rigid-body equilibrium conditions

A rigid body is in equilibrium when, in an inertial frame, it stays at rest or moves with constant linear and constant angular velocity.

Two independent conditions must hold: the vector sum of all external forces is zero, and the sum of the moments of all external forces about any point is zero. In the planar case:

$$\Sigma F_x = 0, \qquad \Sigma F_y = 0, \qquad \Sigma M = 0.$$

The moment of a force is $M = Fd$, where $d$ is the lever arm, the shortest distance from the axis to the line of action of the force; equivalently $M = rF\sin\theta$. The moment is positive when the force tends to rotate the body counterclockwise.

If $\Sigma \vec{F} = 0$, the total moment does not depend on the choice of point, so the point may be chosen so that unknown forces drop out of the equation. The weight of a body acts at its centre of gravity: at the midpoint of a uniform rod and at the centre of a uniform disc.

A uniform ruler weighing $2{,}0$ N is balanced on the tip of a pencil at its midpoint. A load of weight $5{,}0$ N hangs on the left at a distance of $0{,}20$ m from the support. At what distance on the right must a load of weight $4{,}0$ N be hung so that the ruler remains in equilibrium?

A uniform horizontal beam AB of length $3{,}0$ m weighs $150$ N and rests on two supports at points A and B. A $400$ N load is placed on the beam at a distance of $1{,}0$ m from A. Find the forces the beam exerts on the supports.

A uniform rod of length $L = 2{,}0$ m and mass $6{,}0$ kg is attached to a wall by a hinge and held horizontal by a cable fixed to its free end. The cable makes an angle of $40^\circ$ with the rod. Find the tension in the cable and the force the hinge exerts on the rod ($g = 9{,}8$ m/s$^2$).

Work in the PhET "Balancing Act" simulation.

1. Place two masses on the plank and make it horizontal; record $m_1$, $d_1$, $m_2$, $d_2$ and check that $m_1 d_1 = m_2 d_2$.
2. Add a third mass, predict the position of one mass from $\Sigma M = 0$, and only then release the plank — compare the prediction with the result.
3. To test the second condition, $\Sigma F = 0$, observe how the support reading changes when you rearrange the masses while keeping their total weight the same.

For a physical check, rest a ruler on a pencil on a level table, balance it with identical coins and confirm that the product of mass and lever arm on the left equals that on the right. Use only light, safe objects, do not load glass surfaces, and avoid sharp or heavy weights.

Show that equilibrium requires both $\Sigma \vec{F} = 0$ and $\Sigma M = 0$, and that when $\Sigma \vec{F} = 0$ the result does not depend on the point about which moments are taken.

A two-panel diagram.

• Panel 1: a horizontal beam on two supports carrying a load. Arrows show the forces $R_A$, $R_B$, the beam weight at the midpoint and the load weight; dashed lines mark the lever arms — perpendiculars from point A to the lines of action — with the distances labelled.
• Panel 2: the same beam with moments taken about point B; the lever arm of $R_B$ becomes zero, and the moment equation gives the same value of $R_A$.
• Beside them, a small circle with a counterclockwise arrow and a "$+$" sign shows the sign convention for moments.

On a playground the seesaw is a $3{,}0\ \text{m}$ board on a pivot at its centre. A $30\ \text{kg}$ child sits $1{,}5\ \text{m}$ to the left of the pivot and a $45\ \text{kg}$ child sits to the right. Both children have lifted their feet off the ground, and the board stays horizontal, turning neither way.

For the board to stay horizontal both equilibrium conditions hold. Torques about the pivot: $30\cdot 9{,}8\cdot 1{,}5 = 441\ \text{N}\cdot\text{m}$ anticlockwise against $45\cdot 9{,}8\cdot 1{,}0 = 441\ \text{N}\cdot\text{m}$ clockwise, so $\sum \tau = 0$ and the heavier child sits exactly $1{,}0\ \text{m}$ from the pivot. Sum of forces: $\sum F_y = 0$, so the pivot pushes up with $735\ \text{N} = (30+45)\cdot 9{,}8$, the total weight.

A gardener holds a wheelbarrow with $60\ \text{kg}$ of soil. The load's centre of mass is $0{,}40\ \text{m}$ from the wheel axle, and the handles the gardener lifts are $1{,}20\ \text{m}$ from the axle. The barrow is held still, tipping neither up nor down.

The torque of the load's weight about the axle is $60\cdot 9{,}8\cdot 0{,}40 = 235\ \text{N}\cdot\text{m}$; it must be balanced by the torque of the lifting force: $F\cdot 1{,}20 = 235\ \text{N}\cdot\text{m}$, so $F \approx 196\ \text{N}$. The second equilibrium condition — balance of vertical forces: $196\ \text{N} + N_{\text{wheel}} = 588\ \text{Н}$ — leaves the axle carrying about $392\ \text{N}$.

A shop sign hangs on a uniform horizontal beam of mass $8{,}0\ \text{kg}$ and length $1{,}20\ \text{m}$, hinged to the wall at one end. A stay cable runs from the wall to the far end of the beam at $40^\circ$ to the beam. The sign, of mass $3{,}0\ \text{kg}$, is attached at the far end.

The sum of torques about the hinge is zero: the beam's weight $78\ \text{N}$ with lever arm $0{,}60\ \text{m}$ gives $47\ \text{N}\cdot\text{m}$, the sign's weight $29\ \text{N}$ with lever arm $1{,}20\ \text{m}$ gives $35\ \text{N}\cdot\text{m}$, a total of $82\ \text{N}\cdot\text{m}$. The cable tension $T$ has lever arm $1{,}20\sin 40^\circ = 0{,}77\ \text{m}$, so $T = 82/0{,}77 \approx 1{,}1\cdot 10^{2}\ \text{N}$. From $\sum F_x = 0$ the hinge pushes the beam towards the wall with $T\cos 40^\circ \approx 82\ \text{N}$, and from $\sum F_y = 0$ upwards with $107 - T\sin 40^\circ \approx 38\ \text{N}$.