Clockwise moments are taken as negative (‑) and anticlockwise moments as positive (+).
The same convention is used for torques about any axis.
The centre of gravity (CG) of a rigid body is the point at which the total weight \(W=mg\) can be assumed to act.
For a uniform (homogeneous) body the CG coincides with the geometric centre; for irregular bodies it is found by balancing moments (the principle of moments).
For a force \(\mathbf F\) acting at point A, the torque about a reference point O (or about any axis through O) is defined by the cross‑product
\[
\boldsymbol{\tau}= \mathbf r_{OA}\times\mathbf F,
\]
where \(\mathbf r_{OA}\) is the position vector from O to the point of application A.
The magnitude is
\[
\tau = r\,F\sin\theta,
\]
with \(\theta\) the angle between \(\mathbf r\) and \(\mathbf F\).
If the axis of rotation is specified (e.g. the z‑axis), only the component of \(\boldsymbol{\tau}\) along that axis is relevant.
When the axis is not perpendicular to the plane of \(\mathbf r\) and \(\mathbf F\), the torque about the axis \(\mathbf{\hat n}\) is
\[
\tau_{(\mathbf{\hat n})}= (\mathbf r\times\mathbf F)\cdot\mathbf{\hat n}.
\]
Thus the same formula works for a point (any axis through the point) or for a fixed axis not passing through the point.
If several forces \(\mathbf F1,\mathbf F2,\dots,\mathbf F_n\) act on a body, the resultant (net) force is the vector sum
\[
\mathbf R = \sum{i=1}^{n}\mathbf Fi.
\]
Only when \(\mathbf R= \mathbf 0\) does the body experience no translational acceleration. Translational equilibrium (Σ F = 0) must be satisfied before considering rotational equilibrium.
A couple consists of two forces \(\mathbf F1\) and \(\mathbf F2\) such that
Because the forces are equal and opposite, the resultant force is zero:
\[
\mathbf R = \mathbf F1+\mathbf F2 = \mathbf 0,
\]
so the body does not translate. However the pair produces a non‑zero torque (a pure turning effect)
\[
\boxed{\;\tau_{\text{couple}} = F\,d\;}
\]
directed perpendicular to the plane containing the forces (right‑hand rule).
Let the forces act at points A and B with position vectors \(\mathbf rA\) and \(\mathbf rB\) measured from an arbitrary origin O. The torques are
\[
\boldsymbol{\tau}A = \mathbf rA \times \mathbf F,\qquad
\boldsymbol{\tau}B = \mathbf rB \times (-\mathbf F).
\]
The resultant torque of the couple about O is
\[
\boldsymbol{\tau}{\text{couple}} = \boldsymbol{\tau}A+\boldsymbol{\tau}_B
= (\mathbf rA-\mathbf rB)\times\mathbf F.
\]
The vector \(\mathbf rA-\mathbf rB\) is exactly the displacement from B to A, whose magnitude is the perpendicular separation \(d\) and whose direction is normal to the plane of the forces. Hence
\[
\boldsymbol{\tau}_{\text{couple}} = d\,F\,\mathbf{\hat n},
\]
which contains no reference‑point vector; therefore the torque of a couple is the same about any origin O (or any axis parallel to \(\mathbf{\hat n}\)).
For forces \(\mathbf Fi\) acting at points with position vectors \(\mathbf ri\) (relative to a chosen origin O), the total torque is
\[
\boldsymbol{\tau}{\text{result}} = \sum{i}\mathbf ri \times \mathbf Fi.
\]
If all torques are parallel to the same axis (e.g. out of the page), they can be added algebraically using the sign convention.
A rigid body is in static equilibrium when
\[
\sum \mathbf F = \mathbf 0 \qquad\text{and}\qquad \sum \boldsymbol{\tau}= \mathbf 0.
\]
Both conditions must be satisfied simultaneously; neglecting either can lead to an incorrect solution.
A uniform beam 3.0 m long and weight 600 N rests on a hinge at its left end (point A). A tension force of 400 N acts upward at a point 2.0 m from A. Determine the reaction at the hinge.
\[
\begin{cases}
\Sigma Fx: & Rx = 0,\\[4pt]
\Sigma Fy: & Ry + T - W = 0 \;\Longrightarrow\; R_y = W - T = 600-400 = 200\;\text{N (upward)}.
\end{cases}
\]
\[
\tau_A = T(2.0) - W(1.5) = 0 \;\Longrightarrow\; 400(2.0) = 600(1.5).
\]
Both sides equal 800 N·m, confirming the torque balance. Hence the beam is in equilibrium.
| Aspect | Single Force | Force Couple |
|---|---|---|
| Resultant force | Non‑zero → translation (and possibly rotation) | Zero → no translation |
| Resultant torque | Depends on the chosen point or axis | Same magnitude and direction about any point/parallel axis |
| Effect on a rigid body | Translation + possible rotation | Pure rotation |
| Units | Force: N; Torque: N·m | Torque only: N·m |
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