Solid Statics – Resultant of Coplanar Force Systems

Chapter Goals

• Define Force System: any set of forces treated collectively.

• Define Equivalent Force Systems: two force systems that produce the same mechanical effect on a body.

• Define Resultant: a single force that is mechanically equivalent to a given force system.

• Learning objective: calculate resultants for planar (coplanar) force systems and show equivalence.

Vector Representation

Anatomy of a Force Vector

1. Arrow AB represents the line of action.

2. Length of AB (to scale) = magnitude of the force.

3. Direction specified by angle $\theta$ measured counter-clockwise from the positive $x$-axis (standard position).

4. Arrowhead gives the sense (push/pull orientation).

5. Reference axes $x$ and $y$ are placed at the point of application A (tail).

Equal & Negative Vectors

• Equal Vectors: Same magnitude AND same direction; lines of action may differ.

• Negative Vectors: Same magnitude, opposite directions.

Resultant of Concurrent Forces (Graphical Concepts)

Adding Vectors

• The resultant is the geometric (not algebraic) sum.

• Three classical construction rules:

  1. Parallelogram Rule

  • Place tails together, construct opposite sides parallel, diagonal from common tail is $\mathbf R = \mathbf v + \mathbf w$.

  • Properties exploited: opposite sides equal, interior angle relations $(2B+2C)=360^{\circ}$, $\angle A + \angle B = 180^{\circ}$, $\angle A = \angle C$.

  1. Triangle Rule

  • Tip-to-tail of two vectors; resultant spans from first tail to final tip.

  • A “force triangle” visually represents the three forces in equilibrium.

  1. Polygon Rule

  • Generalization for 3+ vectors; tip-to-tail chain forms a polygon. Resultant joins first tail to final tip.

• Commutative: $\mathbf v + \mathbf w = \mathbf w + \mathbf v$ for any construction.

Subtracting Vectors

• To subtract $\mathbf w$ from $\mathbf v$: add the negative, $\mathbf R = \mathbf v + (-\mathbf w)$.

Analytical (Trigonometric) Resultants for Two Forces

• Graphical scale/protractor method is discouraged (“too inaccurate”).

• Adopt trigonometric laws:

  • Law of Cosines: $R^{2}=F<em>{1}^{2}+F</em>{2}^{2}-2F<em>{1}F</em>{2}\cos\beta$ (where $\beta$ is included angle between $F<em>1$ and $F</em>2$).

  • Law of Sines: $\dfrac{\sin\alpha}{F<em>2}=\dfrac{\sin\beta}{R}=\dfrac{\sin\gamma}{F</em>1}.$

Worked Example 1 — Hook With Two Forces

Given $F<em>1=54\text{ N}$ and $F</em>2=60\text{ N}$ at $60^{\circ}$ between them.

1. Draw triangle.

2. Magnitude: $R=98.8\text{ N}$ (Law of Cosines).

3. Direction: $28.3^{\circ}$ from $F_1$ (Law of Sines).

Worked Example 2 — I-Beam Supported by Two Ropes

Weight $W=1\text{ kN}$ acts vertically downward; resultant of two unknown rope tensions $F<em>1, F</em>2$ must equal $1\text{ kN}$ when included angle $\theta=30^{\circ}$. Using sine law on the force triangle yields $F<em>1$ and $F</em>2$ (values derived in slides but not numerically listed).

Exercise 1 (Solved)

Forces: $F<em>1=85\text{ N}$, $F</em>2=150\text{ N}$, angle $30^{\circ}$ between.

• $R=87.4\text{ N}$, direction $120.9^{\circ}$ from $F_1$ (obtained with Law of Cosines, not Sines because angle > $90^{\circ}$).

Exercise 2 (Solved)

Forces: $F<em>1=80\text{ kN}$ at $10^{\circ}$ above +x, $F</em>2=65\text{ kN}$ at $25^{\circ}$ below +x.

• Included angle $35^{\circ}$.

• Resultant $R=45.9\text{ kN}$ at $64.3^{\circ}$ above +x.

Components & Resolution

Definitions

• Components of a Resultant: Individual forces that collectively equal the resultant.

• Rectangular Components: Two perpendicular components usually aligned with axes $x$ (horizontal) and $y$ (vertical).

Why Use Rectangular Components?

• Force effect in each direction is independent (principle of superposition). Problems decompose into two 1-D problems.

Calculating Components (Standard Position)

For a force $F$ making $\theta$ from +x axis:

• $F_x = F\cos\theta$

• $F_y = F\sin\theta$

Reference-Angle Method

When $F$ is not in first quadrant:

1. Determine reference angle $\alpha$ (acute angle between vector and x-axis):

   • Region I: $\alpha = \theta$

   • Region II: $\alpha = 180^{\circ}-\theta$

   • Region III: $\alpha = \theta - 180^{\circ}$

   • Region IV: $\alpha = 360^{\circ}-\theta$

2. Compute magnitude with $\cos \alpha , \sin \alpha$; assign signs by inspection.

Example 3 — Ring With 60 N Force

Method 1 (standard): $\theta=300^{\circ}$ ➔ $F<em>x = +30.0\text{ N}, F</em>y=-52.0\text{ N}$.
Method 2 (reference angle $\alpha=60^{\circ}$): same magnitudes, signs decided by quadrant.

Example 4 — Skater on Inclined Rail

Rail slope 1 vertical to 2 horizontal.

1. $\tan\alpha=2/1 \Rightarrow \alpha=63.43^{\circ}$.

2. Components along rail (x): $W_x = 150\cos63.43^{\circ}=\pm67.1\text{ lb}$.

3. Normal to rail (y): $W_y = 150\sin63.43^{\circ}=\pm134.2\text{ lb}$ (sign depends on chosen rectilinear axis).

Exercise 3 (Solved) — Resolve 75 N at 48°

• $F<em>x = 50.2\text{ N}$, $F</em>y = 55.7\text{ N}$ (signs per quadrant: presentation shows both possibilities).

Re-composition From Components

Given $F<em>x, F</em>y$:

1. Magnitude: $F = \sqrt{F<em>x^{2}+F</em>y^{2}}$.

2. Reference angle: $\alpha=\tan^{-1}\left|\dfrac{F<em>y}{F</em>x}\right|$.

3. Determine quadrant from signs to obtain $\theta$.

Example 5 — Given $F<em>x=-450\text{ N}, F</em>y=-300\text{ N}$

• $F=541\text{ N}$.

• $\alpha=33.7^{\circ}$.

• Vector lies in III (both negative) → $\theta=180^{\circ}+33.7^{\circ}=214^{\circ}$.

Exercise 4 (Solved) — $F<em>x=125\text{ N}, F</em>y=-288\text{ N}$

• $F=314\text{ N}$, $\alpha=66.5^{\circ}$, quadrant IV → $\theta=-66.5^{\circ}\; (\text{or }293.5^{\circ})$.

Resultant by Rectangular Components (Many Forces)

Procedure for $n$ concurrent coplanar forces:

1. Resolve each $F<em>i$ into $F</em>{ix},F_{iy}$.

2. $R<em>x = \sum F</em>{ix}$ (algebraic sum).

3. $R<em>y = \sum F</em>{iy}$.

4. Magnitude $R = \sqrt{R<em>x^{2}+R</em>y^{2}}$.

5. Direction: $\alpha = \tan^{-1}\left|\dfrac{R<em>y}{R</em>x}\right|$ then quadrant via signs.

Example 6 — Two Forces via Components

• $F<em>1 = 3\text{ kN}, \theta</em>1=32^{\circ}$, $F<em>2 = 1.8\text{ kN}, \theta</em>2=105^{\circ}$.

• $F<em>{1x}=+2544\text{ N}, F</em>{1y}=+1590\text{ N}$.

• $F<em>{2x}=-466\text{ N}, F</em>{2y}=+1739\text{ N}$.

• $R<em>x=2078\text{ N}, \, R</em>y=3329\text{ N}$.

• $R=3920\text{ N}$, $\theta=58^{\circ}$ above +x.

Exercise 5A (Solved)

• $F<em>1=150\text{ N}@28^{\circ}$, $F</em>2=100\text{ N}@99^{\circ}$.

• $R=206\text{ N}$ at $55.4^{\circ}$.

Exercise 5B (Assigned)

Three forces: $F<em>1=100\text{ N}$ @ $20^{\circ}$, $F</em>2=200\text{ N}$ @ $40^{\circ}$, $F_3=300\text{ N}$ @ $60^{\circ}$. (Solution not provided in transcript; follow 4-step algorithm.)

Additional Observations & Best Practices

• Always sketch vectors and indicate angles clearly to avoid sign mistakes.

• Prefer rectangular components for >2 forces or when equilibrium equations are required.

• Use Law of Cosines for obtuse included angles; Law of Sines may yield ambiguous results for >90^{\circ}.

• In design/analysis, accurate resolution of forces ensures structures meet safety and serviceability criteria.

• Ethical imperative: mis-calculated resultants can cause catastrophic structural failure, so meticulous step-by-step checking is critical.

Quick Reference Formulas

1. Vector addition (triangle): $\mathbf R = \mathbf v + \mathbf w$.

2. Parallelogram diagonal length (Law of Cosines): $R^{2}=v^{2}+w^{2}-2vw\cos\beta$.

3. Rectangular components: $F<em>x = F\cos\theta$, $F</em>y = F\sin\theta$.

4. Resultant from components: $R = \sqrt{R<em>x^{2}+R</em>y^{2}}$.

5. Direction: $\theta = \begin{cases} \alpha &amp; (\text{I})\ 180^{\circ}-\alpha &amp; (\text{II})\ 180^{\circ}+\alpha &amp; (\text{III})\ 360^{\circ}-\alpha &amp; (\text{IV}) \end{cases}$ with $\alpha=\tan^{-1}|R<em>y/R</em>x|$.

End of Study Notes