Curriculum

Mathematics

Unit 2

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Unit 5

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Text lesson

Parallel Coplanar Forces: Resultant and Equilibrium

Mathematics

Parallel Coplanar Forces: Resultant and Equilibrium

2-1 (b) The resultant of two like parallel forces

Rule

Let $F_{1}$ and $F_{2}$ be like forces at points $A$ and $B$ of a rigid body.
Their resultant $R$ is parallel to them, in the same direction, with magnitude
$R = F_{1} + F_{2}$
Its line of action cuts the line $A B$ at an inner point $C$ which divides $A B$ internally in the inverse ratio of the forces:
$F_{1} \cdot A C = F_{2} \cdot B C or \frac{A C}{B C} = \frac{F_{2}}{F_{1}}$
So $C$ is nearer to the larger force.

Parallel Coplanar Forces: porters carrying a load on a 5 m pole show the resultant of parallel forces: 50 N acting 2 m from one end.

Where the rule comes from

Add two equal and opposite forces $F$ along $A B$ at $A$ and at $B$ ; they do not change the effect.
Combine $F_{1}$ with the force at $A$ , and $F_{2}$ with the force at $B$ , by the parallelogram law. The two resultants meet at a point $O$ .
Resolve them again at O the added horizontal forces cancel and F1, F2 act together along the line through O.
Similar triangles then give $\frac{A C}{O C} = \frac{F}{F_{1}}$ and $\frac{B C}{O C} = \frac{F}{F_{2}}$ , so $F_{1} \cdot A C = F_{2} \cdot B C$ .

✍️ Worked examples

Example 1 Two like forces of 5 N and 7 N act at $A$ and $B$ with $A B = 36$ cm. Find their resultant.
Solution:
$R = 5 + 7 = 12$ N, in the direction of the two forces.
$\frac{A C}{C B} = \frac{7}{5}$ with $A C + C B = 36$ , so $A C = \frac{7}{12} \times 36 = 21$ cm.
The resultant acts 21 cm from $A$ (the end of the 5 N force) and 15 cm from $B$ .
Example 2 Where does the resultant of two equal like forces act?
Solution:
With $F_{1} = F_{2}$ , $A C = B C$ , so the resultant acts at the midpoint of $A B$ with magnitude $2 F$ .

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2-1 (c) The resultant of two unlike parallel forces

Rule

Let $F_{1} > F_{2}$ act in opposite directions at $A$ and $B$ .
The resultant is parallel to them, in the direction of the larger force, with magnitude
$R = F_{1} - F_{2}$
Its line of action cuts $A B$ at a point $C$ outside the segment, on the side of the larger force, and
$F_{1} \cdot A C = F_{2} \cdot B C$
Here $C$ is again nearer to the larger force $F_{1}$ .
If F1=F2 there is no resultant force the pair is a couple (see the next lesson).

✍️ Worked examples

Example 3 Unlike forces of 100 N (at $A$ ) and 40 N (at $B$ ) act at points 240 cm apart. Find the resultant.
Solution:
$R = 100 - 40 = 60$ N in the direction of the 100 N force.
$C$ lies on $B A$ produced beyond $A$ , with $B C = A C + 240$ .
$100 \cdot A C = 40 (A C + 240)$ , so $60 A C = 9600$ and $A C = 160$ cm.
The resultant acts 160 cm from $A$ outside the segment $A B$ .

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2-1 (d) Finding one force from the resultant

Method

Take the unit vector along the resultant and use $R = F_{1} + F_{2}$ with signs.
Take moments about the point where the resultant acts; the sum of moments of the forces about that point is zero.

✍️ Worked examples

Example 4:
One of two parallel forces is 20 N.
The resultant is 35 N, and its line of action is 15 cm from the 20 N force.
Find the other force and its position in each case (a) the 20 N force and the resultant have the same direction, (b) opposite directions.
Solution:
(a) $35 = 20 + F_{2}$ , so $F_{2} = 15$ N, in the same direction.
Moments about C (point of the resultant) 20×15=15×BC, so BC=20 cm. The 15 N force is on the other side of C, so AB=15+20=35 cm.
(b) The resultant (35 N) is opposite to the 20 N force, so $35 = F_{2} - 20$ and $F_{2} = 55$ N in the direction of the resultant.
Moments about C 20×15=55×BC, so BC=6011 cm.
Both $C$ and the 55 N force are on the same side of the 20 N force, so $A B = 15 - \frac{60}{11} = \frac{105}{11}$ cm.

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2-1 (e) The moment theorem (Varignon)

Theorem

The algebraic sum of the moments of a system of parallel coplanar forces about any point in their plane equals the moment of their resultant about the same point.
Write $\sum M_{O} = M_{O} (R)$ .

Draw the common perpendicular through $O$ meeting the lines of action at $A$ , $B$ and $C$ (for $R$ ).
Moment sum about O −F1⋅OA−F2⋅OB (with OA=OC−AC and OB=OC+CB in the appropriate arrangement).
Using $F_{1} \cdot A C = F_{2} \cdot C B$ the sum becomes $- (F_{1} + F_{2}) O C = - R \cdot O C$ , which is the moment of $R$ .
For unlike forces the same method gives $(F_{1} - F_{2}) O C = R \cdot O C$ .
A system of many forces is handled by combining forces in pairs.

If the resultant is zero, the sum of moments about any point is zero or constant (the system is either in equilibrium or equivalent to a couple).
The theorem is the tool for locating the line of action of a resultant: take moments about a convenient point.

✍️ Worked examples

Example 5:
Parallel forces act perpendicular to a line at the points $A$ , $C$ , $E$ , $B$ , which are 0, 4, 9 and 16 cm from $A$ .
The force at $A$ is 5 N, at $C$ it is 6 N, at $E$ it is 4 N in the opposite sense to the force at $C$ , and at $B$ it is 2 N in the same sense as the force at $C$ .
Find the algebraic sum of the moments about $A$ (take the sense of the force at $C$ as positive).
Solution:
The 5 N force at $A$ has zero moment about $A$ .
Moments about A 6×4−4×9+2×16=24−36+32=20 N·cm.

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2-1 (f) The resultant of a system of parallel forces

Procedure

Choose a positive direction (a unit vector $\hat{c}$ along the forces) and add the forces with signs: $R = \sum F$ .
If $R \neq 0$ , take moments about a convenient point $O$ (often the first force):
$R \cdot x = \sum F \cdot d$
where $x$ is the distance from $O$ to the line of action of $R$ , and $d$ the signed distance of each force.
State the magnitude, direction and position of the resultant (where it cuts the line through the points).

✍️ Worked examples

Example 6 $A, B, C, D, E$ lie on a line with $A B : B C : C D : D E = 2 : 3 : 4 : 7$ . Like forces of 30, 50, 20, 70, 40 N act at $A, B, C, D, E$ . Find the resultant.
Solution:
$R = 30 + 50 + 20 + 70 + 40 = 210$ N, in the common direction.
Let $A B = 2 x$ , so the distances from $A$ are $0, 2 x, 5 x, 9 x, 16 x$ .
Moments about A R⋅AO=50(2x)+20(5x)+70(9x)+40(16x)=100x+100x+630x+640x=1470x.
$A O = \frac{1470 x}{210} = 7 x$ , so $\frac{A O}{A E} = \frac{7 x}{16 x} = \frac{7}{16}$ .
The resultant divides AE internally in the ratio 7 16 from A.
Example 7:
Four forces act on points $A, B, C, D$ of a line with $A B = 2$ cm, $B C = 3$ cm, $C D = 5$ cm: 10 N down at $A$ , 25 N up at $B$ , 15 N down at $C$ , 20 N up at $D$ .
Find the resultant.
Solution:
Take upward as positive R=−10+25−15+20=20 N upward.
Distances from A 0,2,5,10 cm.
Moments about A 25(2)−15(5)+20(10)=50−75+200=175 N·cm.
$R \cdot x = 175$ , so $x = \frac{175}{20} = 8.75$ cm from $A$ , on $A D$ .
Example 8:
A light rod $A B$ carries a downward force 8 N at $A$ , an upward 14 N at 3 m from $A$ , and a downward 4 N at $B$ (6 m from $A$ ).
Find the resultant.
Solution:
Upward positive R=−8+14−4=2 N upward.
Moments about A 14(3)−4(6)=42−24=18 N·m.
$x = \frac{18}{2} = 9$ m from $A$ along $A B$ produced beyond $B$ (outside the rod).

Balance (centre of parallel weights)

The weights of particles on a light rod have a resultant equal to the total weight, acting at the balance point.
Example a light rod 1 m long carries 2 N at one end and 3 N at the other; the balance point is 25×100=40 cm from the 3 N end.

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2-1 (g) Theoretical results

Moving one of two like forces

✍️ Worked examples

Example 9:
Like forces $F_{1}$ and $F_{2}$ act at $A$ and $B$ with resultant $R$ .
If $F_{2}$ moves parallel to itself a distance $x$ along $A B$ , prove that the resultant moves a distance $\frac{F_{2}}{F_{1} + F_{2}} x$ in the same direction.
Solution:
Moments about A R⋅AC=F2⋅AB.
After the move the point of action of $F_{2}$ is $B^{'}$ with $A B^{'} = A B + x$ , and the resultant acts at $C^{'}$ : $R \cdot A C^{'} = F_{2} \cdot A B^{'}$ .
Subtract R (AC′−AC)=F2 x, so CC′=F2Rx=F2F1+F2x.

Forces given as vectors

✍️ Worked examples

Example 10 ${\vec{F}}_{1} = 2 \hat{i} - 3 \hat{j}$ acts at $A (1, 3)$ and ${\vec{F}}_{2} = 4 \hat{i} - 6 \hat{j}$ acts at $B (4, 9)$ . Find the resultant and the point where its line of action meets $A B$ .
Solution:
$\vec{R} = {\vec{F}}_{1} + {\vec{F}}_{2} = 6 \hat{i} - 9 \hat{j}$ .
${\vec{F}}_{2} = 2 {\vec{F}}_{1}$ , so the forces are parallel and in the same direction.
ACCB=F2F1=21, so C divides AB in the ratio 2 1 from A.
$C = (\frac{2 (4) + 1 (1)}{3}, \frac{2 (9) + 1 (3)}{3}) = (3, 7)$ .

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2-2 (a) Conditions of equilibrium

The two conditions

Comparison aspect	Main idea	Keywords
•	A rigid body acted on by coplanar forces is in equilibrium if and only if	rigidbodyactedcoplanar
the resultant vector is zero	R⃗=0⃗
the sum of moments about any point is zero	∑M=0

Take $\hat{i}$ and $\hat{j}$ in the plane and $\hat{k}$ perpendicular to it.
$\vec{R} = X \hat{i} + Y \hat{j}$ and the moment is along $\hat{k}$ , so the conditions become
$X = 0, Y = 0, \sum M = 0$
For parallel forces (say vertical), the first reduces to one equation: the forces up equal the forces down.
The moment equation can be taken about any point; the best choice is a point where an unknown force acts, to eliminate it.

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2-2 (b) Beams on supports

Procedure

Draw all forces weights (a uniform rod’s weight acts at its midpoint), loads, reactions of supports, tensions.
Write "vertical forces balance".
Take moments about one support to find the other reaction directly.

✍️ Worked examples

Example 11:
A uniform wooden board, 6 m long, has mass 30 kg per metre.
It rests horizontally on supports at its ends $A$ and $B$ and a box of 240 kg·wt is on the board 4 m from $B$ .
Find the reaction at each support.
Solution:
Mass of board $= 30 \times 6 = 180$ kg, so its weight is 180 kg·wt at the middle (3 m from each end).
Vertical balance RA+RB=180+240=420.
Moments about B RA×6=180×3+240×4=540+960=1500.
$R_{A} = 250$ kg·wt and $R_{B} = 420 - 250 = 170$ kg·wt.
As the box moves closer to $A$ , $R_{A}$ increases and $R_{B}$ decreases.
Example 12:
A uniform rod $A B$ of length 90 cm and weight 60 N is hung horizontally by two vertical strings at its ends.
Where must a 150 N weight be hung so that the tension at $A$ is twice the tension at $B$ ?
Solution:
Let the tension at $B$ be $T$ and at $A$ be $2 T$ .
Vertical balance 2T+T=150+60, so T=70 N.
Moments about A 150x+60(45)=T×90=6300.
$150 x = 3600$ , so $x = 24$ cm from $A$ .
Example 13:
A motorcycle of mass 200 kg has its weight acting midway between the wheels, 1.4 m apart.
A rider of 84 kg·wt sits with the weight acting 1 m behind the front wheel.
Find the ground reactions on each wheel (a) without the rider, (b) with the rider.
Solution:
(a) By symmetry each wheel carries $100$ kg·wt.
(b) Moments about the front wheel Rrear×1.4=200×0.7+84×1=224, so Rrear=160 kg·wt.
Vertical balance Rfront=284−160=124 kg·wt.

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2-2 (c) Tilting and limiting positions

Idea

A beam on two supports is about to tilt about one support when the reaction at the other support becomes zero.
Take moments about the support that remains in contact.

✍️ Worked examples

Example 14:
A uniform beam $A B$ of length 6 m and weight 40 kg·wt rests on supports at $C$ (1 m from $A$ ) and $D$ (1 m from $B$ ).
A man of weight 100 kg·wt walks from the middle towards $B$ .
How far from $A$ is he when the beam is about to tilt?
Solution:
At the limit the reaction at $C$ is zero, so take moments about $D$ (5 m from $A$ ).
The beam’s weight is 2 m to the left of D anticlockwise moment 40×2=80.
The man at distance x from A is (x−5) to the right of D clockwise moment 100(x−5).
$100 (x - 5) = 80$ , so $x = 5.8$ m from $A$ (0.2 m from the end $B$ ).
Example 15 Two men carry a body of 90 kg hanging from a light rod 60 cm long, with the body 20 cm from man $A$ . Find the load each carries.
Solution:
Moments about A RB×60=90×20, so RB=30 kg·wt.
RA=90−30=60 kg·wt the man nearer the load carries more.

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2-2 (d) A rod on a rough surface and a smooth wedge

Method

Forces on the rod weight W, reaction of the wedge S (perpendicular to the rod, since the wedge is smooth), the ground reaction R and friction F at the lower end.
Use $X = 0$ , $Y = 0$ and moments about the foot of the rod.
At the point of slipping, $F = μ R$ .

✍️ Worked examples

Example 16:
A uniform bar $A B$ of weight 5 kg·wt and length 30 cm rests with $A$ on rough ground and touches a smooth wedge at $C$ , 12.5 cm above the ground, at $3 0^{\circ}$ to the horizontal.
It is about to slip.
Find the reaction of the wedge and the coefficient of friction.
Solution:
$A C = \frac{12.5}{\sin 3 0^{\circ}} = 25$ cm. The weight acts 15 cm from $A$ .
Moments about A 5×15cos⁡30∘=S×25, so S=3cos⁡30∘=332≈2.60 kg·wt.
The wedge reaction is perpendicular to the bar, so it makes $3 0^{\circ}$ with the vertical. Horizontal: $F = S \sin 3 0^{\circ} = \frac{3 \sqrt{3}}{4}$ .
Vertical R+Scos⁡30∘=5, so R=5−332⋅32=5−94=114.
$μ = \frac{F}{R} = \frac{3 \sqrt{3} / 4}{11 / 4} = \frac{3 \sqrt{3}}{11} \approx 0.47$ .

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2-2 (e) A ladder against a smooth wall

Setting

Comparison aspect	Main idea	Keywords
•	A uniform ladder of weight $W$ and length $ℓ$ rests with its foot $A$ on rough horizontal ground and its top $B$ against a smooth vertical wall.	uniformladderweightlength
Forces	weight W at the middle vertical ground reaction R and friction F at A the wall reaction S (horizontal because the wall is smooth) at B	weightmiddleverticalground
Conditions	R=W and F=S moments about A give S	momentsgive
Limiting equilibrium	F=μR

✍️ Worked examples

Example 17:
A uniform ladder of weight $W$ makes an angle $θ$ with the ground and is about to slip.
The coefficient of friction at the ground is $μ$ .
Find $\tan θ$ .
Solution:
$R = W$ and $S = F = μ W$ .
Moments about the foot S ℓsin⁡θ=W ℓ2cos⁡θ.
$μ W \sin θ = \frac{W}{2} \cos θ$ , so $\tan θ = \frac{1}{2 μ}$ .
With $μ = 0.5$ the ladder is on the point of slipping at $θ = 4 5^{\circ}$ ; it is safe for larger angles.
Example 18:
A uniform ladder of weight 20 kg·wt rests at $6 0^{\circ}$ to the horizontal against a smooth wall on rough ground with $μ = \frac{1}{2 \sqrt{3}}$ .
Show that a girl of weight 60 kg·wt can climb half the length of the ladder before it slips.
Solution:
Let the ladder have length $ℓ$ and the girl be $x$ along it from the foot.
Vertical R=20+60=80 kg·wt. Limiting friction: F=μR=8023=403, and S=F.
Moments about the foot S ℓsin⁡60∘=20⋅ℓ2cos⁡60∘+60 xcos⁡60∘.
Left side 403⋅32ℓ=20ℓ. Right side: 5ℓ+30x.
$20 ℓ = 5 ℓ + 30 x$ gives $x = \frac{ℓ}{2}$ .
Example 19:
A uniform ladder of weight 30 kg·wt and length 4 m rests on a smooth floor against a smooth wall at $4 5^{\circ}$ , held by a horizontal rope at the foot.
Show the tension grows as a man of weight 80 kg·wt climbs.
Solution:
Floor reaction $R = 30 + 80 = 110$ ; the rope tension $T$ balances the wall reaction $S$ : $T = S$ .
Moments about the foot with the man at distance $x$ along the ladder: $S \times 4 \sin 4 5^{\circ} = 30 \times 2 \cos 4 5^{\circ} + 80 x \cos 4 5^{\circ}$ , so $S = \frac{60 + 80 x}{4} = 15 + 20 x$ .
$T = 15 + 20 x$ increases with $x$ .
If the rope can bear at most 67 kg·wt, then $15 + 20 x \leq 67$ gives $x \leq 2.6$ m.

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2-2 (f) Summary

Point of comparison	Like forces	Unlike forces
Direction of $R$	same as the forces	that of the larger force
Magnitude of $R$	$F_{1} + F_{2}$	$F_{1} - F_{2}$
Position of $R$	between the forces	outside, beyond the larger force
Relation	$F_{1} \cdot A C = F_{2} \cdot B C$	$F_{1} \cdot A C = F_{2} \cdot B C$
Equal forces	resultant at the midpoint	no resultant (a couple)

Checklist

Choose a positive direction before adding forces.
Take moments about a point where an unknown force acts.
Check the signs of clockwise and anticlockwise moments.
Check results the sum of moments about a second point must also be zero.
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Curriculum

Mathematics

Unit 1

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

Parallel Coplanar Forces: Resultant and Equilibrium

🌳 Concept tree

│ ├── Why a new idea is needed

│ └── A balancing experiment

│ ├── Rule

│ └── Where the rule comes from

│ └── Rule

├── 2-1 (d) Finding one force from the resultant

│ └── Method

├── 2-1 (e) The moment theorem (Varignon)

│ ├── Theorem

│ ├── Proof for two like forces

│ └── Consequence

│ ├── Procedure

│ └── Balance (centre of parallel weights)

├── 2-1 (g) Theoretical results

│ ├── Moving one of two like forces

│ └── Forces given as vectors

├── 2-2 (a) Conditions of equilibrium

│ ├── The two conditions

│ └── In components

├── 2-2 (b) Beams on supports

│ └── Procedure

├── 2-2 (c) Tilting and limiting positions

│ └── Idea

│ └── Method

├── 2-2 (e) A ladder against a smooth wall

│ └── Setting

└── 2-2 (f) Summary

├── Comparison

└── Checklist

Why a new idea is needed

Rule

Where the rule comes from

✍️ Worked examples

Rule

✍️ Worked examples

Method

✍️ Worked examples

Theorem

✍️ Worked examples

Procedure

✍️ Worked examples

Balance (centre of parallel weights)

Moving one of two like forces

✍️ Worked examples

Forces given as vectors

✍️ Worked examples

The two conditions

Procedure

✍️ Worked examples

Idea

✍️ Worked examples

Method

✍️ Worked examples

Setting

✍️ Worked examples

Checklist

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