System of Forces Notes FE Sem 1 Engineering Mechanics | MU

Doubtly 18 min read Updated 25 Oct 2026 Save WhatsApp

On this page

This module builds the language of statics: what a force system is, how forces are combined and split into components, and how their turning effect is measured as a moment. You will learn to find the resultant of concurrent, parallel and general coplanar force systems, use couples and Varignon's theorem, and extend the idea to concurrent forces in space. Every later topic, from equilibrium and beam reactions to friction, depends on resolving forces and taking moments correctly. Questions from this module almost always appear as numericals on resultant magnitude, direction and position.

Summary

Classification of force systems: a force is a vector with magnitude, direction, sense and point of application. Force systems are grouped by whether the lines of action lie in one plane (coplanar) or not (non-coplanar or space), and by how they meet: concurrent (all lines pass through one point), parallel (lines parallel, like or unlike), collinear (along one line) and general or non-concurrent non-parallel. Recognising the type decides which method and how many equations are needed.

Principle of transmissibility: for a rigid body, the external effect of a force is unchanged if its point of application is moved anywhere along its own line of action. This lets us slide forces along their lines to a convenient point, but it holds only for external effects on rigid bodies, not for internal stresses or deformable bodies.

Composition and resolution of forces: composition means combining two or more forces into one resultant (parallelogram law, triangle law, polygon law or the method of components). Resolution is the reverse: splitting a force F into components, usually perpendicular, Fx = F cos theta and Fy = F sin theta, where theta is measured from the x axis. Choosing axes along and perpendicular to an inclined surface often simplifies problems.

Resultant of coplanar concurrent forces: resolve each force, add components to get Sum Fx and Sum Fy, then R = square root of (Sum Fx squared + Sum Fy squared) and the direction from tan alpha = Sum Fy / Sum Fx, with the quadrant fixed by the signs of the sums. The resultant passes through the common point of concurrency.

Resultant of parallel forces: the resultant is the algebraic sum of the forces, R = Sum F, acting parallel to them. Its position is found by equating the moment of R about any point to the sum of moments of the individual forces. If the forces reduce to zero net force but a non-zero moment, the system reduces to a couple.

Moment of a force, couples and Varignon's theorem: the moment of a force about a point is M = F x d, where d is the perpendicular distance from the point to the line of action; it is measured in N m and has a sense (clockwise or anticlockwise). A couple is a pair of equal, opposite, non-collinear parallel forces; its moment F x d is the same about every point (a free vector) and it cannot be balanced by a single force. Varignon's theorem states that the moment of a force about a point equals the algebraic sum of the moments of its components about the same point, which makes moment calculation easy using x and y components.

Resultant of a general coplanar force system: find Sum Fx, Sum Fy and the sum of moments Sum M_O about a chosen point O. The system is equivalent to a single force R at O plus a couple M = Sum M_O (an equivalent force-couple system). The actual line of action of R is located at a perpendicular distance d = Sum M_O / R from O, or by its x-intercept x = Sum M_O / Sum Fy, placed so that R produces the same sense of moment.

Resultant of non-coplanar concurrent (space) forces: each force is written as a vector F = F times the unit vector along its line, using the coordinates of two points on the line (unit vector = position vector / its magnitude). Adding components gives R = Sum Fx i + Sum Fy j + Sum Fz k, the magnitude R = square root of the sum of squares, and direction cosines cos theta_x = Sum Fx / R, cos theta_y = Sum Fy / R, cos theta_z = Sum Fz / R.

Mind map

System of Forces

Classification
- Coplanar and non-coplanar
- Concurrent
- Parallel (like, unlike)
- Collinear
- General (non-concurrent)
Basic principles
- Transmissibility
- Composition
- Resolution into components
- Parallelogram, triangle, polygon laws
Coplanar resultants
- Concurrent: R and angle
- Parallel: R and position
- General: R, angle, line of action
Moments
- M = F x d
- Sign convention
- Varignon's theorem
- Couple and its moment
Equivalent systems
- Force-couple system
- Shifting a force to another point
Space forces
- Unit vector along line
- Vector components
- Direction cosines

Method to find the resultant of a general coplanar force system

Draw the body, choose x-y axes and a reference point O (often a corner or the origin).
Resolve every inclined force into x and y components with correct signs.
Add components: Sum Fx and Sum Fy (include only forces, not couples).
Compute R = square root of (Sum Fx^2 + Sum Fy^2) and angle alpha = arctan of |Sum Fy / Sum Fx|; fix the quadrant from the signs.
Take moments of all forces (using Varignon's theorem with components) and all applied couples about O to get Sum M_O with sign.
State the equivalent force-couple system at O: R at O plus couple Sum M_O.
Locate the line of action: d = |Sum M_O| / R, or x-intercept x = Sum M_O / Sum Fy, ensuring R gives the same sense of moment about O.
Show the final resultant on a neat sketch with magnitude, angle and position.

Important definitions

Term	Meaning
Force	An external agency that changes or tends to change the state of rest or motion of a body; a vector quantity measured in newtons (N).
Force system	A group of forces acting on a body at the same time.
Coplanar force system	A force system in which the lines of action of all forces lie in one plane.
Concurrent force system	A force system in which the lines of action of all forces pass through a single point.
Parallel force system	Forces whose lines of action are parallel; like parallel forces act in the same sense, unlike parallel forces in opposite senses.
Resultant	A single force (or force and couple) that produces the same external effect on a rigid body as the given force system.
Principle of transmissibility	The external effect of a force on a rigid body is unchanged if its point of application is shifted along its line of action.
Resolution of a force	Splitting a force into two or more components that together produce the same effect.
Moment of a force	The turning effect of a force about a point, equal to force times perpendicular distance from the point to its line of action.
Couple	Two equal and opposite parallel forces with different lines of action; it causes pure rotation and its moment is the same about every point.
Varignon's theorem	The moment of a force about any point equals the algebraic sum of the moments of its components about that point.
Force-couple system	A force acting at a chosen point together with a couple, equivalent to a given force or force system.

Formulas

Formula	Meaning
\( F_x = F\cos\theta,\quad F_y = F\sin\theta \)	Rectangular components of force F (N) making angle theta with the x axis.
\( R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2},\quad \tan\alpha = \frac{\sum F_y}{\sum F_x} \)	Magnitude and direction of the resultant of coplanar forces; fix the quadrant of alpha from the signs of the sums.
\( R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta},\quad \tan\alpha = \frac{Q\sin\theta}{P + Q\cos\theta} \)	Parallelogram law for two forces P and Q with included angle theta; alpha is measured from P.
\( M_O = F\,d \)	Moment of force F about O, d = perpendicular distance (m); unit N m. Anticlockwise usually taken positive.
\( M_O = x F_y - y F_x \)	Varignon's theorem in component form for a force with components Fx, Fy acting at point (x, y) relative to O; positive means anticlockwise.
\( M = F\,d \) (couple)	Moment of a couple, d = arm (perpendicular distance between the two forces); independent of the reference point.
\( R = \sum F,\quad \bar{x} = \frac{\sum M_O}{R} \)	Resultant of parallel forces and its distance from O.
\( d = \frac{\|\sum M_O\|}{R},\quad x = \frac{\sum M_O}{\sum F_y} \)	Perpendicular distance of the resultant from O and its x-intercept for a general coplanar system.
\( \vec{F} = F\,\frac{(x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}} \)	Vector form of a space force of magnitude F directed from point 1 to point 2.
\( R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2 + (\sum F_z)^2},\quad \cos\theta_x = \frac{\sum F_x}{R} \)	Resultant of concurrent space forces and its direction cosines (similarly for y and z); cos^2 theta_x + cos^2 theta_y + cos^2 theta_z = 1.

Key points to remember

Always state the sign convention (for example right and up positive, anticlockwise moment positive) before solving.
Transmissibility applies only to rigid bodies and only to external effects.
The resultant of a concurrent system passes through the point of concurrency, so no moment equation is needed to locate it.
A couple cannot be replaced by a single force; it can only be balanced by another couple of equal and opposite moment.
The moment of a couple is the same about every point in its plane, so it can be moved freely in its plane.
A force F at A is equivalent to the same force F at B plus a couple of moment equal to the moment of F about B.
If Sum Fx = 0 and Sum Fy = 0 but Sum M is not zero, the resultant is a pure couple.
Varignon's theorem avoids finding perpendicular distances: resolve the force at a convenient point on its line and take moments of the components.
For space forces, find the unit vector from the coordinates of two points on the line of action; direction cosines of the resultant must satisfy the sum-of-squares-equals-one check.

Solved examples

Example 1. Four coplanar concurrent forces act at a point: 20 N at 0 degrees, 30 N at 60 degrees, 40 N at 150 degrees and 50 N at 225 degrees, all angles measured anticlockwise from the positive x axis. Find the magnitude and direction of the resultant.

Resolve each force (F cos theta, F sin theta). 20 N: (20.00, 0). 30 N: (15.00, 25.98). 40 N: (-34.64, 20.00). 50 N: (-35.36, -35.36). Sum Fx = 20 + 15 - 34.64 - 35.36 = -35.00 N. Sum Fy = 0 + 25.98 + 20.00 - 35.36 = 10.63 N. R = square root of (35.00^2 + 10.63^2) = 36.57 N. tan alpha = 10.63/35.00 gives alpha = 16.89 degrees. Since Sum Fx is negative and Sum Fy is positive, R lies in the second quadrant: R = 36.57 N at 16.89 degrees above the negative x axis, that is 163.11 degrees from the positive x axis.

Example 2. A horizontal beam AB carries vertical loads: 10 kN downward at A, 20 kN downward at 2 m from A, 15 kN upward at 4 m from A and 25 kN downward at 6 m from A. Find the resultant and its position from A.

Taking downward positive: R = 10 + 20 - 15 + 25 = 40 kN downward. Moments about A (clockwise positive): 10 x 0 + 20 x 2 - 15 x 4 + 25 x 6 = 0 + 40 - 60 + 150 = 130 kN m clockwise. By Varignon's theorem, R x x = 130, so x = 130/40 = 3.25 m. The resultant is 40 kN acting vertically downward at 3.25 m to the right of A.

Example 3. A rectangular plate ABCD has A(0,0), B(4,0), C(4,3) and D(0,3) in metres. Forces acting are: 15 kN at A at 30 degrees above the positive x axis, 10 kN at B vertically upward, 20 kN at C horizontally towards D, and an anticlockwise couple of 12 kN m. Find the resultant and the point where its line of action cuts line AB (extended if needed).

Components: 15 kN at A gives Fx = 15 cos 30 = 12.99 kN, Fy = 15 sin 30 = 7.50 kN. 10 kN at B: (0, +10). 20 kN at C towards D: (-20, 0). Sum Fx = 12.99 - 20 = -7.01 kN. Sum Fy = 7.50 + 10 = 17.50 kN. R = square root of (7.01^2 + 17.50^2) = 18.85 kN. tan alpha = 17.50/7.01 gives alpha = 68.17 degrees with the negative x axis (second quadrant), i.e. 111.83 degrees from the positive x axis. Moments about A (anticlockwise positive), using M = xFy - yFx: force at A gives 0; 10 kN at B: 4 x 10 = +40; 20 kN at C: -(3)(-20) = +60; couple +12. Sum M_A = 112 kN m anticlockwise. Let R cut AB at (x, 0): moment of R about A = x times Sum Fy = 17.5x = 112, so x = 6.4 m. The resultant 18.85 kN at 111.83 degrees cuts AB produced at 6.4 m to the right of A (2.4 m beyond B).

Example 4. A force of 500 N acts at the point (3 m, 4 m) at 30 degrees to the positive x axis. Using Varignon's theorem, find its moment about the origin O and the perpendicular distance of its line of action from O.

Fx = 500 cos 30 = 433.01 N, Fy = 500 sin 30 = 250 N. By Varignon's theorem, M_O = xFy - yFx = 3(250) - 4(433.01) = 750 - 1732.05 = -982.05 N m, i.e. 982.05 N m clockwise. Perpendicular distance d = M/F = 982.05/500 = 1.964 m.

Example 5. Three forces act at the origin O and are directed towards the points shown: F1 = 100 N towards A(2, 3, 6), F2 = 140 N towards B(-3, 2, 6), F3 = 60 N towards C(6, -2, 3) (coordinates in m). Find the resultant in vector form, its magnitude and its direction angles.

Lengths: OA = square root of (4 + 9 + 36) = 7, OB = square root of (9 + 4 + 36) = 7, OC = square root of (36 + 4 + 9) = 7. F1 = (100/7)(2i + 3j + 6k) = 28.57i + 42.86j + 85.71k N. F2 = (140/7)(-3i + 2j + 6k) = -60i + 40j + 120k N. F3 = (60/7)(6i - 2j + 3k) = 51.43i - 17.14j + 25.71k N. R = 20.00i + 65.71j + 231.43k N. Magnitude R = square root of (20^2 + 65.71^2 + 231.43^2) = 241.41 N. cos theta_x = 20/241.41 = 0.0828 so theta_x = 85.25 degrees; cos theta_y = 65.71/241.41 = 0.2722 so theta_y = 74.20 degrees; cos theta_z = 231.43/241.41 = 0.9587 so theta_z = 16.53 degrees. Check: 0.0828^2 + 0.2722^2 + 0.9587^2 = 1.000.

Example 6. A couple consists of two forces of 50 N each with an arm of 0.4 m. Find the moment of the couple. If it is to be replaced by an equivalent couple whose forces act 0.25 m apart, find the forces.

Moment of couple M = F x d = 50 x 0.4 = 20 N m. An equivalent couple must have the same moment and sense: F' x 0.25 = 20, so F' = 80 N. Two equal, opposite, parallel forces of 80 N each, 0.25 m apart, in the same rotational sense, give the equivalent couple.

Example 7. A vertical downward force of 400 N acts at point A, and its line of action is 0.3 m to the right of point O. Replace it by an equivalent force-couple system at O.

Shifting a force to a parallel position requires adding a couple equal to the moment of the force about the new point. Force at O: 400 N downward (same magnitude and direction). Couple: M_O = 400 x 0.3 = 120 N m clockwise (a downward force to the right of O turns clockwise about O). Equivalent system at O: 400 N vertically downward plus a 120 N m clockwise couple.

Important questions and answers

Q1. State and explain the principle of transmissibility with a sketch. State its limitation. (3 marks)

The principle of transmissibility states that the point of application of a force acting on a rigid body may be shifted to any other point on its line of action without changing its external effect. For example, pushing a cart at the back or pulling it at the front with the same force along the same line gives the same motion. Limitation: it applies only to rigid bodies and external effects; internal effects such as stresses and deformation change when the point of application changes.

Q2. Classify force systems with neat sketches. (4 marks)

Coplanar forces lie in one plane; non-coplanar (space) forces do not. Coplanar systems: (1) collinear, all forces along one line; (2) concurrent, lines meet at a point; (3) parallel, lines parallel, like (same sense) or unlike (opposite sense); (4) general or non-concurrent non-parallel. Non-coplanar systems are similarly concurrent, parallel or general in three dimensions. Sketches show each case with arrows on a body.

Q3. State and prove Varignon's theorem. (4 marks)

Statement: the moment of a force about any point equals the algebraic sum of the moments of its components about the same point. Proof: let force R at point A(x, y) have components Fx and Fy, with R making angle theta with the x axis. Moment of Fx about O is -yFx and moment of Fy about O is xFy, so sum = xFy - yFx = R(x sin theta - y cos theta). The perpendicular distance from O to the line of action of R is d = x sin theta - y cos theta (projection of OA perpendicular to R). Hence moment of R about O = R d = xFy - yFx, which equals the sum of the moments of the components. The theorem extends to any number of concurrent forces.

Q4. Define a couple and state its characteristics. (4 marks)

A couple is a pair of equal, opposite and parallel forces whose lines of action are different. Characteristics: (1) the resultant force is zero, so it produces only rotation; (2) its moment equals one force times the arm and is the same about every point in the plane; (3) it cannot be balanced by a single force, only by an equal and opposite couple; (4) couples in the same plane can be added algebraically; (5) a couple can be moved anywhere in its plane or to a parallel plane without changing its effect.

Q5. Explain how a force can be replaced by an equivalent force-couple system at another point. (3 marks)

Let F act at A. At point B apply two equal and opposite forces F and -F parallel to the original; this does not change the effect. The original F at A and -F at B form a couple of moment F x d, where d is the perpendicular distance from B to the line of F. What remains is F at B plus a couple M = F d. Thus any force equals the same force at another point together with a couple equal to the moment of the original force about that point.

Q6. Find the resultant of two forces 60 N and 80 N acting at a point with an included angle of 60 degrees. (3 marks)

R = square root of (60^2 + 80^2 + 2 x 60 x 80 cos 60) = square root of (3600 + 6400 + 4800) = square root of 14800 = 121.66 N. tan alpha = 80 sin 60 / (60 + 80 cos 60) = 69.28/100 = 0.6928, so alpha = 34.72 degrees with the 60 N force.

Q7. Four parallel forces 20 N up, 50 N down, 30 N up and 40 N down act on a beam at distances 0, 1, 2.5 and 4 m from the left end. Find the resultant and its location. (4 marks)

Taking upward positive: R = 20 - 50 + 30 - 40 = -40 N, i.e. 40 N downward. Moments about the left end, clockwise positive: 50 x 1 - 30 x 2.5 + 40 x 4 = 50 - 75 + 160 = 135 N m clockwise. The 40 N downward resultant at distance x gives clockwise moment 40x = 135, so x = 3.375 m from the left end.

Q8. Explain the method to find the resultant of a non-concurrent coplanar force system. (5 marks)

Resolve all forces into x and y components and sum them: Sum Fx, Sum Fy. Magnitude R = square root of (Sum Fx^2 + Sum Fy^2); direction from tan alpha = Sum Fy / Sum Fx with the quadrant from the signs. Take moments of all forces and couples about a reference point O to obtain Sum M_O. The system equals R at O plus a couple Sum M_O. The position of R is located by d = Sum M_O / R from O, or by intercepts x = Sum M_O / Sum Fy on the x axis and y = -Sum M_O / Sum Fx on the y axis, choosing the side so that R gives the same sense of moment about O.

The resultant is the single force that produces the same effect as all the forces of a system. The equilibrant is the single force that, when added to the system, keeps the body in equilibrium. The equilibrant is equal in magnitude to the resultant, acts along the same line of action but in the opposite sense.

Q10. A force of 1000 N passes from A(1, 2, 3) to B(5, -2, 5) m. Write the force in vector form and find its direction cosines. (4 marks)

Position vector AB = 4i - 4j + 2k, |AB| = square root of (16 + 16 + 4) = 6 m. Unit vector = (4i - 4j + 2k)/6. F = 1000(0.6667i - 0.6667j + 0.3333k) = 666.67i - 666.67j + 333.33k N. Direction cosines: l = 0.6667, m = -0.6667, n = 0.3333 (check: 0.4444 + 0.4444 + 0.1111 = 1).

Exam tips

Draw a neat sketch with axes and show components of each force; examiners award marks for the diagram.
Write the sign convention at the top of the solution and use it consistently for forces and moments.
Always give the direction of the resultant with the quadrant and draw it, not just the angle from a calculator.
When locating a resultant, check that R placed at the found position gives the same sense of moment as Sum M.
For space forces, check that direction cosines satisfy l^2 + m^2 + n^2 = 1 to catch arithmetic errors.
Keep units (N, kN, N m, kN m) in every line and the final answer.

FAQs

Is the prerequisite (Lami's theorem, parallelogram law) examined?

The syllabus says no questions are set directly on the prerequisite topics, but they are used inside problems, so you must be comfortable with them.

What is the difference between moment and couple?

A moment is the turning effect of one force about a point and depends on that point. A couple is a pair of equal, opposite, parallel forces; its moment is the same about every point.

How do I decide the quadrant of the resultant?

Use the signs of Sum Fx and Sum Fy: (+,+) first, (-,+) second, (-,-) third and (+,-) fourth quadrant. Find the acute angle with arctan of the absolute ratio and place it accordingly.

Which space force problems are in the syllabus?

Only the resultant of concurrent non-coplanar forces: writing forces in vector form, adding them and finding magnitude and direction cosines.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Engineering Mechanics) and were prepared with AI help and checked.

More in Engineering Mechanics

See all →