Forces and equilibrium: vectors, resultants, equilibrium of a particle

Cambridge IGCSE/A‑Level Mathematics 9709 – Mechanics (M1) – Revised Notes

4.1 Forces and Vectors

1. Vector fundamentals

• A vector has a magnitude and a direction.
• Component (Cartesian) form
   $$\mathbf{v}= \begin{pmatrix}v_x\\ v_y\end{pmatrix}=v_x\mathbf{i}+v_y\mathbf{j}$$
• Magnitude: $$|\mathbf{v}|=\sqrt{v_x^{2}+v_y^{2}}$$
• Direction (angle $\theta$ measured anticlockwise from the positive $x$‑axis): $$\theta=\tan^{-1}\!\frac{v_y}{v_x}$$
• Basic operations
  • Addition: $\displaystyle\mathbf{u}+\mathbf{v}=(u_x+v_x)\mathbf{i}+(u_y+v_y)\mathbf{j}$
  • Subtraction: $\displaystyle\mathbf{u}-\mathbf{v}=(u_x-v_x)\mathbf{i}+(u_y-v_y)\mathbf{j}$
  • Scalar multiplication: $\displaystyle k\mathbf{v}= (kv_x)\mathbf{i}+(kv_y)\mathbf{j}$

2. Resolving a force into components

For a force $F$ acting at an angle $\alpha$ from the positive $x$‑axis:

$$F_x = F\cos\alpha,\qquad F_y = F\sin\alpha$$

Use the sign convention of the chosen coordinate system (right/up positive).

3. Resultant of several forces

1. Component method – resolve each force into $x$ and $y$ components, sum the components, then recombine into a single vector.
2. Parallelogram (or triangle) rule – place the vectors tip‑to‑tail; the diagonal of the parallelogram (or the closing side of the triangle) is the resultant.

Force	Magnitude (N)	Direction (° from + $x$)	$x$‑component (N)	$y$‑component (N)
$\mathbf{F}_1$	40	0	40	0
$\mathbf{F}_2$	30	120	$30\cos120^\circ=-15$	$30\sin120^\circ=25.98$
$\mathbf{F}_3$	50	210	$50\cos210^\circ=-43.30$	$50\sin210^\circ=-25.00$
Resultant $\mathbf{R}$			$40-15-43.30=-18.30$	$0+25.98-25.00=0.98$

Resultant magnitude and direction

$$|\mathbf{R}|=\sqrt{(-18.30)^2+(0.98)^2}\approx18.33\text{ N}$$ $$\theta_R=\tan^{-1}\!\left(\frac{0.98}{-18.30}\right)\approx177^\circ\quad(\text{from }+x)$$

4. Newton’s Laws of Motion (Particle level)

1. First law (inertia) – a particle remains at rest or moves with constant velocity unless acted on by a net external force.
2. Second law – the vector sum of the forces on a particle equals the mass times its acceleration: $$\sum\mathbf{F}=m\mathbf{a}$$ In component form: $\displaystyle\sum F_x=m a_x,\;\;\sum F_y=m a_y$.
3. Third law – for every action force there is an equal and opposite reaction force acting on a different body.

5. Equilibrium of a particle

A particle is in equilibrium when its acceleration is zero, i.e. the resultant force is zero.

$$\sum\mathbf{F}= \mathbf{0}\;\Longleftrightarrow\;\begin{cases} \sum F_x =0\\[2mm] \sum F_y =0 \end{cases}$$

For planar problems a third equation is often obtained from moments about a convenient point $O$:

$$\sum M_O =0\qquad\text{where}\qquad M_O =F\,d_{\perp}$$

• $d_{\perp}$ = perpendicular distance from the line of action of the force to point $O$.
• Take moments anticlockwise as positive (or adopt any consistent sign convention).

6. Limiting friction and smooth contact

• On a rough surface the contact reaction is split into a normal component $R_N$ (perpendicular to the surface) and a frictional component $F$ (parallel to the surface).
• Limiting (maximum) static friction: $$F_{\max}= \mu_s R_N$$ where $\mu_s$ is the coefficient of static friction.
• Kinetic (sliding) friction: $$F_k= \mu_k R_N$$ where $\mu_k$ is the coefficient of kinetic friction.
• For a smooth (frictionless) contact $F=0$ and only $R_N$ acts.

7. Triangle of forces & Lami’s theorem (optional)

• If three concurrent forces are in equilibrium they can be represented by the sides of a closed triangle (triangle of forces).
• Lami’s theorem: for three concurrent forces $F_1,F_2,F_3$ in equilibrium, $$\frac{F_1}{\sin\alpha}= \frac{F_2}{\sin\beta}= \frac{F_3}{\sin\gamma}$$ where $\alpha,\beta,\gamma$ are the angles opposite the respective forces.

8. Worked example – Block on a rough inclined plane

A block of mass $m=5\,$kg rests on a $30^\circ$ incline. Coefficient of static friction $\mu_s=0.4$.

1. Weight $W=mg=5\times10=50\,$N.
2. Resolve $W$:
   • Parallel to the plane: $W_{\parallel}=W\sin30^\circ=25\,$N (down the slope).
   • Perpendicular to the plane: $W_{\perp}=W\cos30^\circ=43.3\,$N.
3. Normal reaction $R_N=W_{\perp}=43.3\,$N.
4. Maximum static friction $F_{\max}= \mu_sR_N=0.4\times43.3=17.3\,$N (up the slope).
5. Since $W_{\parallel}=25\,$N $>F_{\max}$, the block will start to slip.
6. Net force down the plane: $F_{\text{net}}=W_{\parallel}-F_{\max}=7.7\,$N.
7. Resulting acceleration: $a=F_{\text{net}}/m=7.7/5=1.54\,$m s⁻² down the slope.

9. Checklist – Solving equilibrium problems

• Draw a clear free‑body diagram (FBD) and label every force.
• Choose a convenient coordinate system (usually along and perpendicular to a plane).
• Resolve each force into components (or use normal + friction components).
• Write the equilibrium equations:
  • $\displaystyle\sum F_x=0$
  • $\displaystyle\sum F_y=0$
  • Optional: $\displaystyle\sum M_O=0$ for an extra equation.
• Solve the simultaneous equations for the unknowns.
• Check that the resultant of all forces is zero (or that the moment sum vanishes).

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4.2 Kinematics of Motion in a Straight Line

1. Key definitions

• Displacement $s$ – vector change in position; sign indicates direction.
• Velocity $v$ – rate of change of displacement; $v=\dfrac{ds}{dt}$.
• Speed $|v|$ – magnitude of velocity (scalar).
• Acceleration $a$ – rate of change of velocity; $a=\dfrac{dv}{dt}$.

2. Constant‑acceleration equations (the “SUVAT” set)

$$\begin{aligned} v &= u + at \\[2mm] s &= ut + \tfrac12 at^{2} \\[2mm] v^{2} &= u^{2}+2as \end{aligned}$$

where $u$ = initial velocity, $v$ = final velocity, $a$ = (constant) acceleration, $t$ = time, $s$ = displacement.

3. Graphical interpretation

• $s$–$t$ graph: slope at any point = instantaneous velocity.
• $v$–$t$ graph: slope = instantaneous acceleration; area under the curve = displacement.
• $a$–$t$ graph: area under the curve = change in velocity.

4. Calculus approach (A‑Level requirement)

• Given $a(t)$, integrate to obtain velocity: $$v(t)=\int a(t)\,dt + C_1$$
• Integrate $v(t)$ to obtain displacement: $$s(t)=\int v(t)\,dt + C_2$$
• Constants $C_1$, $C_2$ are determined from the initial conditions (usually $v(0)=u$, $s(0)=0$).

5. Worked example – Particle thrown vertically upwards

Initial speed $u=20\,$m s⁻¹, $g=10\,$m s⁻² (downward). Take upward as positive.

1. Acceleration: $a=-g=-10\,$m s⁻².
2. Time to highest point ($v=0$): $0=u+at\;\Rightarrow\;t=\dfrac{-u}{a}=2\,$s.
3. Maximum height: $s=ut+\tfrac12at^{2}=20(2)+\tfrac12(-10)(2)^{2}=40-20=20\,$m.
4. Total time of flight (up + down) $=2t=4\,$s.

6. Worked example – Motion down a smooth $15^\circ$ incline

Particle starts from rest, slides down an incline of length $L=5\,$m.

1. Component of gravity along the plane: $a=g\sin15^\circ=10\sin15^\circ=2.59\,$m s⁻².
2. Using $v^{2}=u^{2}+2as$ with $u=0$, $s=5\,$m: $$v=\sqrt{2(2.59)(5)}\approx5.1\,$m s⁻¹.

7. Checklist – Straight‑line kinematics

• State a clear sign convention (choose a positive direction).
• Identify whether acceleration is constant; if not, write $a(t)$.
• Select the appropriate SUVAT equation(s) or integrate if $a$ varies.
• Check units, and verify that the answer is physically reasonable (e.g., speed cannot be negative when a magnitude is required).

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4.3 Momentum, Impulse & Collisions

1. Linear momentum

$$\mathbf{p}=m\mathbf{v}$$

Momentum is a vector; its direction is the same as the velocity.

2. Impulse

$$\mathbf{J}=\int \mathbf{F}\,dt = \Delta\mathbf{p}$$

For a constant force $F$ acting for a time $\Delta t$: $\displaystyle J=F\Delta t$.

3. Conservation of momentum

In an isolated system (no external net force) the total momentum remains constant:

$$\sum\mathbf{p}_{\text{initial}} = \sum\mathbf{p}_{\text{final}}$$

4. One‑dimensional collisions

• Elastic collision – both momentum and kinetic energy are conserved.
• Perfectly inelastic collision – bodies stick together; only momentum is conserved.

5. Example – Elastic collision of two carts

Cart A: $m_A=2\,$kg, $u_A=3\,$m s⁻¹ (right).
Cart B: $m_B=3\,$kg, $u_B=0\,$m s⁻¹.

1. Momentum: $2(3)+3(0)=2v_A+3v_B\;\Rightarrow\;6=2v_A+3v_B$.
2. Kinetic energy: $\tfrac12(2)(3)^2=\tfrac12(2)v_A^{2}+\tfrac12(3)v_B^{2}\;\Rightarrow\;9=v_A^{2}+1.5v_B^{2}$.
3. Solving gives $v_A=1\,$m s⁻¹ (right) and $v_B=\dfrac{4}{3}\,$m s⁻¹ (right).

6. Example – Perfectly inelastic collision (bullet embeds in block)

Bullet: $m_b=0.02\,$kg, $u_b=300\,$m s⁻¹.
Block: $m_B=2\,$kg, $u_B=0\,$m s⁻¹.
After impact the bullet sticks in the block.

1. Conserve momentum: $$0.02(300)+2(0)= (0.02+2)v\;\Rightarrow\;v=\frac{0.02\times300}{2.02}\approx2.97\text{ m s}^{-1}.$$
2. Kinetic energy is not conserved – the loss appears as heat, deformation, etc.

7. Checklist – Momentum problems

• Draw a clear FBD for each object before and after the interaction.
• Choose a consistent sign convention (e.g., right = positive).
• Write the momentum conservation equation(s). Include kinetic‑energy equation only for elastic collisions.
• Solve for the required unknown(s) and, where appropriate, check that kinetic‑energy values satisfy the elastic‑collision condition.

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4.4 Summary of Key Formulae

Topic	Formula
Vector magnitude	$|\mathbf{v}|=\sqrt{v_x^{2}+v_y^{2}}$
Vector direction	$\theta=\tan^{-1}(v_y/v_x)$
Resultant (components)	$R_x=\sum F_x,\;R_y=\sum F_y$ $\displaystyle |\mathbf{R}|=\sqrt{R_x^{2}+R_y^{2}},\;\theta_R=\tan^{-1}(R_y/R_x)$
Moment of a force	$M=F\,d_{\perp}$ (anticlockwise = positive)
Newton’s 2nd law	$\sum\mathbf{F}=m\mathbf{a}$
Static friction	$F_{\max}= \mu_s R_N$
Kinetic friction	$F_k= \mu_k R_N$
Constant‑acceleration equations	See “SUVAT” set above
Impulse	$\mathbf{J}= \Delta\mathbf{p}= \int\mathbf{F}\,dt$
Momentum	$\mathbf{p}=m\mathbf{v}$
Elastic‑collision equations	Momentum + kinetic‑energy conservation
Perfectly inelastic	Momentum conservation only
Lami’s theorem

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4.5 How to Use These Notes Effectively

1. Read the relevant theory section first (e.g., vectors, Newton’s laws, kinematics).
2. Work through the worked examples, then attempt the practice questions at the end of your textbook.
3. When solving a new problem:
   • Identify the physical situation and draw a neat free‑body diagram.
   • Choose a convenient coordinate system and resolve all forces.
   • Write down the appropriate equilibrium, motion, or momentum equations.
   • Solve algebraically, then check units and sign.
4. Use the “Checklist” boxes as a self‑assessment tool before moving on.