Learning Lab

Translational Equilibrium

Translational equilibrium means that the sum of the forces (or net force) acting on the system is zero, ie, sum of the forces is zero \(\sum F\) \(=0,\)we divide all forces into horizontal and vertical components or:

Forces up = Forces down

Forces left = Forces right

Example 1

Consider a hanging set of traffic lights of mass \(m\) suspended above an intersection by a pair of wires of negligible mass, as shown below. It is easiest to break up the forces to vertical and horizontal forces.

Since this structure is in equilibrium the magnitude of the components of forces upwards must equal the magnitude of the components of forces downwards, or

Forces up = Forces down (there is no vertical motion)

Resolving \(T_{1}\) and \(T_{2}\) vertically in the diagram above:

\[\begin{align*} \textrm{forces Up} & =\textrm{forces Down}\\ T_{1}\sin30+T_{2}\sin60 & =W\\ T_{1}\sin30+T_{2}\sin60 & =mg \end{align*}\]

Alternatively, we could resolve \(T_{1}\) and \(T_{2}\) horizontally.

Forces left = Forces right (there is no horizontal motion)

Resolving \(T_{1}\) and \(T_{2}\) horizontally in the diagram above:

Open imageNote that vertical weight force W (=mg) when resolved horizontally equals zero or \(W\) \(\cos90=0\)

\[\begin{align*} \textrm{forces left} & =\textrm{forces right}\\ T_{1}\cos30^{\circ} & =T_{2}\cos60^{\circ}. \end{align*}\]

Rotational Equilibrium

Torque or Moment of a Force is the turning effect or torque \(\tau\) of a force F acting at a distance\(r\) from the axis of rotation of a rigid body is given by

\(\tau=F\times r\)

where \(F\) and \(r\) are mutually perpendicular (at right angles) to each other.

For rotational equilibrium to occur the sum of the torques (or net torque) acting on the structure is zero, ie \(\sum\tau=0\), (sum of \(\tau\) = 0) or

Clockwise Torques = Counterclockwise Torques

Example 2

Consider the seesaw example below.

For translational equilibrium: \(\Sigma F=0\) or \[\begin{align*} \textrm{Forces Up} & =\textrm{Forces Down}\\ F_{R} & =F_{1}+F_{2} \end{align*}\]

For rotational equilibrium: \(\Sigma\tau=0\) (when a seesaw is not moving) or

\[\begin{align*} \textrm{Clockwise Torque} & =\textrm{CounterClockwise Torque}\\ F_{2}\times x_{2} & =F_{1}\times x_{1} \end{align*}\] An example of a system that is not in rotational equilibrium is a moving see-saw. If you are sitting on one end and an elephant (!) decides to sit on the other end, then it will rotate!!