Torque - Equilibrium
A body is said to be in equilibrium if it continues its state of rest or its state of uniform motion. Equilibrium can be categorized in two ways:
1) Static equilibrium: If a body is at rest and remains at rest, then the equilibrium is said to be static equilibrium.
2) Dynamic equilibrium: If a body is initially moving with some velocity and it continues its motion rectilinearly with the same velocity, or if the body is rotating with some initial angular velocity and the angular velocity remains constant, then the body is said to be in dynamic equilibrium.
If the velocity and angular velocity of the body, either zero or non-zero, remain constant with time, then acceleration and angular acceleration are zero.
For rectilinear motion, if the acceleration is zero, then by Newton’s second law the net force is also zero.
Similarly, for rotatory motion, if angular acceleration is zero, then by Newton’s second law for rotation net torque is also zero.
Equilibrium can further be classified as follows:
1) Translational equilibrium: If the net force acting on a body is zero, then the body is said to be in translational equilibrium. In such a case, the center of mass of the body remains either at rest or moves rectilinearly with constant velocity.
2) Rotational equilibrium: If the net torque acting on a body is zero, then the body is said to be in rotational equilibrium. In such a case, the angular velocity of the body remains constant.
NOTE:
An object may be rotating, even rotating at a changing rate, but may be in translational equilibrium if the acceleration of the center of mass of the object is still zero.
In the diagram shown above, the net force acting on the rod is zero and the acceleration of the center of mass also equals zero. About center of mass both forces produce the torque in the same sense that is anticlockwise. Thus, the net torque is non-zero and the body is only in translation equilibrium.
Similarly, an object may be translating with variable speed and still be in rotatory equilibrium.
In the diagram shown, a force is applied at the center of the rod. Because of this force, the rod translates and the center of mass of rod accelerates, but the rod does not rotate and remains in rotational equilibrium. In this case, the torque about the center of mass is zero.
A rod of mass M and length L is hinged at one end and the other end is held by applying a force F. Find the force needed to keep the rod in equilibrium.
Three forces act on the rod:
1) the gravitational force at the center of gravity
2) the hinge force
3) the force F at the end of the rod.
To keep the rod in equilibrium, the net torque should be zero. As a result, writing torque about the hinge, torque of gravitational force and force F are in opposite directions. The torque of hinge force about hinge is zero as it passes through the hinge itself. Therefore, \[\begin{align} Mg\frac{L}{2} - FL &= 0\\ F &= \frac{{Mg}}{2}. \end{align}\]
The above shows a \(3\) m rod fixed by the rotation axis at the point where the internal division ratio is \(1:2.\) If three forces indicated by the three red arrows--\(20\) N, \(40\) N, F--are acting and the rod does not rotate, what is the magnitude of \(F\)?