Chapter 12 The conditions for Equilibrium of a Rigid Body

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12.0 Equilibrium conditions of a rigid object

A rigid body, such as chair, a bridge, or a building is said to be in a mechanical equilibrium if as viewed from a initial reference frame.

Equilibrium occurs when an object is not accelerated and is not rotating. In order for an object to be in equilibrium, it must satisfy the conditions for equilibrium:

§ First condition of equilibrium is satisfied when the vector sum of the forces acting on an object equal zero, i.e., .

If the line of action of each forces acting on the object passes through a common point, the forces are concurrent and is the only condition needed to solve the problem.

§ Second condition of equilibrium is satisfied when the sum of all torques acting on an object about any axis perpendicular to the plane of the forces equals zero

If the forces are not concurrent or the line of action of each force does not pass through a common point, then a net torque will cause a change in the state of rotation of the object. For non-concurrent force problems, both conditions of equilibrium must mutually be applied to solve the problem.

If the forces all lie in the xy-plane, then all the torque vectors are parallel to the z-axis. In calculating the torque, you end up with a single scalar component equation where the sum of all the torque in the z plane is equal to zero.

12.1 Static Equilibrium occurs when an object at equilibrium is at rest

When an object is said to be in static equilibrium, that object is neither in translation motion nor in rotational motion. Therefore, in this special case, the total momentum, both linear and angular are equal to zero.

Conditions for static equilibrium:
§ No translational motion: Þ

§ No rotational motion: Þ , about any axis of rotation.

The two above conditions: implies there is no motion of any kind including rotational or translational motions.

Stable equilibrium: if displaced slightly, an object will return to its original position.

Unstable equilibrium: if displaced slightly, it moves farther away from its original position.

Examples Static Equilibrium:

Forces must balance to zero, , 1^st condition of equilibrium.
Torques must balance to zero, , 2^nd condition of equilibrium.
Note: (1 does not automatically imply 2)
Action of opposing forces causes stress.
Stress leads to strain (deformation)
Near Earth Gravity acts at the center of mass; center of mass and center of gravity are at same location

12.2 Solved Examples

Example 12.1 – A Seesaw with unequal masses

A uniform beam of length L meters and weight M kg supports two unequal masses each of mass m₁ and m₂ respectively, balanced at the opposite ends of the seesaw, as shown in the figure below,

Draw the free-body-diagram.

(a) Calculate the normal force N exerted on the beam by the support.

(b) Developed the equations for net force and net torque for the system using equilibrium conditions for a rigid body.

Solution

An important technique in solving problems of this type is the idea that we may choose any axis to sum the torques from. If we choose the point A, we see that

Text Box:

From the FBD, the net force can be written as:

(, system under equilibrium condition)

x-component: (no forces acting in the x-direction)

y-component:

We see that the normal force, N, is given by:

In addition, we may write the net torque using the second condition for equilibrium as:

Example 12.2 - The Seesaw

A uniform beam of weight 40 N supports two unequal animals, a monkey weighing 300 N and a bear weighing 800 N, balanced at the opposite ends of the seesaw, a shown in Fig. . If the support (called fulcrum) is placed under the center of gravity of the beam, and the monkey is 3m from the center,

(a) Calculate the normal force N exerted on the beam by the support.

(b) Determine where the bear should be in order to balance the system.

Text Box: Fig. 12.3 A Seesaw with unequal masses.

Solution: From the free-body-diagram, Fig. (b), we can observed that in additional to the normal force , the external forces acting on the board are the weights of the animals and the weight of the beam, all of which act downward. Further, we can assume that the center of gravity of the beam is at its geometric center because we are told that the board is uniform. Also, since the system is in equilibrium, the upward force must be balance all the downward forces, or the net force, , i.e.,

(1^st condition of equilibrium)

x-component: (no forces acting in the x-direction)

y-component:

b) To find the position of the bear must be to balance the system, we must apply the second condition for equilibrium , and taking the center of gravity of the beam as the axis of rotation, i.e.,

Exercise: If the beam was not uniform and the fulcrum did not lies under the center of gravity of the beam, what other information would need to solve the problem.