- Note
- Things to remember

A rigid body will be in equilibrium if the following two conditions are met.

- The sum of the forces acting on the body must be zero.
- The net torque acting on the body must be zero.

For an object to be in translational equilibrium the vector sum of the forces acting on it must be zero. Starting that the vector sum of forces acting on an object is zero is equivalent to:

$$ \sum F_x = 0, \: \: \: \sum F_y = 0, \: \: \: \sum F_z = 0. $$

Where F_{x} , F_{y} and F_{z} are the components of a force F in three perpendicular directions. It means that all the force along the x-axis add to zero. Same is true for forces along y and z axis. From Newton’s second law of motion,

\begin{align*} F &= ma \\ \text {For translational motion} , F = 0, so \\ \text{or,} \: 0 &= ma \\ \text{or,} \: a &= 0 \\ \text{or,} \: v &= \text {constant} \end{align*}

Thus, when a body is in transitional equilibrium, it ie either at rest or it is moving with constant velocity.

For an object to be in rotational equilibrium, the net torque acting on it must be zero.

$$\text {Net torque,} \: \tau = 0 $$

Stating that the net torque acting on an object is zero means that angular acceleration of the body about any axis of rotation is zero. So, the body is either at rest or moving with constant angular velocity about the axis.

**Stable equilibrium**

A body is in stable equilibrium if it returns to its equilibrium position after it has been displaced slightly as in the figure. In this equilibrium, C.G. of the body lies low and when the body is displaced from its equilibrium position, the C.G at this position is higher than before.

**Unstable equilibrium**

A body is in unstable equilibrium if it does not return to its equilibrium position after it has been displaced slightly as in the figure. In this equilibrium, C.G. of the body lies high and when the body is displaced from its equilibrium position, the C.G at this position is lower than before.

**Neutral equilibrium**

A body is in neutral equilibrium if it always stays in the displaced position after it has been displayed slightly as in the figure. In this equilibrium height of the C.G. of the body does not change but remains at the same height from the base in all displaced position.

**Condition for the body in stable equilibrium**

The conditions for a body in stable equilibrium are:

- The C.G. of the body should lie as low as possible.

G.G. of the body should lie as low as possible for a body to be in the stable state. Due to this reason, the bottom of the ship is made heavy and the cargo is always kept at its base. This makes the ship more stable. - The base of the body should be as large as possible.

The body is in stable equilibrium when its base is as large as possible. AN animal with four legs is more stable than the animal with two legs. - C.G. should lie within the base of the body on displaced position.

A man carrying a bucket of water in his right-hand leans towards the left-hand side so that vertical line through the C.G will pass through the base. Due to the same reason, a man carrying a load on his back has to bend forward. If he wants to carry the load straight upon his back, the vertical line through the C.G. falls outside the base and cannot carry the road.

It states that if a particle is in an equilibrium state under the action of three concurrent forces, then each force is proportional to the sine of the angle between the other two.

Consider three forces P, Q and R are acting on the particle A such that the particle is in equilibrium. If α, β, and γ are the angles between the three forces as shown in the figure. From lami's theorem we have

$$ \frac {P}{\sin \alpha} = \frac {Q}{\sin \beta} = \frac {R}{\sin \gamma} $$

Since the particle is in equilibrium under the action of three forces, they can be represented in magnitude and direction by three sides DABC as shown in the figure. i.e. forces P, Q, and R are represented by side AB, BC and CA respectively. Also ∠A = 180^{o} – β, ∠ B = 180^{o} – γ and ∠ C = 180^{o} – α.

From sine law of triangle ABC, we have

\begin{align*} \frac {AB}{\sin C} = \frac {BC}{\sin A} = \frac {CA}{\sin B} \frac {P}{\sin (180^o -\alpha)} &= \frac {Q}{\sin (180^o -\beta)} = \frac {R}{\sin 180^o -\gamma)}\\ \therefore \frac {P}{\sin \alpha} & = \frac {Q}{\sin \beta} = \frac {R}{\sin \gamma} \\ \end{align*}

Hence Lami’s therorem is proved.

For an object to be in translational equilibrium the vector sum of the forces acting on it must be zero.

For an object to be in rotational equilibrium, the net torque acting on it must be zero.

Stating that the net torque acting on an object is zero means that angular acceleration of the body about any axis of rotation is zero.

A body is in stable equilibrium if it returns to its equilibrium position after it has been displaced slightly as in the figure.

A body is in unstable equilibrium if it does not return to its equilibrium position after it has been displaced slightly as in the figure.

A body is in neutral equilibrium if it always stays in the displaced position after it has been displayed slightly as in figure.

It states that if a particle is in equilibrium state under the action of three concurrent forces, then each force is proportional to the sine of the angle between the other two.

.-
## You scored /0

## ASK ANY QUESTION ON Equilibrium of Rigid Bodies

No discussion on this note yet. Be first to comment on this note