Equilibrium (Dynamics) | Complete notes | Important Questions | Short Questions and Numerical Problem Solutions | Class 11 (Physics)

Equilibrium

Image by Suraj Rai: Stone in Equilibrium

Equilibrium is the state of rest or balance due to the equal action of opposing forces.

Requirements of Equilibrium

The vectors sum of all the external forces that act on the body must be zero. $\vec{F_{net}}=0$
The vector sum of all external torques that act on the body, measured about any possible point, must also be zero. $\vec{\tau_{net}}=0$

Moments

When the steering wheel of a car is turned, the applied force is said to exert a moment , or turning-effect, about the axle attached to the wheel. The magnitude of the moment of a force F about a point O is defined as the product of the force F and the perpendicular distance OA from O to the line of action of F (fig.). Thus

$moment= \vec{F}\times OA $ The moment of a force is also called a torque(τ).The magnitude of the moment is expressed in newton metre (Nm) where F is in Newton and OA is in metre. We shall take an anticlockwise moment as positive in sign and a clockwise moment as negative in sign.

Parallel Forces and Equlibrium

When a number of forces act on a body and their directions are parallel, they are called parallel forces. If the parallel forces are acting in the same direction, they are called like parallel forces and when they are acting in the opposite direction, they are called unlike parallel forces . as in fig.below.

If a rod carries loads of 10, 20, 30, 15, and 25 N at O, A, B, C, D respectively as in fig., the resultant R of the weights, which are parallel forces, is given by,

R= 10+20+30+15+25= 100 N

It is found from empirical observation and theoretical observation that the moment of the resultant of a number of forces about any point is equal to the algebraic sum of the moments of the individual forces about the same point . This result enables us to find where the resultant R acts. Taking moments about O for all the forces in fig., above, we have,
20 ×0.6+30×0.9+15×1.3+25×2.1
If x is the distance of R from O, the moment of R about O=R×x =100×x .

100 ×x
So, x=1.1 m

The resultant of number of forces in equilibrium is zero; and the moment of the resultant about any point is hence zero. It therefore follows that the algebraic sum of the moments of all the forces about any point is zero when those forces are in equilibrium. This means that,

total clockwise moment of the forces about any point=the total anticlockwise moment of the remaining forces about the same point.

Torque due to a couple

Two equal and opposite forces whose line of action ( line where the force is acting ) do not coincide are said to form a couple. The two forces always have a turning effect, or moment, called a torque , which is defined as,

torque ( τ )=one force ( F ) × perpendicular distance between forces (r)

Centre of gravity (c.g.)

The gravitational force F on a body effectively acts at a single point, called the ( center of gravity ), c.g. of the body The term "effectively" suggests that if the gravitational forces on the individual elements of a body were somehow turned off and the gravitational force F at the c.g. were turned on, the net force and the net torque (about any point) acting on the body would not change.
In the simple case of a ruler, the c.g. is the point of support where the ruler is balanced. The c.g. of the curved surface of a hollow cylinder acts at the mid-point of the cylinder axis. It is also the position of the c.g. of a uniform solid cylinder. The c.g. of a triangular plate or lamina is two-thirds of the distance along a median from corresponding point of the triangle. The center of gravity of a symmetrical body (body having uniform mass density) lies at the geometrical centre of the body.

In some cases, such as hollow bodies or irregularly shaped objects, the centre of gravity (or centre of mass) may occur in space at a point external to the physical material—e.g., in the centre of a tennis ball or between the legs of a chair.

Centre of mass (c.m.)

Take a smooth uniform rod on a horizontal surface with negligible friction such as ice. What do you see when you struck the rod by a force near one end? Does the rod rotate? Does it accelerate? Now, struck the same rod at the centre. What happens now? In the first case, you will find that the rod will rotate as it accelerates. However, in the second case, the rod will accelerate without rotation. In this case, you have strucked at the centre of mass of the rod. Centre of mass of an object may be defined as the point in which an applied force produces acceleration but no rotation.

Consider the figure above, which shows the particles of masses m₁, m₂, ..., which together forms the object of total mass M. If x₁, x₂,..., are the respective x-cordinates of the particles relative to axes O_x, O_y, then generally the co-ordinate $\color{blue}\bar{x}$ of the centre of mass C is defined by,

$\bar{x}=\frac{m_1x_1+m_2x_2+...}{m_1+m_2+...}=\frac{\Sigma mx}{M}$ Similarly, the distance $\bar{y}$ of the centre of mass C from O_x is given by, $\bar{y}=\frac{\Sigma my}{M}$

States of equlibrium

Stable equilibrium : If the body comes back to it's original position even after slight displacement (fig.(a)), then the body is said to be in stable equilibrium. In this case, the centre of gravity (C.G.) after displacement rises little higher than it's initial position. Take a coca-cola bottle with it's base lying on the table and displace it slightly from its position without giving it a translational motion, the bottle will come back to its original position. It is an example of stable equilibrium.
Unstable equilibrium : If the body does not return to it's original equilibrium position after slight displacement (fig.(b)), the body is said to be in unstable equilibrium. In this case, C.G. after displacement is lower than before.Take the same coca-cola bottle but now with it's mouth lying on the table and displace it slightly. Now, the bottle will not return to it's initial position : falls down!
Neutral equilibrium : If the body always stays in the displaced position, then the body is said to be in neutral equilibrium (fig.(c)). In this case, C.G. does not change. Take the same coca-cola bottle taken in above examples. Now, lie it in the table and displace it slightly, it will now remain in the displaced position. It is an example of neutral equilibrium.

Conditions of stable equilibrium

C.G. of the body should lie as low as possible : Due to this reason, the base of a vehicle, the bottom of a ship, etc. is made heavy. When the C.G. of the body rises, ther is high chance of toppling. For. eg., the truck loaded with iron rod (C.G. is lowered!) is stable however with cotton loaded on it (C.G. is raised!) has high chance of toppling.
The base of a body should be as large as possible : Four wheeler vehicles are more stable compared with 2 wheeler vehicles as 4 wheelers have larger base. Due to the same reason, cow is more stable than man.
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. Doing this, the vertical line through the C.G. will pass through the base.

During pregnancy, women often develop back pains from leaning backward while walking. Why do they have to walk this way?
For a body to be in stable equilibrium, the vertical line passing through the center of gravity should lie within the base of the body on the displaced position. During pregnancy, C.G. shifts forward and thus the body will be unstable and then she might topple forward and fall on her stomach. However, when the pregnant woman leans backward, the vertical line through C.G. will lie within the base of the body and will make her stable.

Click on Equilibrium_questions to find the important questions.
Click on Equilibrium_solutions to find the solutions to short questions and numerical problems.
Jump to Rotational Dynamics .

Physics in Depth