Chap. 14: Static equilibrium

We start here on Monday April 16 and finish this material on Wednesday, April 18.

Newton's first law says that there is no change in the motion of an object on which no forces are acting.
We now know about torques and rotation, so we could say:

Revised first law: an object at rest remains at rest and an object remains in uniform motion (translation and rotation) if the sum of the forces and torques is zero.

This situation is so common that it has a special name.

Static equilibrium: A body is in static equilibrium when it is stationary and the net external forces and torques are zero.

A static equilibrium requires: S_i F_i = 0 and S_i t_i = 0

A problem that you might think you have is that if its not rotating, you don't know where the axis is so you cannot figure out and sum the torques. It doesn't matter! For an object not already rotating, you can use any axis to find the torques and if the sum is zero the object will not begin to rotate.

Example 1: Mass on a spring

For this problem we need only: S_i F_i = 0.

The spring force: F_s = -ky. (The y direction is up.)

The gravity force: F_g = -mg.

Equilbrium: F_s + F_g = -mg-ky = 0.

Then we must have: y = -mg/k.

This says that the mass will hang in a static equilibrium if the spring is stretched so that the y spring force just cancels gravity which requires that the spring be stretched by the distance –mg/k. If we start off with the spring stretched more or less than this the mass will bounce up and down!

Example 2: Mass hanging from beam of length L supported by a rope at the end

What is the tension in the rope T because of the hanging mass mg?

We neglect the mass of the beam.

The obvious choice for the pivot point is the left side of the beam.

The force -mg acts clockwise and is a negative torque -mgL.

The torque from the tension acts at an angle q. We must find the projection of this force onto the direction perpendicular to the beam.

The torque is rFsin q = LT sin q.

The net torque = LT sin q - mgL = 0.

From this we get: T = mg/(sin q).

How hard is the beam pushing against the wall? The beam is not accelerating horizontally so the force of the wall is exactly counteracting the horizontal component of the tension and is T cos q.

Example 3: The mass and the rope are attached at different places

Let's suppose the rope is attached at L/3 at the mass is attached at L.

The net torque = (L/3)T sin q - mgL = 0.

From this we get: T = 3mg/(sin q).

What would we get for the horizontal force exerted by the wall?

The role of the center of gravity

We know that we should carry an 8 foot piece of lumber holding on to the middle. If we don't, the longer section will tend to rotate downward and hit the ground. This is because gravity exerts a torque on the 8 foot lumber when it is NOT carried at the center. The best place to carry a sledgehammer is near the head because the iron head is so much heavier than the wooden handle. The "best place" is the center of gravity, where the sum of the torques due to gravity are zero.

We can express this idea mathematically. The torque from gravity is:

t = S_i r_i x F_i = S_i r_i x m_i g = S_i m_ir_i x g .

Let's represent the total mass by M and the downward force of gravity by Mg. (g is in the -y direction)

Then we can rewrite our equation by multiplying and dividing the right side by M:

t = [S_i m_ir_i / M] x Mg = r_com x Mg.

where r_com is the vector pointing to the center of mass.

Usually we get a torque by multiplying the force by the lever arm. So the torque of gravity we could get by using the force of gravity on the whole mass and the lever arm given by r_com = [S_i m_ir_i / M]. This vector points from the origin (the axis about which we want to know the torque) to the center of mass of the object.

Example: A mass M on a board supported by two scales

The scale on the right reads more weight than the one on the left. The board is 4 m long.

First we notice that the weight is sitting still (not accelerating). The board neither accelerates up or down so the sum of the forces up by the scales is equal to the force of gravity down. So if M is 100 lb, then the sum of the readings on the scales is 100 lb.

Second, we notice that the board is not rotating about any axis so the sum of the torques is zero.

That means that the force upward of the right scale multiplied by 1 meter is equal to the force of the left scale multiplied by 3 meters. So the scale on the right reads three times the scale on the left.

Thinking a bit, that means that the right scale reads 3/4 of the weight and the left scale reads 1/4 of the weight.
Let's use a different axis. We'll put the axis at the left scale. Gravity down times 3 meters is equal to the force of the right scale times 4 meters. Again, the force of the right scale is 3/4 of the weight of the object.

Stable and unstable equilibrium

A yardstick standing vertically on a table is in equilibrium. The force up by the table equals the weight. If the yardstick is perfectly vertical, the torque is zero because the force of gravity points exactly along the yardstick to the axis of rotation which is where the yardstick touches the tabletop.

We know the yardstick will fall over.

If it moves slightly in any direction, then there is a torque that tends to pull it in that direction. So once it starts, it moves faster and faster in the original direction. This is called an unstable equilibrium.

Unstable equilibrium: a point where the forces are zero and a displacement results in a force which is away from the equilibrium point.

Example: ball on a hilltop

Stable equilibrium: a point where the forces are zero and a displacement results in a (restoring) force which is toward the equilibrium point.

Example: ball at the bottom of a bowl.

How about a pencil balanced on the eraser?

Look at the drawings. On the left, the pencil is tilted slightly. The gravity force vector passes to the left of the pivot point (bottom right) so the torque (rF sin q) very small but not zero. This torque tends to make the pencil return upright. The pencil on the right, on the other hand, is tilted over more. The force vector passes to the right of the pivot point, so the pencil will continue to fall. So falling over depends on the location of the center of gravity relative to the pivot point.

Equilibria and energy conservation:

We have learned that if the potential energy goes down, the kinetic energy goes up.

What is the gravitational potential energy? It is Mgz for an object at height z.

If M is composed of many small masses, we can see that

Gravitational potential energy = U = S_i m_i g z_i = [S_i m_i z_i /M] Mg = M g (r_com)_z.

The symbol (r_com)_z means the z component (height) of the center of mass

This means that if the object moves, and the height of the center of mass goes down, then gravity will tend to make the object keep moving in the same direction (unstable).

But gravity will tend to stop motion causing the center of mass to rise.

The force is -dU/dz. To see this, recall that work is force times distance: W = F Dz.

Also recall that the work that gravity does is the decrease in the gravitational potential energy: W = -mg Dz = -DU, where again U is the potential energy.

Then W = F Dz = -DU, so F = -DU/Dz = -dU/dz.

If there is no force, we are in equilibrium, so we can say

Equilibrium for z displacement means: dU/dz = 0.

The plot of U(z) could be a straight line. For the straight line, d²U/dz² is also zero. To decide stable or unstable, we have to examine the second derivative.

Stable equilibrium means: d²U/dz² > 0 .

This means that the potential energy goes up if the particle is displaced. For example, the ball in a bowl. The curve at left is the function U(z). The horizontal scale is the displacement z and the vertical scale is the potential energy.

Unstable equilibrium means: d²U/dz² < 0 .

This means that the potential energy goes down is the particle is displaced. For example, the ball on top of a hill.

Metastable equilibrium: the pencil balanced on the eraser (again)

The curve at left is the potential energy as a function of diplacement. When the pencil is tilted a little bit its center of gravity is raised up because the pivot point is not under the center of gravity. The pivot point shifts from the left to the right as the pencil rocks from the left to the right. Once the center of gravity has moved so it is beyond the pivot point, the center of gravity gets lower as the pencil tilts more. Then it is unstable. This kind of equilibrium in which a small displacement is stable and a large displacement is unstable is called metastable.

Concept review:

1. A static equilibrium has S_i F_i = 0 and S_i t_i = 0.

2. The gravitational torque on a complex object can be calculated by placing the gravitational force vector mg at the center of mass.

3. A stable equilibrium is a point where the forces and torques are zero and a small displacement results in a restoring force.

4. An unstable equilibrium is a point where the forces and torques are zero and a small displacement results in a force driving the system further from equilibrium.

5. If U is the potential energy as a function of displacement x, dU/dx = 0 is an equilibrium point, and if

d²U/dx² > 0, the equilibrium is stable, and if

d²U/dx² < 0, the equilibrium is unstable.