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.
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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
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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 |
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
The torque about the pivot is (taking counterclockwise to be positive)
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/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
while the x component yields
Similarly, from the torque equation, we have
Thus,
So, the total force from the floor is
