Moment of a force
Forces in Action - OCR A-Level Physics
Key Definition
Moment of a force
The turning effect of a force about a pivot. Moment = force multiplied by the perpendicular distance from the pivot to the line of action of the force.
The turning effect of a force about a pivot. Moment = force multiplied by the perpendicular distance from the pivot to the line of action of the force.
$$M = Fd$$
$$\text{moment} = F \times d$$
Key Definition
Principle of moments
For a body in rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point.
For a body in rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point.
- Unit: $\text{N m}$. Moments are vectors; the direction is either clockwise or anticlockwise.
- If the force acts at an angle $\theta$ to the lever, use $\text{moment} = Fx\sin\theta$, where $x$ is the distance from the pivot to where the force is applied.
- A force whose line of action passes through the pivot has zero perpendicular distance and so produces no moment about that pivot.
- Real example: a $30 \text{ N}$ push on a $0.25 \text{ m}$ spanner produces a moment of $30 \times 0.25 = 7.5 \text{ N m}$ about the bolt.
Diagram pending
Horizontal lever pivoted at its midpoint. Downward force on the left at distance $d_1$ producing an anticlockwise moment; downward force on the right at distance $d_2$ producing a clockwise moment. Labelled pivot, perpendicular distances and force arrows.
Will be replaced with a GeoGebra SVG in stream 2.
Common Mistake
MEDIUM
Students often: Don't use the distance along the beam or rod instead of the perpendicular distance to the line of action of the force.
Instead: The distance in the moment equation MUST be the perpendicular distance from the pivot to the line of action. If the force is at an angle theta to the beam, use $moment = Fd \sin(\theta)$.
Instead: The distance in the moment equation MUST be the perpendicular distance from the pivot to the line of action. If the force is at an angle theta to the beam, use $moment = Fd \sin(\theta)$.
Worked Example [4 marks]
A uniform beam of length 4.0 m and weight 200 N is supported at its centre. A 50 N weight is placed 1.5 m from the left end. Where must a 30 N weight be placed to balance the beam?
Show Solution
1
The beam is supported at its centre (2.0 m from each end). The beam's weight acts at the pivot, so it produces no moment
[1]2
The $50 \text{ N}$ weight is $2.0 - 1.5 = 0.5 \text{ m}$ to the left of the pivot. Anticlockwise moment $= 50 \times 0.5 = 25 \text{ N m}$.
[1]3
For equilibriumA state in which the resultant force AND the resultant moment on an object are both zero.
Clockwise moment $=$ anticlockwise moment, so $30 \times d = 25$.
[1]4
$d = \frac{25}{30} = 0.83 \text{ m}$ to the right of the pivot $= 2.0 + 0.83 = 2.83 \text{ m}$ from the left end.
[1]Answer
The $30 \text{ N}$ weight must be placed $2.83 \text{ m}$ from the left end ($0.83 \text{ m}$ to the right of the pivot).
Examiner Tips and Tricks
- Take moments about the point where an unknown force acts; that force has zero moment, so it drops out of the equation.
- State your pivot explicitly: "Taking moments about the hinge, ..." Markers expect to see this.