Translational & Rotational Equilibrium

For a whole rigid body, Newton's second law is ΣF⃗ = Ma⃗_cm: internal forces come in action–reaction pairs and cancel, so only external forces can move the centre of mass. About the centre of mass the rotational law is Στ⃗ = dL⃗/dt, and because each internal pair acts along the line joining its two points, internal torques cancel too. Equilibrium forbids both kinds of acceleration, so it needs both sums to vanish, and they are genuinely separate. Two 40 N forces pointing opposite ways along parallel lines 0.15 m apart add to zero force yet exert 6.0 N m of torque: the body spins up while its centre of mass stays put. A single 40 N push aimed through the centre of mass has zero torque about that point yet accelerates the body. The equations cannot tell resting from drifting, since a puck gliding over frictionless ice satisfies both sums; static equilibrium adds the requirement that the body starts at rest.

Move the reference point from the origin O to a point P at r⃗_P. Every position vector becomes r⃗ᵢ − r⃗_P, so τ⃗_P = Σ(r⃗ᵢ − r⃗_P) × F⃗ᵢ = τ⃗_O − r⃗_P × ΣF⃗. Each lever arm changes, but all the changes collect into one term proportional to the net force. When the forces balance, that term vanishes and the torque is the same about every point. Test it on a rod pushed 50 N up at x = 0 and 50 N down at x = 0.40 m. About x = 0: (0.40 m)(−50 N) = −20 N m. About x = 1.00 m: (−1.00 m)(50 N) + (−0.60 m)(−50 N) = −20 N m. Remove the downward push and the two answers split, 0 against −50 N m. So zero torque about one point proves nothing alone; add zero net force and it holds about all of them. In the figure, the torque about P plotted against P's position is a straight line of slope −ΣF_y, flat only when the forces balance.

For forces in the xy-plane, ΣFₓ = 0, ΣF_y = 0 and Στ = 0 about one point are three independent scalar equations. A torque equation about a second point is not a fourth: by the result above it is the first torque equation plus multiples of the two force equations, so it can check the algebra but never find a new unknown. Special geometry lowers the count. If every force is parallel to y, ΣFₓ = 0 holds automatically and two equations remain. If every line of action passes through one point, the torque about that point is automatically zero, again leaving two. A body loaded in three dimensions has six: three force components and three torque components. The planar three may be swapped but not extended. ΣFₓ with torques about A and B is an independent set provided AB is not perpendicular to x, and torques about A, B and C form one provided the three points are not collinear. Counting unknowns against that number is the first line of every statics solution.

A force whose line of action passes through the axis has no lever arm, so it drops out of that torque equation whatever its size and whatever its direction. A pin or axle exerts a force of unknown magnitude and direction; take torques about it and both unknown components vanish together. More generally, any point where two unknown lines of action cross removes both, and that point need not lie on the body. Take a wheelbarrow whose loaded tray weighs 540 N, with its line of action 0.45 m behind the axle, lifted at handles 1.35 m behind the axle. Torques about the axle give F(1.35 m) = (540 N)(0.45 m), so F = 180 N without the wheel's force ever appearing; ΣF_y = 0 then says the wheel carries 360 N. When an unknown's direction is not obvious, guess it, keep one sign convention throughout, and read a negative answer as the algebra reversing your arrow, not as a mistake.

A square table on four legs, loaded with 400 N at its centre, has four unknown leg forces but only three useful equations: vertical force balance and torque balance about two horizontal axes. Symmetry suggests 100 N per leg, but 150, 50, 150 and 50 N going round the corners satisfies every equation just as well, because each side's pair of legs still totals 200 N. Equilibrium fixes those pair totals and leaves a diagonal shift free, which is why a table on an uneven floor rocks between diagonals. Such a body is statically indeterminate, and no choice of axis rescues it, because the missing equation is not about balance. It is about deformation: the legs, top and floor compress by amounts set by their stiffness, and those compatibility conditions decide the split. That is exactly where the rigid-body model hands over to elasticity, later in this unit.

The two conditions describe a state, not its future. A pencil balanced exactly on its point has zero net force and zero net torque, yet the smallest tilt creates a torque that increases the tilt; whether a disturbance is restored or amplified is a separate question about how the torques change with position. The conditions also stop applying the moment the body accelerates. A car braking hard has ΣF⃗ = Ma⃗ pointing backwards, so the torque now depends on the axis. About the centre of mass the torques still balance, since the car does not pitch over, and that balance is what loads the front tyres more heavily. About a rear tyre's contact point they do not, and writing Στ = 0 there returns the parked car's front-to-rear split, missing the load transfer entirely. Accelerating bodies need ΣF⃗ = Ma⃗_cm with torques about the centre of mass, not the statics freedom to pick any axis.