Moments and centre of gravity flashcards

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Moment of a force
Force × perpendicular distance from the point (pivot) to the line of action of the force. Unit N m.
Line of action of a force
The straight line through the point where the force acts, along the direction of the force, extended both ways.
Perpendicular distance (for a moment)
The shortest distance from the pivot to the line of action — measured at 90° to the line of action.
Principle of moments
For a body in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point.
Centre of gravity
The single point at which the whole weight of a body can be considered to act.
Uniform (beam, rod, plank)
Mass evenly spread along it, so its centre of gravity is at its midpoint.
Light (beam, rod, string)
Its weight is negligible and is left out of the calculation.
Equilibrium of an extended body
Zero resultant force and zero resultant moment about any point.
Plumb line
A thread with a small weight on the end; it hangs vertically and marks the vertical line through its support.
Equation for the moment of a force
moment = Fx, with x the perpendicular distance from the pivot to the line of action. (Yes)
Moment of force F applied at distance r from the pivot, at angle θ to the lever
moment = F r sin θ (perpendicular distance r sin θ, or perpendicular component F sin θ). (No — derive it)
Conditions for equilibrium, written as equations
ΣF = 0 (up = down, left = right) and Σ clockwise moments = Σ anticlockwise moments. (No — learn it)
Weight acting at the centre of gravity
W = mg, with g = 9.81 N kg⁻¹. (Yes)
Free-body diagram of a uniform beam on two supports
Beam as a line; weight as one arrow down from the midpoint; an upward arrow at each support; distances marked.
Where is the weight drawn on a free-body diagram of a non-uniform plank?
One arrow straight down from its centre of gravity, which is nearer the heavier end.
Diagram: perpendicular distance for a force at an angle
Extend the line of action, then draw the perpendicular from the pivot to it; that length is x.
Diagram: why a tilted block topples
Draw the vertical through G. If it falls outside the base edge the block is pivoting about, it topples.
Suspension method for a lamina: what does the diagram show?
Card hanging from a pin with a plumb line; lines from two holes cross at the centre of gravity.
Why does a force through the pivot have no moment?
Its perpendicular distance from the pivot is zero.
Why take moments about the point where an unknown force acts?
That force has zero moment there, so it drops out of the equation.
Why does a lamina's centre of gravity lie on the plumb line from the pin?
At rest the resultant moment about the pin is zero, so the weight's line of action passes through the pin.
Finding the centre of gravity of a rod by balancing
Balance it on a knife-edge; G is directly above the knife-edge. Repeat and take a mean.
What makes an object more stable?
A low centre of gravity and a wide base: it must tilt further before the weight's line passes outside the base.
Why does a tower crane have a counterweight?
To balance the moment of the load and jib about the tower, so the tower is not bent over.
Beam on two supports: which support carries more?
The one nearer to the heavy load.
30 N acts at 90° to a lever, 0.40 m from the pivot. Moment?
30 × 0.40 = 12 N m. (Yes)
50 N acts at 40° to a 0.30 m spanner, at its end. Moment about the nut?
50 × 0.30 sin 40° = 9.6 N m.
500 N sits 1.2 m left of a see-saw pivot. Where must 400 N sit to balance?
400x = 500 × 1.2 → x = 1.5 m right of the pivot.
Uniform 3.0 m beam, weight 200 N, supports at both ends; 600 N load 2.0 m from end A. Support forces?
About A: R_B × 3.0 = 200 × 1.5 + 600 × 2.0 → R_B = 500 N; R_A = 800 − 500 = 300 N.
Uniform 2.0 m beam, weight 100 N, hinged to a wall, held horizontal by a cable from its far end at 30° to the beam. Tension?
T sin 30° × 2.0 = 100 × 1.0 → T = 100 N.
A rod's centre of gravity is 0.20 m to one side of a knife-edge. A 100 g mass hung 0.30 m on the other side balances it. Mass of the rod?
M × 0.20 = 0.100 × 0.30 → M = 0.15 kg (g cancels).
Defining moment as “force × distance”
Incomplete. Say force × perpendicular distance from the pivot to the line of action.
“On the point of tipping”
The force at the other support is zero. State it, then take moments about the edge it tips over.
Forgetting the beam's own weight
Unless the beam is “light”, include its weight at its centre of gravity (midpoint if uniform).
Distances measured from the end of the beam
Every distance in a moment is measured from the chosen pivot, not from the end of the beam.
Using the whole tension of a cable at an angle
Only T sin θ (θ between cable and beam) turns the beam about the hinge.
Moments balance, so the job is done?
Also check that forces balance: the pivot or supports carry the total weight.
Unit of moment
N m — never N m⁻¹, and never written as J.

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