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Forces and Moments overview

Topic 4 of 4

Combine force and moment balance

Equilibrium requires both zero resultant force and zero resultant torque. A body can satisfy one condition while failing the other.

ΣFx = 0 and ΣFy = 0

Σmoments = 0

For forces in a plane, resolve along two perpendicular axes and add signed components. For rotation, choose a point and one positive turning sense. The principle of moments states that in rotational equilibrium, total clockwise moment equals total anticlockwise moment about the chosen point.

A couple shows why zero resultant force is insufficient. Likewise, balanced moments alone do not establish zero resultant force. Equilibrium does not require forces to be absent, and force balance alone does not show that the body is at rest. The worked arrangements below are explicitly stationary.

A beam with supports away from its ends

A uniform rigid beam is 1.20 m long and weighs 10 N. Its two resting supports are at x = 0.20 m and x = 1.00 m, measured from the left end. A 20 N downward load acts at x = 1.10 m. Both support forces act vertically upward.

Balance the beam's forces and moments

A uniform beam on two non-central supportsThe beam is 1.20 metres long. Positions use a common scale of 240 drawing units per metre from its left end. Upward supports are at 0.20 and 1.00 metres. Its own ten-newton weight acts at its midpoint, 0.60 metres; a twenty-newton load is at 1.10 metres. Arrows use six drawing units per newton: the left reaction is 2.5 newtons and the right reaction 27.5 newtons. The beam is stationary. Position labels are separated using plain leader lines where close together; they are coordinates, not moment arms. From the left support the weight, right support and load have arms 0.40, 0.80 and 0.90 metres.Left: 2.5 NRight: 27.5 N10 N20 NBeam's weightAdded load0.000.200.601.001.101.20Position from the left end / m

All force arrows share one scale, so the small left reaction is deliberately short. Taking moments about the left support, use arms 0.40 m for the beam's weight, 0.80 m for the right reaction and 0.90 m for the added load. These differ from the labelled positions.

The beam's own weight acts at x = 0.60 m. Positions measured from the left end are distinct from moment arms measured from a chosen support. Both reactions and both downward forces belong in the beam model.

Worked support forces

Choose a point that removes one unknown moment

  1. Place the beam's weight: its centre of gravity is at the midpoint, x = 0.60 m.
  2. Take moments about the left support: the left reaction has zero moment there. The right reaction's arm is 1.00 - 0.20 = 0.80 m; the weight's arm is 0.40 m; the load's arm is 0.90 m.
  3. Balance the moments: Rright(0.80) = 10(0.40) + 20(0.90). Thus Rright = 27.5 N.
  4. Balance vertical forces: Rleft + 27.5 = 10 + 20. Thus Rleft = 2.5 N.

Check about the right support: 10(0.40) = 2.5(0.80) + 20(0.10). Both sides are 4.0 N m.

The right reaction can exceed the added 20 N load because it also helps support the beam's own weight and balance the turning effect. The reactions need not be equal just because there are two supports.

A negative answer for an assumed upward reaction would mean that a downward force is needed under the proposed conditions. A simple resting support cannot pull the beam down; reconsider whether contact can be maintained, rather than accepting the assumed arrangement unchanged.

Check the model against a measurement

Mark the beam's centre of gravity and measure load and support positions from one fixed reference. Keep applied loads vertical, record the support readings with their units, and compare their sum with the total weight. Use the same points to calculate perpendicular moment arms. Finite support widths or a shifted load position affect the assumed lines of action; address that alignment rather than only repeating the force readings.

Optional check A uniform 1.20 m beam weighs 10 N and rests on supports at x = 0.20 m and 1.00 m. Its 20 N load is moved to x = 0.90 m. What are the left and right support forces in equilibrium?
A uniform 1.20 m beam weighs 10 N and rests on supports at x = 0.20 m and 1.00 m. Its 20 N load is moved to x = 0.90 m. What are the left and right support forces in equilibrium?

A stationary knot and a closed force triangle

A light knot supports a 50 N load through a third string. The knot's own weight is neglected. The third string pulls the knot down with 50 N; do not also add the load's gravitational force to the knot's diagram.

The left string points four parts left for three parts up. The right string points three parts right for four parts up. Each direction uses a 3-4-5 triangle: horizontal and vertical components are the corresponding fractions of the tension.

The spatial arrangement and the force triangle answer different questions

1. The strings set the force directions

A light stationary knot supporting a fifty-newton loadFrom the knot, the left string points four left for three up and the right string three right for four up. A third string hangs vertically down to a fifty-newton load. The solved tensions are thirty newtons on the left and forty newtons on the right. The physical string lengths are schematic and are not the magnitudes of the forces. The knot is light and stationary.50 NTL = 30 NTR = 40 N4 left3 up3 right4 upKnotString lengths do not scale the forces.

2. Place the force arrows head to tail

The thirty, forty and fifty-newton force arrows form a closed triangleEqual horizontal and vertical scales use five drawing units per newton. Starting at the lower right, the thirty-newton left tension points twenty-four newtons left and eighteen up. From its tip the forty-newton right tension points twenty-four right and thirty-two up. The fifty-newton downward force then returns to the starting point. Arrow lengths are in the ratio three to four to five. The head-to-tail triangle closes, showing zero resultant force on the knot. Its positions are force coordinates, not physical string positions.30 N40 N50 NStart / finishOne force scale in both directions.

The three vectors are (-24, +18) N, (+24, +32) N and (0, -50) N. Closing the triangle establishes force balance. For an extended body, also check the moments and lines of action.

The spatial diagram gives string directions; the separate force triangle gives force magnitudes and directions. Its arrows are (-24, +18) N, (+24, +32) N and (0, -50) N, placed head to tail with equal scales on both axes.

Worked tensions

Balance components before comparing magnitudes

Take right and up as positive. Write the tensions as Tleft and Tright.

Horizontal balance:
0.8Tleft = 0.6Tright.

Vertical balance:
0.6Tleft + 0.8Tright = 50.

The first equation gives Tleft = 0.75Tright. Substituting gives 1.25Tright = 50, so Tright = 40 N and Tleft = 30 N.

The components are (-24, +18) N and (+24, +32) N. Horizontally, -24 + 24 = 0. Vertically, 18 + 32 - 50 = 0.

Placed head to tail, the three force vectors form a closed triangle: the final tip returns to the initial tail. The tensions are unequal, but their horizontal components are equal and opposite.

Here all forces act at the same point, so the light-knot model has no remaining couple. For an extended body, a closed force triangle establishes zero resultant force but does not by itself establish zero resultant moment. Check the actual positions and lines of action too.

Optional check At a light stationary knot, the left tension has components (-24, +18) N and the right tension (+24, +32) N. A third string pulls down with 50 N. What do these values show?
At a light stationary knot, the left tension has components (-24, +18) N and the right tension (+24, +32) N. A third string pulls down with 50 N. What do these values show?

Use sine and cosine rules for a non-right triangle

A force triangle need not be right-angled. For any triangle with side lengths a, b and c opposite interior angles α, β and γ:

a/sin α = b/sin β = c/sin γ

c2 = a2 + b2 - 2ab cos γ

Match each side to its opposite angle. In the cosine rule, γ is the interior angle between the sides a and b. In a force triangle the side lengths represent force magnitudes, not physical string lengths.

Worked non-right triangle

Two 20 N tensions support a load

A different stationary light knot has two equal 20 N tensions, each directed 60° above the horizontal, one to the left and one to the right. A third string pulls down with the load force.

The tensions are 60° apart when drawn from the same origin. When drawn head to tail, their closed force triangle has a 120° interior angle between the two 20 N sides and two other interior angles of 30°.

Calling the downward load magnitude W, the cosine rule gives:

W2 = 202 + 202 - 2(20)(20) cos 120°
= 1200 N2.

Therefore W = 20√3 N ≈ 34.6 N.

The sine rule gives the same value:
20/sin 30° = W/sin 120°, so W = 20 sin 120°/sin 30° ≈ 34.6 N.

Use the interior angles of the closed force triangle

The common-origin sixty-degree angle becomes a one-hundred-and-twenty-degree triangle angleAt upper left, two equal twenty-newton tensions drawn from a common origin point sixty and one hundred and twenty degrees anticlockwise from the positive horizontal; the angle between them is sixty degrees. The main head-to-tail triangle retains those force directions. It uses twelve drawing units per newton equally in horizontal and vertical directions: the first twenty-newton arrow points left and up, the second right and up, and the downward load closes the triangle. The interior angle at the left vertex is one hundred and twenty degrees and those at the upper and lower vertices are each thirty degrees. The downward side W is twenty times the square root of three newtons, approximately 34.6. Grey leaders attach the two small thirty-degree angle labels; they are not forces.60°20 N eachCommon originClosed triangle20 N20 NW120°30°30°Start / finish

The downward side is W = 20√3 N, approximately 34.6 N. Its opposite interior angle is 120°. Each 20 N side is opposite a 30° angle. The 60° between the common-origin forces is not the interior angle to insert into the subtraction form of the cosine rule.

The 60-degree angle between tensions in the spatial arrangement becomes a 120-degree interior angle in the head-to-tail force triangle. The side opposite 120 degrees is the 20 square-root-3 N load force.

The subtraction form of the cosine rule uses the triangle's interior angle. Substituting the 60° tail-to-tail angle into that form would give the wrong load.

Finally, keep the system boundary consistent. If a body and an attached load are analysed together, forces between them are internal to that combined system. If analysing one alone, those interaction forces cross its boundary and must be included.