M1.6 - Moments

Learning objectives

M1.6.1Moment of a forceMoment of a force.; Simple problems involving coplanar parallel forces acting on a body and conditions for equilibrium in such situations.

The moment of a force about a point measures its turning effect. It equals the force magnitude multiplied by the perpendicular distance from the point to the force's line of action. A force whose line of action passes through the point has zero moment about that point.

$M=F d_\perp\qquad\text{with units N m}$

Choose one rotational sense as positive and use it consistently; for example, anticlockwise moments positive and clockwise moments negative. For parallel vertical forces on a horizontal body, $d_\perp$ is the horizontal separation. The distance is measured to the line of action, not along the body unless that length is perpendicular to the force.

$\sum F=0,\qquad \sum M_P=0\quad\text{about any point }P$

Model statement	Consequence in the force diagram
uniform rod or beam	its weight acts at the geometric midpoint
non-uniform rod or beam	its weight acts at the stated or unknown centre of mass
particle attached to a body	its weight acts at the attachment point
support contact is maintained	its normal reaction is non-negative
body is on the point of tilting about a support	the reaction at the other support is zero; take moments about the pivot

First draw every parallel force at its correct position. Resolve vertically to relate the reactions, weights and tensions. Then take moments about a point that removes an unwanted force—usually a support whose reaction is unknown. Keep each term dimensionally as force times distance, solve, and check that all required contact reactions are non-negative.

A uniform $4$ m beam of mass $10$ kg is supported at its ends $A$ and $B$ . A $6$ kg particle is attached $1$ m from $A$ . If the upward reactions are $R_A$ and $R_B$ , vertical equilibrium gives $R_A+R_B=16g.$ Taking moments about $A$ gives $4R_B=(10g)(2)+(6g)(1),$ so $R_B=6.5g$ N and $R_A=9.5g$ N. Both are positive, so both supports can remain in contact.

Force equilibrium alone does not prevent rotation, and moment equilibrium alone does not prevent translation: a body needs both conditions. Do not use the full rod length when the perpendicular distance is different, attach a non-uniform body's weight to its midpoint, or keep a positive reaction at a support that has just lost contact at limiting tilt.

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