Statics of rigid bodies flashcards

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Moment of a force about a point
The magnitude of the force × the perpendicular distance from the point to the line of action of the force. Unit N m; give the sense (clockwise or anticlockwise).
Line of action of a force
The straight line through the point of application in the direction of the force. A force can slide along it without changing its moment.
Rigid body
A body whose size matters but which does not bend or stretch: the distances between its points stay fixed (e.g. a rod, beam or ladder).
Equilibrium of a rigid body
Resultant force zero AND total moment about any point zero.
Limiting equilibrium
Equilibrium in which friction has reached its maximum value, F = μR, so the body is on the point of slipping.
Smooth hinge
A frictionless joint that can exert a force in any direction on the rod (two unknown components) but no turning effect.
Smooth peg
A fixed support that touches a rod without friction; its force on the rod is perpendicular to the rod.
Uniform rod
A rod whose mass is spread evenly, so its weight acts at its midpoint.
Moment of force F acting at distance l from a point, at angle θ to the line joining them
Moment = F l sin θ (= F × perpendicular distance). (No — learn it)
Moment (anticlockwise) of Xi + Yj acting at xi + yj relative to the point
xY − yX. (No — derive it from the components)
Conditions for equilibrium of a rigid body in a plane
ΣF_x = 0, ΣF_y = 0, ΣM = 0 about any one point (three equations). (No — learn it)
Friction at a rough contact
F ≤ μR; F = μR when limiting (about to slip). (No — learn it)
Magnitude and direction of a hinge force with components X and Y
R = √(X² + Y²), at tan⁻¹(Y/X) to the horizontal (say above or below). (No — derive it)
Least μ for a uniform ladder alone at angle α, smooth wall, rough ground
μ ≥ 1/(2 tan α). (No — derive it)
Forces on a ladder: smooth wall, rough ground
Weight down at the centre; S horizontal from the wall; R up at the foot; friction at the foot towards the wall.
Forces on a ladder: rough wall, rough ground, about to slip
As for a smooth wall, plus wall friction μ₂S acting UP the wall; friction at the foot μ₁R towards the wall.
Forces on a rod hinged to a wall and held by a string
Weight(s) down; tension along the string towards its fixed point; hinge force as components X (horizontal) and Y (vertical).
Forces on a rod resting on a smooth peg with its foot on rough ground
Weight down at the centre; peg reaction perpendicular to the rod; R up and friction horizontal at the foot, towards the peg.
A beam about to tilt about support D
The reaction at the other support is zero; all the upward force acts at D.
Why take moments about a hinge or the foot of a ladder?
Unknown forces acting there have zero moment about that point, so the moments equation contains fewer unknowns.
Why does friction at the foot of a ladder act towards the wall?
If the ladder slipped, the foot would slide away from the wall; friction opposes that tendency.
Why does the ground reaction on a ladder fall below the total weight when the wall is rough?
Upward wall friction supports part of the weight, so R = weight − wall friction.
Is the force at a hinge along the rod?
Not in general. Only for a light rod with forces applied at its two ends. Find X and Y and combine them.
Three non-parallel forces keep a body in equilibrium. What must be true of them?
Their lines of action meet at a single point (they are concurrent).
Why does a person climbing a ladder make it more likely to slip?
The person's moment about the foot grows, so S, and so the friction needed, grows; R stays the same.
How many unknowns can equilibrium of one rigid body in a plane determine?
Three: two resolving equations and one moments equation (or equivalent).
A 30 N force acts at the end of a 0.5 m spanner at 60° to it. Moment about the nut?
30 × 0.5 sin 60° = 13.0 N m.
Force (2i + 5j) N acts at the point (i + 2j) m. Moment about the origin?
xY − yX = 1 × 5 − 2 × 2 = 1 N m anticlockwise.
Uniform ladder, smooth wall, rough ground, at 60° to the ground. Least μ?
μ ≥ 1/(2 tan 60°) = 0.289 (exactly √3/6).
Uniform rod, weight 50 N, hinged at A, horizontal, string at B at 30° to the rod. Tension?
M(A): T sin 30° × l = 50 × l/2, so T = 50 N.
Uniform 6 m plank, 20 kg, on supports 1 m and 4 m from A. How far from A can a 50 kg person stand before it tilts?
Tilt about the support at 4 m: 20g × 1 = 50g(x − 4), so x = 4.4 m.
Uniform 2 m rod, 10 kg, foot on the ground, resting on a smooth peg 1.5 m from the foot, at 45°. Peg force?
M(foot): N × 1.5 = 10g × 1 × cos 45°, so N = 46.2 N, perpendicular to the rod.
Hinge components X = 30 N (horizontal), Y = 40 N (up). Hinge force?
√(30² + 40²) = 50 N at 53.1° above the horizontal.
Using force × distance along the rod for a force at an angle
Wrong: use the perpendicular distance, or the perpendicular component of the force.
Writing F = μR at a contact that is not slipping
Use F ≤ μR unless the question says limiting / about to slip at that contact.
Giving a hinge force direction as just an angle
State the reference: e.g. “18° above the horizontal” or “at 72° to the wall”.
When to substitute g = 9.8 in statics
Keep g as a symbol; it often cancels (μ, angles, distances). Substitute only for a force, then give 2–3 s.f.
A ladder with rough wall and rough ground about to slip
Both frictions are limiting at once: F₁ = μ₁R and F₂ = μ₂S.

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