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.
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 F_{ext} 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 cm^{3} and that of the rod is 12p cm^{3}. 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
x_{1} = 1 cm
x_{2} = 8 cm
and so
_{}
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 m_{1}
= 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
m_{1}g
+ m_{2}g  P = 0
The torque about the pivot
is (taking counterclockwise to be positive)
m_{1}g(l_{0})
– m_{2}g(l – l_{0}) = 0
Solving this for m_{2}
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  N_{w}
= 0
y
component: N_{f}
 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: N_{w}lsinq  mg(l/2)cosq = 0
where l is the
length of the ladder. So we need to
find N_{w}, N_{f} and f. From the y component equation, we see
that
N_{f} = mg
while the x
component yields
f
= N_{w}
Similarly, from the
torque equation, we have
F_{1} = mg/2tanq
Thus,
_{}
f
= 113.2 N
_{}
So, the total force from
the floor is
_{}