Mechanical Equilibrium

We are now ready to consider objects in equilibrium. There are two conditions to equilibrium for most objects. The first condition is stated by Newton's first law:

 

 

The second condition of equilibrium is

 

 

Basically, we say that a body is in equilibrium if the vector sum of the forces and torques are zero.

Usually, we will need to look at both translational equilibrium and rotational equilibrium. The rule of thumb is that if the body is a point source, or all of the forces act at the same point on the body, then we do not need to consider rotational equilibrium. If the forces act at different points on the body, then we must take rotation into account.

Center of Mass

In order to simplify our calculations, it is convenient to talk in terms of the center of mass, or center of gravity, of a body. We define the center of mass to be the weighed average of the components of the body

 

 

(9.1)

 

If the object is a continuous mass distribution, we replace the summation with an integral. Also, notice that the center of mass is a vector.

The usefulness of the center of mass is that we will often need to calculate the torque of the center of mass about some point. It can be shown that in terms of the center of mass, the equilibrium conditions reduce to

 

 

(9.2)

 

where Fext is all of the forces external to the body, and R is the distance from the center of mass.

 

Example:

Locate the center of mass of the machine part in the diagram below

 

 

By symmetry, the center of mass lies along the axis and the center of mass of each part is midway between its ends. The volume of the disk is 8p cm3 and that of the rod is 12p cm3. Since the weights of the two parts are proportional to their volumes,

 

 

Taking the origin to be at the left face of the disc on the axis, we have

 

x1 = 1 cm x2 = 8 cm

 

and so

 

 

Examples of Equilibrium

Example:

A 5 m long rigid rod whose own weight is negligible is pivoted at a point 2 m from the left end. A mass of m1 = 25 kg is attached to the left end. What must the mass be of a block attached to the right end so that the rod is in equilibrium? What is the force of the pivot on the rod?

 

 

There are no forces in the x direction. The forces in the y direction are

 

m1g + m2g - P = 0

 

The torque about the pivot is (taking counterclockwise to be positive)

 

m1g(l0) m2g(l l0) = 0

 

Solving this for m2 yields

 

 

Thus, the force of the pivot must be

 

 

Example:

A 40 kg ladder is 10 m long. It leans in equilibrium against a frictionless vertical wall and makes an angle of 60 with the horizontal. Find the magnitude and direction of the force that the floor and wall exert against the ladder.

 

 

Since the wall is frictionless, the force from the wall is horizontal. The force from the floor consists of two parts: a normal component that is vertical, and a friction force that is horizontal. Setting up the force components

 

x component: f - Nw = 0

 

y component: Nf - mg = 0

 

Similarly, since the forces act at different locations on the ladder, there will also be a torque. Taking the torque to be positive in the counterclockwise direction around the point on the floor, the torques are

 

torques: Nwlsinq - mg(l/2)cosq = 0

 

where l is the length of the ladder. So we need to find Nw, Nf and f. From the y component equation, we see that

 

Nf = mg

 

while the x component yields

 

f = Nw

 

Similarly, from the torque equation, we have

 

F1 = mg/2tanq

 

Thus,

 

 

f = 113.2 N

 

 

So, the total force from the floor is