*Notes to a video lecture on http://www.unizor.com*

__Equilibrium of Solids__

As we know, if a vector sum of forces, acting on a point-object, equals

to null-vector, this point-object is at rest or in a state of uniform

motion along a straight line in any inertial frame of reference.

Alternatively, we can say that a frame of reference associated with this

point-object is

*inertial*.

The complexity of a concept of equilibrium of solids lies in the fact

that, even if vector sum of forces, acting on a solid, equals to

null-vector, this solid might not be in the state of rest in any

inertial system. In a simple case of two forces, equal in magnitude and

opposite in direction, applied to two different points of a solid, will

rotate it, making a frame of reference associated with it

*non-inertial*.

Any rotation of a solid, however complex, can be represented as a

combination of three rotations around three coordinate axes. Therefore, a

solid in a state of equilibrium should be in equilibrium relative to

each coordinate axis, and, to study equilibrium of a solid in

three-dimensional space, it is sufficient to study it relative to one

axis.

As we know, in case of rotation around an axis, the main factor replacing a concept of force for translational movement, is

*torque*. Since

*torque*is a cause of rotation, the necessary condition to prevent a rotation is

*balancing of torques*.

Torque is a vector equal to a vector (cross) product of a radius-vector

**from the axis of rotation to a point of application of force (perpendicularly to the axis) by a vector of force:**

*r*

*τ = r×F*Using this concept of torque, we can formulate the following necessary condition for an equilibrium of a solid.

**Sum of torques of all forces acting on a solid relative to each coordinate axis must be equal to null-vector**.

In the previous lecture we introduced a concept of a degree of freedom

and, in particular, mentioned six degrees of freedom of a solid: three

degrees of freedom along each coordinate axis and three degrees of

freedom of rotation around each coordinate axis.

If a vector sum of all forces acting on a solid is a null-vector, there

will be a point that is not moving anywhere or, more precisely, a frame

reference associated with this point is inertial. This condition assures

an equilibrium by three degrees of freedom associated with

translational motion (a motion

*along*coordinate axes).

If a vector sum of all torques relatively to each of the three

coordinate axes is a null-vector, there will be no rotation and the

system will be in equilibrium by three other degrees of freedom

associated with rotational motion (a motion

*around*coordinate axes).

**So, for a solid to be in equilibrium, the necessary conditions are**

all of the above - sum of all the forces and sum of all the torques must

be null-vectors.

all of the above - sum of all the forces and sum of all the torques must

be null-vectors

Consider the following example.

A weightless rod of a length

**is hanging horizontally on a thread fixed at its midpoint.**

*2r*On both ends of a rod there are equal weights

**hanging down.**

*W*To balance the translational movement of a system vertically down the force of tension

**on a thread that holds the rod must balance a sum of forces of weight acting in opposite direction:**

*T*

*T + W + W = 0*

*T = −2W*To balance the rotational movement of a system around a midpoint of a

rod the torque of one weight should be equal in magnitude to the torque

of another weight since the directions of these two torques are

opposite:

*τ*_{blue}= r ×W

*τ*_{red}= −r ×W

*τ*_{red}+ τ_{blue}= 0Consider a less symmetrical case below.

Here a rod of the length

**is hanging by a thread fixed not in the middle, but at a distance**

*3r***from the left edge and, therefore, at a distance**

*r***from the right edge.**

*2r*At the same time, the weight on the left is twice as heavy as the one on the right.

Tension on the thread holding a rod must be greater to hold more weight:

*T + W + 2W = 0*

*T = −3W*From the rotational viewpoint, this system is in equilibrium because the

torques on the left of a rod and on the right are equal in magnitude,

while opposite in direction:

*τ*_{blue}= 2r ×W

*τ*_{red}= −r ×2W

*τ*_{red}+ τ_{blue}= 0