Concurrent Forces

Statics · FE Reference Handbook section

Statics

60 formulas

10 exam-style examples

~60 min

All Statics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Concurrent Forces within Statics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what concurrent forces describes physically and when it applies.
State every one of the 60 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: keep force in lbf or kN and distance in ft or m consistently.

Lecture

Why this section exists. Concurrent Forces is the part of Statics that lets you connect a determinate frame, truss or beam in equilibrium to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as one free body, three equilibrium equations, one unknown reported. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. keep force in lbf or kN and distance in ft or m consistently. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: concurrent forces.
Capstone Studio instructional photograph

Statics — Concurrent Forces: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a determinate frame, truss or beam in equilibrium. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 60 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

A	Quantity produced by "A = bh/2" — read its definition and unit from the handbook line directly above the equation.
I yc	Quantity produced by "I yc = b 3h/36 ry2c = b 2 18 Ixc yc = Abh 36 = b 2 h 2 72" — read its definition and unit from the handbook line directly above the equation.
rx2	Quantity produced by "rx2 = h 2 6 Ixy = Abh 4 = b 2 h 2 8" — read its definition and unit from the handbook line directly above the equation.
yc	Quantity produced by "yc = h/3 Ix = bh3/12" — read its definition and unit from the handbook line directly above the equation.
ry2	Quantity produced by "ry2 = b 2 2" — read its definition and unit from the handbook line directly above the equation.
xc	Quantity produced by "xc = (a + b)/3" — read its definition and unit from the handbook line directly above the equation.
Iyc	Quantity produced by "Iyc = bh b − ab + a 2" — read its definition and unit from the handbook line directly above the equation.
ry2c	Quantity produced by "ry2c = b 2 − ab + a 2 ) 18 [ ]" — read its definition and unit from the handbook line directly above the equation.
I x	Quantity produced by "I x = bh 12 rx2 = h 2 6 I xy" — read its definition and unit from the handbook line directly above the equation.
x yc	Quantity produced by "x yc = h/3" — read its definition and unit from the handbook line directly above the equation.
I y	Quantity produced by "I y = bh b + ab + a 2" — read its definition and unit from the handbook line directly above the equation.
I xc y c	Quantity produced by "I xc y c = 0" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

A concurrent-force system is one in which the lines of action of the applied forces all meet at one point.
A two-force body in static equilibrium has two applied forces that are equal in magnitude, opposite in direction, and collinear.
Figure Area & Centroid Area Moment of Inertia (Radius of Gyration)2 Product of Inertia
C h
[ ( 2
)] 36 (
[ ( 2
)] 12 (
b x
y a
h 2 a 2 + 4ab + b 2 )
36(a + b ) 18(a + b )
C h h(2a + b )
3(a + b ) h 3 (3a + b ) h (3a + b )
12 6(a + b )
b x
( )
2 2 2
( )
b x
( 2 2
Housner, George W., and Donald E. Hudson, Applied Mechanics Dynamics, D. Van Nostrand Company, Inc., Princeton, NJ, 1959. Table reprinted by permission of G.W. Housner & D.E. Hudson.
Figure Area & Centroid Area Moment of Inertia (Radius of Gyration)2 Product of Inertia
4 4

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Tensions in a two-cable suspension — Concurrent Forces

A 1731 lb sign hangs from two cables inclined 46° and 44° above horizontal on opposite sides. Find both cable tensions.

Given

W = 1731 lb
θ₁ = 46°
θ₂ = 44°

Find

T₁ and T₂

Start with the thinking

Concurrent force system — two equations, two unknowns.
Write both x and y equilibrium before solving.

Step-by-step solution

Horizontal — ΣF_x = 0: T₁cos θ₁ = T₂cos θ₂
Vertical — ΣF_y = 0: T₁sin θ₁ + T₂sin θ₂ = W
Solve — T₂ = W·cos θ₁/sin(θ₁ + θ₂) = 1731·cos 46°/sin 90° = 1,202 lb
Back-substitute
$T_{1} = T_{2} cos θ_{2} / cos θ_{1} = 1, 245 l b$
Check
$T_{1} sin θ_{1} + T_{2} sin θ_{2} = 1, 731 l b \approx W ✓$

Answer: T₁ ≈ 1,245 lb, T₂ ≈ 1,202 lb

Why the other options are there

865.5 lb each (symmetry wrongly assumed)
2,406 lb (second cable ignored)

Reference: FE Reference Handbook — Statics → Concurrent Forces

Example 2

Resolution of a force system into a single resultant — Concurrent Forces

Two forces act at a gusset plate: 21 kN at 25° and 53 kN at 142° from the positive x-axis. Resolve each force into components and determine the magnitude and direction of the resultant force.

Given

F₁ = 21 kN at 25°
F₂ = 53 kN at 142°

Find

Resultant force magnitude R and its direction

Start with the thinking

A force system is resolved by summing x- and y-components separately.
The resultant of concurrent forces is the vector sum, never the arithmetic sum of magnitudes.

Step-by-step solution

Formula
$F_{x} = F cos θ, F_{y} = F sin θ$
F₁ components
$21 cos 2 5^{\circ} = 19.03 k N, 21 s in 2 5^{\circ} = 8.87 k N$
F₂ components
$53 cos 14 2^{\circ} = - 41.76 k N, 53 s in 14 2^{\circ} = 32.63 k N$
Sums — ΣFₓ = -22.73 kN, ΣF_y = 41.51 kN
Formula — R = √(ΣFₓ² + ΣF_y²)
Substituting
$R = (- 22.7 3^{2} + 41.5 1^{2}) = 47.32 k N a t 118. 7^{\circ}$

Answer: R = 47.32 kN acting at 118.7° from the x-axis

Why the other options are there

74 kN (magnitudes added)
22.73 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Concurrent Forces

Example 3

Tensions in a two-cable suspension — Concurrent Forces (2)

A 2108 lb sign hangs from two cables inclined 52° and 47° above horizontal on opposite sides. Find both cable tensions.

Given

W = 2108 lb
θ₁ = 52°
θ₂ = 47°

Find

T₁ and T₂

Start with the thinking

Concurrent force system — two equations, two unknowns.
Write both x and y equilibrium before solving.

Step-by-step solution

Horizontal — ΣF_x = 0: T₁cos θ₁ = T₂cos θ₂
Vertical — ΣF_y = 0: T₁sin θ₁ + T₂sin θ₂ = W
Solve — T₂ = W·cos θ₁/sin(θ₁ + θ₂) = 2108·cos 52°/sin 99° = 1,314 lb
Back-substitute
$T_{1} = T_{2} cos θ_{2} / cos θ_{1} = 1, 456 l b$
Check
$T_{1} sin θ_{1} + T_{2} sin θ_{2} = 2, 108 l b \approx W ✓$

Answer: T₁ ≈ 1,456 lb, T₂ ≈ 1,314 lb

Why the other options are there

1,054 lb each (symmetry wrongly assumed)
2,675 lb (second cable ignored)

Reference: FE Reference Handbook — Statics → Concurrent Forces

Example 4

Resolution of a force system into a single resultant — Concurrent Forces (2)

Two forces act at a gusset plate: 41 kN at 73° and 55 kN at 155° from the positive x-axis. Resolve each force into components and determine the magnitude and direction of the resultant force.

Given

F₁ = 41 kN at 73°
F₂ = 55 kN at 155°

Find

Resultant force magnitude R and its direction

Start with the thinking

A force system is resolved by summing x- and y-components separately.
The resultant of concurrent forces is the vector sum, never the arithmetic sum of magnitudes.

Step-by-step solution

Formula
$F_{x} = F cos θ, F_{y} = F sin θ$
F₁ components
$41 cos 7 3^{\circ} = 11.99 k N, 41 s in 7 3^{\circ} = 39.21 k N$
F₂ components
$55 cos 15 5^{\circ} = - 49.85 k N, 55 s in 15 5^{\circ} = 23.24 k N$
Sums — ΣFₓ = -37.86 kN, ΣF_y = 62.45 kN
Formula — R = √(ΣFₓ² + ΣF_y²)
Substituting
$R = (- 37.8 6^{2} + 62.4 5^{2}) = 73.03 k N a t 121. 2^{\circ}$

Answer: R = 73.03 kN acting at 121.2° from the x-axis

Why the other options are there

96 kN (magnitudes added)
37.86 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Concurrent Forces

Example 5

Tensions in a two-cable suspension — Concurrent Forces (3)

A 1386 lb sign hangs from two cables inclined 27° and 32° above horizontal on opposite sides. Find both cable tensions.

Given

W = 1386 lb
θ₁ = 27°
θ₂ = 32°

Find

T₁ and T₂

Start with the thinking

Concurrent force system — two equations, two unknowns.
Write both x and y equilibrium before solving.

Step-by-step solution

Horizontal — ΣF_x = 0: T₁cos θ₁ = T₂cos θ₂
Vertical — ΣF_y = 0: T₁sin θ₁ + T₂sin θ₂ = W
Solve — T₂ = W·cos θ₁/sin(θ₁ + θ₂) = 1386·cos 27°/sin 59° = 1,441 lb
Back-substitute
$T_{1} = T_{2} cos θ_{2} / cos θ_{1} = 1, 371 l b$
Check
$T_{1} sin θ_{1} + T_{2} sin θ_{2} = 1, 386 l b \approx W ✓$

Answer: T₁ ≈ 1,371 lb, T₂ ≈ 1,441 lb

Why the other options are there

693.0 lb each (symmetry wrongly assumed)
3,053 lb (second cable ignored)

Reference: FE Reference Handbook — Statics → Concurrent Forces

Example 6

Resolution of a force system into a single resultant — Concurrent Forces (3)

Two forces act at a gusset plate: 68 kN at 57° and 28 kN at 157° from the positive x-axis. Resolve each force into components and determine the magnitude and direction of the resultant force.

Given

F₁ = 68 kN at 57°
F₂ = 28 kN at 157°

Find

Resultant force magnitude R and its direction

Start with the thinking

A force system is resolved by summing x- and y-components separately.
The resultant of concurrent forces is the vector sum, never the arithmetic sum of magnitudes.

Step-by-step solution

Formula
$F_{x} = F cos θ, F_{y} = F sin θ$
F₁ components
$68 cos 5 7^{\circ} = 37.04 k N, 68 s in 5 7^{\circ} = 57.03 k N$
F₂ components
$28 cos 15 7^{\circ} = - 25.77 k N, 28 s in 15 7^{\circ} = 10.94 k N$
Sums — ΣFₓ = 11.26 kN, ΣF_y = 67.97 kN
Formula — R = √(ΣFₓ² + ΣF_y²)
Substituting
$R = (11.2 6^{2} + 67.9 7^{2}) = 68.90 k N a t 80. 6^{\circ}$

Answer: R = 68.90 kN acting at 80.6° from the x-axis

Why the other options are there

96 kN (magnitudes added)
11.26 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Concurrent Forces

Example 7

Tensions in a two-cable suspension — Concurrent Forces (4)

A 2230 lb sign hangs from two cables inclined 42° and 27° above horizontal on opposite sides. Find both cable tensions.

Given

W = 2230 lb
θ₁ = 42°
θ₂ = 27°

Find

T₁ and T₂

Start with the thinking

Concurrent force system — two equations, two unknowns.
Write both x and y equilibrium before solving.

Step-by-step solution

Horizontal — ΣF_x = 0: T₁cos θ₁ = T₂cos θ₂
Vertical — ΣF_y = 0: T₁sin θ₁ + T₂sin θ₂ = W
Solve — T₂ = W·cos θ₁/sin(θ₁ + θ₂) = 2230·cos 42°/sin 69° = 1,775 lb
Back-substitute
$T_{1} = T_{2} cos θ_{2} / cos θ_{1} = 2, 128 l b$
Check
$T_{1} sin θ_{1} + T_{2} sin θ_{2} = 2, 230 l b \approx W ✓$

Answer: T₁ ≈ 2,128 lb, T₂ ≈ 1,775 lb

Why the other options are there

1,115 lb each (symmetry wrongly assumed)
3,333 lb (second cable ignored)

Reference: FE Reference Handbook — Statics → Concurrent Forces

Example 8

Resolution of a force system into a single resultant — Concurrent Forces (4)

Two forces act at a gusset plate: 43 kN at 38° and 45 kN at 102° from the positive x-axis. Resolve each force into components and determine the magnitude and direction of the resultant force.

Given

F₁ = 43 kN at 38°
F₂ = 45 kN at 102°

Find

Resultant force magnitude R and its direction

Start with the thinking

A force system is resolved by summing x- and y-components separately.
The resultant of concurrent forces is the vector sum, never the arithmetic sum of magnitudes.

Step-by-step solution

Formula
$F_{x} = F cos θ, F_{y} = F sin θ$
F₁ components
$43 cos 3 8^{\circ} = 33.88 k N, 43 s in 3 8^{\circ} = 26.47 k N$
F₂ components
$45 cos 10 2^{\circ} = - 9.36 k N, 45 s in 10 2^{\circ} = 44.02 k N$
Sums — ΣFₓ = 24.53 kN, ΣF_y = 70.49 kN
Formula — R = √(ΣFₓ² + ΣF_y²)
Substituting
$R = (24.5 3^{2} + 70.4 9^{2}) = 74.64 k N a t 70. 8^{\circ}$

Answer: R = 74.64 kN acting at 70.8° from the x-axis

Why the other options are there

88 kN (magnitudes added)
24.53 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Concurrent Forces

Example 9

Tensions in a two-cable suspension — Concurrent Forces (5)

A 2362 lb sign hangs from two cables inclined 49° and 47° above horizontal on opposite sides. Find both cable tensions.

Given

W = 2362 lb
θ₁ = 49°
θ₂ = 47°

Find

T₁ and T₂

Start with the thinking

Concurrent force system — two equations, two unknowns.
Write both x and y equilibrium before solving.

Step-by-step solution

Horizontal — ΣF_x = 0: T₁cos θ₁ = T₂cos θ₂
Vertical — ΣF_y = 0: T₁sin θ₁ + T₂sin θ₂ = W
Solve — T₂ = W·cos θ₁/sin(θ₁ + θ₂) = 2362·cos 49°/sin 96° = 1,558 lb
Back-substitute
$T_{1} = T_{2} cos θ_{2} / cos θ_{1} = 1, 620 l b$
Check
$T_{1} sin θ_{1} + T_{2} sin θ_{2} = 2, 362 l b \approx W ✓$

Answer: T₁ ≈ 1,620 lb, T₂ ≈ 1,558 lb

Why the other options are there

1,181 lb each (symmetry wrongly assumed)
3,130 lb (second cable ignored)

Reference: FE Reference Handbook — Statics → Concurrent Forces

Example 10

Resolution of a force system into a single resultant — Concurrent Forces (5)

Two forces act at a gusset plate: 49 kN at 72° and 71 kN at 130° from the positive x-axis. Resolve each force into components and determine the magnitude and direction of the resultant force.

Given

F₁ = 49 kN at 72°
F₂ = 71 kN at 130°

Find

Resultant force magnitude R and its direction

Start with the thinking

A force system is resolved by summing x- and y-components separately.
The resultant of concurrent forces is the vector sum, never the arithmetic sum of magnitudes.

Step-by-step solution

Formula
$F_{x} = F cos θ, F_{y} = F sin θ$
F₁ components
$49 cos 7 2^{\circ} = 15.14 k N, 49 s in 7 2^{\circ} = 46.60 k N$
F₂ components
$71 cos 13 0^{\circ} = - 45.64 k N, 71 s in 13 0^{\circ} = 54.39 k N$
Sums — ΣFₓ = -30.50 kN, ΣF_y = 101.0 kN
Formula — R = √(ΣFₓ² + ΣF_y²)
Substituting
$R = (- 30.5 0^{2} + 101. 0^{2}) = 105.5 k N a t 106. 8^{\circ}$

Answer: R = 105.5 kN acting at 106.8° from the x-axis

Why the other options are there

120 kN (magnitudes added)
30.50 kN (y-component dropped)

Reference: FE Reference Handbook — Statics → Concurrent Forces

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a determinate frame, truss or beam in equilibrium, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Concurrent Forces contains 60 relations; you must be able to find this page in under 15 seconds.
Exam style: one free body, three equilibrium equations, one unknown reported.
Unit rule: keep force in lbf or kN and distance in ft or m consistently.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

keep force in lbf or kN and distance in ft or m consistently
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.