A: (86.6, 50)
Sum Fx: 86.6 + (-86.6) = 0 ✓
Sum Fy: 50 + 50 + (-100) = 0 ✓
A solid is in equilibrium under the action of three non-parallel forces if and only if:
When these conditions are met, the three forces form a closed triangle when placed head-to-tail (the dynamic triangle).
To give you a taste of what’s inside the exclusive PDF, here is a classic problem fully corrected.
A uniform rod AB of length 2 m and weight 100 N is hinged at A to a vertical wall. The rod is held horizontally by a string attached at B and inclined at 30° to the horizontal.
Find:
![Diagram description: Vertical wall left, rod horizontal, hinge at A left end, string from B right end up-left to wall, angle 30° with horizontal.]
Since the rod is in equilibrium under three forces, their lines of action must intersect at one point.