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2.5.7.1. Conditions for Equilibrium

Table of Contents

Balance of Forces and Torques

Static equilibrium describes a situation where an object remains at rest and does not start moving or turning. For this to happen, two separate requirements must be satisfied. The object must have no overall tendency to translate, and it must have no overall tendency to rotate.

The first requirement concerns forces. If the forces acting on an object do not cancel, the object will accelerate according to Newton's second law. The second requirement concerns torques. Even if the forces cancel, the object can still rotate if their turning effects do not cancel.

So the conditions for equilibrium are twofold. The net force must be zero, and the net torque must be zero.

For static equilibrium, both conditions must hold:
$$\sum \vec{F} = 0$$
and
$$\sum \tau = 0$$
If either one is not satisfied, the object is not in equilibrium.

Translational Equilibrium

Translational equilibrium means that the center of mass of the object does not accelerate. In simple terms, the pushes and pulls in each direction balance out.

In two dimensions, this means

$$\sum F_x = 0, \qquad \sum F_y = 0$$

In three dimensions, there is also

$$\sum F_z = 0$$

These equations are just the component form of the vector equation $\sum \vec{F} = 0$.

For example, if a book rests on a table, gravity pulls downward and the table pushes upward with an equal normal force. Since these vertical forces cancel, the book does not move up or down.

Rotational Equilibrium

Rotational equilibrium means that the object has no angular acceleration. The clockwise and counterclockwise turning effects balance each other.

Torque depends on the force, the distance from the pivot, and the angle at which the force acts. In many equilibrium problems, it is enough to treat clockwise torque as negative and counterclockwise torque as positive, or the reverse, as long as the sign convention is used consistently.

Thus rotational equilibrium requires

$$\sum \tau = 0$$

This means that the total clockwise torque equals the total counterclockwise torque.

A common mistake is to check only forces. An object can have
$$\sum \vec{F} = 0$$
but still rotate if
$$\sum \tau \ne 0$$
Equilibrium requires both conditions at the same time.

Choosing a Pivot Point

Torque is calculated about a chosen axis or pivot point. In equilibrium, the net torque is zero about any point. This is very useful, because you may choose the point that makes the calculation easiest.

A smart choice of pivot often removes unknown forces from the torque equation. If a force acts exactly through the pivot, its lever arm is zero, so it produces no torque about that point.

Consider a beam supported at one end, with a load hanging farther out. If we take torques about the support, the support force does not appear in the torque equation because its lever arm is zero.

Beam in static equilibrium

Force and Torque Conditions Together

Most static equilibrium problems require solving both the force equations and the torque equation. Each gives independent information.

For a rigid body in a plane, the usual set of equations is

$$\sum F_x = 0$$

$$\sum F_y = 0$$

$$\sum \tau = 0$$

These three equations are often enough to determine up to three unknown quantities.

Example of a Balanced Beam

Suppose a horizontal beam is supported at one point. A weight $W_1$ acts at distance $d_1$ on one side, and another weight $W_2$ acts at distance $d_2$ on the other side. If the beam is in rotational equilibrium, then the torques must cancel:

$$W_1 d_1 = W_2 d_2$$

This is the basic balance condition for many lever problems.

For a simple lever in equilibrium,
$$\text{clockwise torque} = \text{counterclockwise torque}$$
For perpendicular forces, this becomes
$$F_1 d_1 = F_2 d_2$$

This relation shows that a smaller force can balance a larger one if it acts farther from the pivot.

Sign Conventions

When writing torque equations, you must choose a sign convention. A common choice is that counterclockwise torques are positive and clockwise torques are negative. Another choice also works, provided it is used consistently throughout the problem.

Forces in the horizontal and vertical directions also need signs. Usually, rightward and upward are taken as positive.

The following table shows a common convention.

QuantityPositive direction
$x$ forceRight
$y$ forceUp
Torque $\tau$Counterclockwise

With this convention, the equilibrium equations in two dimensions become

$$\sum F_x = 0, \qquad \sum F_y = 0, \qquad \sum \tau = 0$$

Why Both Conditions Matter

It is useful to see why force balance alone is not enough. Imagine two equal and opposite forces acting on opposite sides of an object. Their sum is zero, so there is no translational acceleration. But if they act at different points, they can form a couple and make the object rotate.

Likewise, torque balance alone is not enough. Torques could cancel while the net force is not zero, causing the object to accelerate without rotating.

So equilibrium is a complete balance of motion, both linear and rotational.

Equal and opposite forces causing rotation

In this case,

$$\sum \vec{F} = 0$$

but

$$\sum \tau \ne 0$$

so the body rotates.

Practical Problem Solving

When solving equilibrium problems, the general method is simple. First identify all forces acting on the object. Then write the force balance equations in each direction. After that, choose a convenient pivot and write the torque balance equation. Solve the equations together.

Often, the hardest part is not the algebra, but choosing the correct distances and signs for torques.

Summary of the Conditions

Static equilibrium for a rigid body means complete mechanical balance. The object remains at rest only if there is no linear acceleration and no angular acceleration.

In two-dimensional problems, the full set of conditions is

$$\sum F_x = 0$$

$$\sum F_y = 0$$

$$\sum \tau = 0$$

The conditions for equilibrium are the central result of this chapter:
$$\sum \vec{F} = 0$$
and
$$\sum \tau = 0$$
In two dimensions, this is usually written as
$$\sum F_x = 0,\qquad \sum F_y = 0,\qquad \sum \tau = 0$$

These equations form the foundation of all static equilibrium calculations for beams, ladders, bridges, signs, and many other rigid bodies.

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