## PHYSICS

### §5. Equilibrium of a body with a fixed axis of the rotation in the planar case. Moment of the force.

Let's limit the motion of a solid body by the rotation around the motionless fixed axis O (see Figure 6) and prohibit the motion of this body along that axis. At Figure 6 the O axis is perpendicular to the plane of the figure. It should be recalled that the solid body is the body which deformations from the acting forces may be neglected. Let's consider the planar case, that is the case when all acting to the body forces are situated in the same plane perpendicular to the axis of rotation. Only two forces , applied at the points K and E, are shown to make the figure less complicated. We are interested in the conditions of the absence of body's rotation, that is, when the body should be in the equilibrium.

Let's draw two perpendiculars to the point O on the line of forces' action and designate the distances from the axis of rotation to this lines as . Let's displace the points of force's application along their lines of action to points .

If you do not have a sufficiently good knowledge of the concept of the work of a force, you may pass through the next paragraph of the text containing the proof of the conditions of the equilibrium and begin the reading from the equation (5).

Let's make the experiment in your mind. Let the body to rotate infinitely slow to infinitely small angle a clockwise in the result of negligible disturbance of the equilibrium, for example, a very small increase of the force , or a very weak push. It should result in the movement of the points of application of the forces to the small distances .

During that the force makes on the body the negative work and the force makes on the body the positive . The work of all forces on the body should be equal to the change of its kinetic energy. But during infinitesimally slow rotation of the body the kinetic energy was and remains equal to zero. So, we can obtain, taken into account that other forces, in addition to above mentioned two forces, act to the body:

The last equality is the condition of the equilibrium.We can see that the quantitative value, responsible for the equilibrium of the body and characterizing the possibility of the separate force to rotate the body, is not the force itself, but the value, equal to the product of the force's modulus and the distance from the axis of rotation to the line of force's action.

*The distance from the axis of rotation to the line of force's action is called the arm of the force.*

For the planar case *the value of the product force's modulus and the force's arm is called the moment of force relative to the axis*: . The absolute value of the moment of the force is determined from this formula.

The condition of the equilibrium (5) takes the form , if *the moments causing the clockwise rotation receive plus sign (or they are considered as positive), and the moments causing the counterclockwise rotation receive minus sign (or they are considered as negative).*

*So, the solid body with the fixed axis of rotation is in equilibrium in some inertial frame of reference if the algebraic sum of the moments of external forces acting to the body relative to this axis is equal to zero.*

In other words it is possible to say that the equilibrium is achieved then the sum of the moments rotating the body clockwise is equal to then the sum of the moments rotating the body counterclockwise. If the forces acting to the body has the resultant force then the condition of the equilibrium can be formulated as follows. The moment of the resultant force relative to the axis of rotation is equal to zero, that is, the line of action of the resultant force (if it is not equal to zero) crosses the axis of rotation.

The formulated condition of the equilibrium is necessary but it is not sufficient. Really, if the sum of the moments of forces acting to the body is equal to zero then the body can be not only motionless but also rotating with constant angular velocity.

It should be noted that the body is influenced by the force from the side of the axis of rotation, to which it sticks. It is clear that independently from the direction of this force its moment relative to the axis of rotation is equal to zero because of the equality to zero of the force's arm. Therefore the reaction force acting to the body from the side of the axis is sometimes completely excluded from the consideration.

Let's describe another convenient from the practical point of view method of the determination of the moment of the force relative to the axis O (see Figure 7); the point C of application of this force is situated at the distance r from the axis. Let's decompose the force vector to force with the line of action, crossing the O axis, and the force perpendicular to , that is, change a single force to an equivalent system of two forces. The moment of the force relative to O axis is equal to zero and the arm of the force is equal to r. Therefore the moment M of the force is equal to the moment of the force relative to O axis, that is .

## PROBLEM 3

The homogenous beam of mass (see Figure 8) is attached with a joint in the lower point A and is held out by the light rope BC for the upper end; the angle between the rope and the beam is equal to . The weight of mass is suspended at the point B. The angle between the beam and the horizontal is equal to . Let's find the tension force of the rope.

## SOLUTION

Several forces act to the beam: the tension force of the rope , the force from the weight , the gravity force , applied in the center of the beam, and the reaction force from a joint. Let's designate the length of the beam as L. Let's write the expressions for the moments of the forces relative to the axis A. The moment of the force is equal to zero. The moment of the force is equal to . The moment of the force is equal to . The arm of the force . Its moment if equal to . It is possible to determine the moment by the different method by the decomposition of the force to two components , and to note, that .

Let's write the condition of the equilibrium of the beam with the fixed axis A: In that way, . From this it follows that .

### INDEX

- §1. Introduction.
- §2. Force. Equivalence of forces. Resultant force. Addition of forces. Decomposition of force.
- §3. Equilibrium of a material point.
- §4. Equilibrium of a body in the absence of the rotation.
- §5. Equilibrium of a body with a fixed axis of the rotation in the planar case. Moment of the force.
- Test questions.
- Problems.