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Statics — Forces, Moments, and Beams

24 min Engineering Math

Statics

For a structure to be in equilibrium, all forces and moments must balance — a system of linear equations.

Equilibrium Conditions

$$\Sigma F_x = 0, \quad \Sigma F_y = 0, \quad \Sigma M = 0$$

Sum of forces in every direction is zero, and sum of moments about any point is zero.

Moment (Torque)

$$M = F \times d$$

Force times perpendicular distance from the pivot.

Example 1

A 6 m beam is supported at both ends. A 600 N load sits 2 m from the left end. Find the support reactions.

Let $R_L$ and $R_R$ be the support forces.

$\Sigma F_y = 0$: $R_L + R_R = 600$.

$\Sigma M_L = 0$: $600(2) - R_R(6) = 0$ → $R_R = 200$ N.

$R_L = 600 - 200 = 400$ N.

Example 2

A seesaw: 40 kg child sits 2 m from pivot. Where must a 30 kg child sit to balance?

$40(2) = 30(d)$ → $d = 80/30 = 2.67$ m from pivot.

Example 3

Two forces act on a door: 20 N at 0.8 m from the hinge and 10 N at 0.4 m, both perpendicular. Net moment?

If same direction: $20(0.8) + 10(0.4) = 16 + 4 = 20$ N·m.

1. A 4 m beam with 800 N load at center. Find each support reaction.

2. Moment of 50 N force applied 3 m from pivot?

3. A beam: 500 N at 1 m from left, 300 N at 3 m from left, beam is 5 m. Find $R_L$.

4. Seesaw: 60 kg at 1.5 m. Where does 45 kg sit to balance?

5. What does $\Sigma M = 0$ mean physically?

6. If a beam has two equal loads placed symmetrically, what are the reactions?

Show Answer Key

1. $R_L = R_R = 400$ N (symmetric loading)

2. $M = 150$ N·m

3. $\Sigma M_R = 0$: $R_L(5) = 500(4) + 300(2) = 2600$ → $R_L = 520$ N

4. $d = 60(1.5)/45 = 2$ m

5. No rotational tendency — the structure is in rotational equilibrium.

6. Each reaction is half the total load.