M2.5 - Statics of rigid bodies

The moment of a force about a point measures the force's turning effect about that point. Its magnitude is the force multiplied by the perpendicular distance from the point to the force's line of action—not necessarily the distance to the point where the force is applied.

Choose and state a sign convention, such as anticlockwise positive. A force whose line of action passes through the chosen point has $d_{\perp}=0$ and hence zero moment about that point. Moment has unit $\mathrm{N\,m}$; it is not a force and should not be labelled in newtons.

A horizontal rod is pivoted at $A$. A $30$ N downward force acts $2$ m from $A$, and an upward force $P$ acts $5$ m from $A$. Taking anticlockwise as positive, equilibrium of moments about $A$ gives$$5P-30(2)=0,$$so $P=12$ N. Any reaction at $A$ contributes no moment about $A$, which is why this pivot makes the equation efficient.

Before calculating, extend each force mentally into its line of action and find the shortest distance from the pivot. Check that every term has dimensions force $\times$ distance and that forces on opposite sides or with opposite turning effects receive opposite signs.

A rigid body in coplanar equilibrium has no translational acceleration and no angular acceleration. Draw an isolated free-body diagram, replacing every contact by the forces it can exert, then use two independent force balances and one moment balance.

Friction opposes the impending or actual relative motion at a contact. In ordinary equilibrium, $|F|\le\mu R$; write $F=\mu R$ only when the body is stated to be in limiting equilibrium or on the point of slipping. A smooth contact has no friction, not no reaction.

A reliable order is: (1) mark every weight, reaction, friction and tension; (2) choose axes that simplify components; (3) take moments about a point through which the most unknown forces act; (4) resolve horizontally and vertically; (5) solve and check directions, non-negative reactions and any friction inequality. Other pivots give equivalent equations but may keep more unknowns.

A uniform $5$ m ladder of weight $W$ rests at angle $\theta$ on rough horizontal ground and against a smooth vertical wall, where $\sin\theta=4/5$ and $\cos\theta=3/5$. Let the wall reaction be $H$, and let the ground supply vertical reaction $R$ and horizontal friction $F$. Taking moments about the foot,$$H(5\sin\theta)=W\left(\frac52\cos\theta\right),$$so $4H=\tfrac32W$ and $H=3W/8$. Force balance gives $F=H=3W/8$ and $R=W$. Therefore equilibrium requires $\mu\ge F/R=3/8$; if the ladder is on the point of slipping, $\mu=3/8$.

When a supported body is on the point of tilting about one contact, the reaction at the other contact has fallen to zero. Take moments about the remaining contact and keep only forces still acting. This is a different limiting condition from impending sliding, although a problem may require both ideas.