Moments 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; state clockwise or anticlockwise.
Rigid body
A body whose size matters but which does not bend or change shape, so the distances between points on it stay fixed.
Equilibrium of a rigid body
The resultant force is zero AND the total moment about any point is zero.
Uniform rod
A rod whose mass is evenly spread, so its weight acts at its midpoint.
Non-uniform rod
A rod whose mass is not evenly spread: its centre of mass G is at an unknown point, usually written AG = x.
Light rod
A rod whose mass is negligible, so it has no weight in the equations.
On the point of tilting about support D
The body is still in equilibrium but about to lose contact with the other support C, so the reaction at C is zero.
Centre of mass
The point at which the whole weight of a body can be taken to act.
Moment of a force F at perpendicular distance d
Moment = F × d (N m). (No — learn it)
Equilibrium under parallel vertical forces (two equations)
Total upward force = total downward force; Σ clockwise moments = Σ anticlockwise moments about any point. (No — learn it)
Reaction at D for a uniform rod of weight W on supports C and D (no loads), G a distance a from C
Moments about C: R_D × CD = W × a, so R_D = Wa / CD. (No — derive it)
Balance point of two particles m₁ and m₂ on a light rod of length L (distance from m₁)
m₁x = m₂(L − x), so x = m₂L / (m₁ + m₂). (No — derive it)
Reactions at two supports as a person walks along a plank: shape of the graphs
Two straight lines: one reaction falls as the other rises; their sum is constant (total weight). Equilibrium only while both are ≥ 0.
What to draw on a diagram for a moments question
Every force (weights at centres of mass, reactions or tensions at their points) and every distance, measured from one end.
Where does the weight go on the diagram of a non-uniform rod?
At G, an unknown point marked with a letter (AG = x), not at the midpoint.
Direction of the force at a support or a vertical string
A support pushes up on the body; a string pulls up on it. A support or string can never give a negative force.
Why is zero resultant force not enough for a rigid body to be in equilibrium?
Two equal, opposite forces with different lines of action give zero resultant force but still turn the body.
Why take moments about the point where an unknown force acts?
That force has zero moment about the point, so it does not appear in the equation.
Why can moments be taken about any point for a body in equilibrium?
With zero resultant force, the total moment is the same about every point; in equilibrium it is zero about all of them.
What does a negative reaction in your working tell you?
That position or load is impossible: the body would have tilted before reaching it.
How do you show that a plank does NOT tilt?
Find the reaction at the support that would lose contact and show that it is positive.
A rod balances on a single knife-edge. Where is its centre of mass?
Directly above the knife-edge: otherwise the weight would have a moment about it.
Light see-saw pivoted at its centre: 30 kg child 1.6 m from the pivot. Where does a 40 kg child sit to balance?
30g × 1.6 = 40g × d → d = 1.2 m on the other side.
Uniform rod AB, 2.0 m, weight 60 N, on supports at A and B; 40 N load 0.5 m from A. Reactions?
M(A): R_B × 2.0 = 60 × 1.0 + 40 × 0.5 → R_B = 40 N; R_A = 100 − 40 = 60 N.
Uniform 24 kg plank; its centre is 1.0 m from support D. How far beyond D can a 60 kg person stand?
About to tilt about D (other reaction 0): 24g × 1.0 = 60g × y → y = 0.4 m.
Non-uniform 15 kg rod hangs from strings at A and C (AC = 2.5 m); tension at C = 2 × tension at A. Find AG.
3T = 15g → T = 5g; M(A): 10g × 2.5 = 15g × AG → AG = 5/3 ≈ 1.67 m.
Light rod: 20 N up at 0.9 m from A and 8 N down at 2.4 m from A. Resultant moment about A?
20 × 0.9 = 18 N m anticlockwise; 8 × 2.4 = 19.2 N m clockwise → 1.2 N m clockwise.
Rod balances at C (AC = 0.9 m); with 0.4 kg at A it balances at D (AD = 0.7 m). Mass of rod?
G is at C. M(D): 0.4g × 0.7 = mg × 0.2 → m = 1.4 kg.
A 6 kg rod and a 6 kg load hang from two vertical strings; one string is slack. Tension in the other?
Resolve: T = 12g = 118 N.
Measuring distances from the end A when taking moments about C
Wrong: every distance in a moments equation is measured from the point you take moments about.
Forgetting the rod's own weight
Unless the rod is light, its weight (at its centre of mass) appears in every equation.
On the point of tilting: which reaction is zero?
The one at the support the body is lifting off — NOT the one it pivots about.
When to substitute g = 9.8
Keep g as a letter until the final line; it often cancels. Give final answers to 2 or 3 s.f.
Labelling your equations
Head each one “moments about C” or “resolving vertically”, so the method is clear.
Finding a range of positions
Write each force in terms of x, then apply every limit: each reaction ≥ 0, each tension ≤ its maximum, x on the rod.

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