Unit M1: Mechanics 1

A mechanics model replaces a complicated real object or contact with a simpler idealisation. Each modelling word is an assumption, and the assumption tells you which features may be ignored and which mathematical consequences may be used.

Translate every adjective before forming equations. For example, a block modelled as a particle may have its weight applied at one point; an inextensible string connecting two particles constrains their motions; and a smooth pulley lets the same tension be used on both sides of one continuous light string. These conclusions come from different assumptions and should not be interchanged.

A model is judged by purpose, not by whether it copies reality perfectly. Start with the stated idealisations, solve the simplified problem, then identify which omitted feature could materially change the result. Allowing string mass or extension, pulley friction, body dimensions, deformation, air resistance, measurement error, or changing acceleration can make a particular model more realistic, but only when that feature is relevant to the situation.

A vector has both magnitude and direction. In a plane, write $\mathbf v=a\mathbf i+b\mathbf j$, where the perpendicular unit vectors $\mathbf i$ and $\mathbf j$ define the positive coordinate directions. The signed components $a$ and $b$ say how much of the vector acts in each direction.

A direction angle can be found from the component triangle, but the signs of $a$ and $b$ must determine the correct quadrant. For a bearing, measure clockwise from north and give a three-figure angle. A calculator value such as $\tan^{-1}(b/a)$ gives a reference angle only; it does not by itself determine the direction.

To resolve a vector of magnitude $V$ at an angle $\theta$ to the positive $\mathbf i$ direction, use$$\mathbf v=(V\cos\theta)\mathbf i+(V\sin\theta)\mathbf j.$$Change signs to match the actual quadrant. If the angle is measured from the $\mathbf j$ direction, the sine and cosine roles interchange.

Forces $10\mathbf i+\mathbf j$ N and $-15\mathbf i+6.5\mathbf j$ N have resultant$$\mathbf R=-5\mathbf i+7.5\mathbf j\text{ N},\qquad |\mathbf R|=\sqrt{(-5)^2+7.5^2}=\frac{5\sqrt{13}}2\text{ N}.$$Its components place it north-west. The reference angle west of north satisfies $\tan\alpha=5/7.5$, so its bearing is $360^\circ-\alpha\approx326^\circ$.

Displacement, velocity, acceleration and force are vectors: each must keep its components, direction and units. The same component rules apply to all four quantities, but their meanings are different.

Use a consistent time unit before dividing or multiplying. For example, velocity may be in $\mathrm{m\,s^{-1}}$ or $\mathrm{km\,h^{-1}}$, acceleration in $\mathrm{m\,s^{-2}}$ or $\mathrm{km\,h^{-2}}$, and force in newtons. Speed is the magnitude $|\mathbf v|$; it is a scalar and has no direction.

A particle starts at $\mathbf r_0=2\mathbf i+5\mathbf j$ m and moves with constant velocity $\mathbf v=3\mathbf i-2\mathbf j$ m s$^{-1}$. After $4$ s,$$\mathbf r=(2\mathbf i+5\mathbf j)+4(3\mathbf i-2\mathbf j)=14\mathbf i-3\mathbf j\text{ m}.$$Its displacement is $12\mathbf i-8\mathbf j$ m, while the distance from its starting point is $\sqrt{12^2+(-8)^2}=4\sqrt{13}$ m.

For two moving particles, form one relative vector consistently:$$\overrightarrow{AB}(t)=\mathbf r_B(t)-\mathbf r_A(t).$$They meet only if both components are zero at the same time. Their separation is $|\overrightarrow{AB}(t)|$; minimising its square gives the same closest time without an unnecessary square root.

If velocity changes from $-\mathbf i+4\mathbf j$ to $5\mathbf i-8\mathbf j$ m s$^{-1}$ in $3$ s at constant acceleration, then$$\mathbf a=\frac{(5\mathbf i-8\mathbf j)-(-\mathbf i+4\mathbf j)}3=2\mathbf i-4\mathbf j\text{ m s}^{-2}.$$Subtraction order fixes the direction of the change.

For motion in one straight line with constant acceleration, choose one positive direction and keep every velocity, acceleration and displacement signed. Use $u$ for velocity at the start of the interval, $v$ at the end, $a$ for the constant acceleration, $s$ for signed displacement and $t\ge0$ for elapsed time. Distance and speed are non-negative magnitudes; they are not interchangeable with $s$ and $v$.

Before substituting: (1) isolate one interval in which $a$ is constant; (2) state the positive direction; (3) list the five quantities with signs and units; (4) choose a relation containing the target and no unnecessary unknown; (5) reject roots that conflict with time, direction or the described stage. For distance travelled during the $n$th second, calculate $s(n)-s(n-1)$ within the same constant-acceleration interval.

For free vertical motion near Earth's surface, take upward as positive so $a=-g$. A particle projected upward at $14.7\,\mathrm{m\,s^{-1}}$ reaches instantaneous rest when$$0=14.7-9.8t,$$so $t=1.5$ s. Its greatest displacement above projection is$$s=14.7(1.5)-\frac12(9.8)(1.5)^2=11.025\text{ m}.$$After the top, velocity is negative while speed is positive. At a return to the projection level, $s=0$ even though the total distance travelled is twice the greatest height.

On a velocity-time graph, equal positions occur when the signed areas accumulated from the same starting time are equal; catching up means equal displacements, not equal velocities. On a speed-time graph, all area contributes positively to distance. Vertical segments would represent an instantaneous change and should appear only when the model explicitly permits one.

A force is a vector interaction that can change a particle's velocity. Draw or list only the forces acting on the chosen particle, combine them as a resultant, then apply Newton's second law in a fixed coordinate system.

Common forces include weight $m\mathbf g$ vertically downward, a normal reaction perpendicular to a contact surface, tension along a taut string, friction along a rough contact, and stated driving or resistance forces. Resolve each force along the chosen axes before adding; never insert a force merely because an object is moving.

A particle of mass $2$ kg is acted on by forces $(-2\mathbf i+3\mathbf j)$ N and $(4\mathbf i+5\mathbf j)$ N. Hence$$\sum\mathbf F=2\mathbf i+8\mathbf j\text{ N},\qquad \mathbf a=\mathbf i+4\mathbf j\text{ m s}^{-2}.$$Its acceleration magnitude is $\sqrt{17}\,\mathrm{m\,s^{-2}}$; dividing the force magnitude by mass gives the same value.

A connected-particle problem becomes manageable when the body or system is chosen before equations are written. For every stage: identify the active connections and contacts, choose a positive direction, resolve forces, apply $\sum F=ma$, then use kinematics only after the acceleration is known.

Masses $3$ kg and $2$ kg hang on opposite sides of a smooth pulley, with the $3$ kg mass moving downward. For the separate particles,$$3g-T=3a,\qquad T-2g=2a.$$Adding eliminates the internal tension: $g=5a$, so $a=g/5$. Substitution gives $T=12g/5$ N. The same $a$ must be used because the taut inextensible string constrains both motions.

Use a whole-system equation to find acceleration efficiently, then a single-particle equation to recover an internal tension or coupling force. On an incline, resolve perpendicular first when a reaction or friction is needed, then resolve parallel to the motion. Every term in an equation must be a force on the selected body and every $ma$ term must use its mass.

When a particle hits the ground, a string becomes slack or breaks, contact is lost, or an applied force changes, the force model changes immediately. End the first stage, carry forward the boundary velocity and position, draw a new force set, find the new acceleration, and begin a new constant-acceleration interval.

Linear momentum is signed in one dimension: for mass $m$ moving with velocity $v$, $p=mv$. Choose one positive direction before a collision and give every initial and final velocity its sign. Speed alone is insufficient when a particle reverses.

Impulse is the change in momentum of one particle and has units $\mathrm{N\,s}=\mathrm{kg\,m\,s^{-1}}$. The impulses two colliding particles exert on each other are equal in magnitude and opposite in direction. Report $|I|$ when a magnitude is requested.

During a short direct collision, the two internal impulses cancel when both particles are treated as one system. If external impulse is negligible over that interval, total linear momentum is conserved. This is a signed equation, not an equation between total speeds or kinetic energies.

A $2$ kg particle moving at $4\,\mathrm{m\,s^{-1}}$ collides directly with a $3$ kg particle moving at $-1\,\mathrm{m\,s^{-1}}$. If the second particle leaves at $1\,\mathrm{m\,s^{-1}}$, conservation gives$$2(4)+3(-1)=2v+3(1),$$so $v=1\,\mathrm{m\,s^{-1}}$. The impulse on the first particle is $2(1-4)=-6\,\mathrm{N\,s}$, hence magnitude $6\,\mathrm{N\,s}$.

For a particle moving along a rough surface in this syllabus, friction acts along the contact surface opposite the relative motion. Its magnitude is the coefficient of friction times the normal reaction.

The coefficient $\mu$ is dimensionless. The reaction $R$ must be found from forces perpendicular to the surface; it equals $mg$ only on a horizontal surface with no other force having a perpendicular component. A pull angled upward reduces $R$, while a push angled downward increases it.

A $5$ kg block is pulled across a horizontal floor by a $20$ N force at $30^\circ$ above the horizontal. If $\mu=0.2$, perpendicular balance gives$$R=5g-20\sin30^\circ=39\text{ N}.$$Thus $F=0.2(39)=7.8$ N. Horizontally,$$20\cos30^\circ-7.8=5a,$$so $a=1.90\,\mathrm{m\,s^{-2}}$ to three significant figures.

On an incline with no other perpendicular force, $R=mg\cos\theta$. Friction still opposes motion: it acts down the plane while the particle moves up, and up the plane while it moves down. Recalculate $R$ whenever the applied-force geometry changes.

A force has magnitude and direction, so treat it as a vector. Choose fixed perpendicular axes, resolve every force into signed components, and add corresponding components. The resultant is the single vector with the same combined effect as all the forces.

These component formulae assume that $\theta$ is measured from the positive $x$-direction. If the angle is measured from another axis, use the adjacent component with cosine and the opposite component with sine, then assign signs from the actual directions rather than from the diagram alone.

Forces of $10$ N east and $6$ N at $60^\circ$ north of west act on a particle. Taking east and north as positive,$$\mathbf R=(10-6\cos60^\circ)\mathbf i+(6\sin60^\circ)\mathbf j=7\mathbf i+3\sqrt3\mathbf j.$$Thus $|\mathbf R|=\sqrt{76}=8.72$ N. Its direction is $\tan^{-1}(3\sqrt3/7)=36.6^\circ$ north of east, equivalent to bearing $053.4^\circ$.

A particle is in equilibrium when the vector resultant of all forces acting on it is zero. It may be at rest or moving with constant velocity. Draw the force diagram for one chosen particle, choose two independent axes, and resolve the equilibrium condition along both.

Horizontal and vertical axes are often convenient, but axes parallel and perpendicular to an inclined surface can remove unnecessary components. With three non-parallel forces in equilibrium, their vectors form a closed triangle; resolving remains the safest general method. With only two forces, they must be equal, opposite and collinear.

A $4$ kg particle rests on a smooth plane inclined at $30^\circ$ to the horizontal and is held by a light string parallel to the plane. Perpendicular equilibrium gives$$R-4g\cos30^\circ=0,$$so $R=2\sqrt3g$ N. Parallel equilibrium gives$$T-4g\sin30^\circ=0,$$so $T=2g=19.6\text{ N}.$$The smooth plane contributes no friction.

After solving, verify that each unknown force has a physically possible direction and magnitude. A negative result means the assumed direction was reversed or the proposed contact cannot supply that force. A normal reaction is not automatically $mg$: it is determined by all components perpendicular to the surface.

For a particle in equilibrium on a rough surface, static friction is a response force. Its magnitude adjusts to the value needed to prevent relative motion, up to a maximum set by the coefficient of friction $\mu$ and the normal reaction $R$.

Only when the particle is on the point of sliding is friction limiting, so $F=\mu R$. Its direction opposes the impending motion. If no impending direction is stated, first solve the other equilibrium equations: the sign of the required friction reveals its direction.

A $5$ kg block rests on a horizontal rough floor. It is pulled by a force $P$ at $30^\circ$ above the horizontal and $\mu=0.4$. At impending motion to the right,$$R=5g-P\sin30^\circ,\qquad P\cos30^\circ=0.4R.$$Hence $P(\cos30^\circ+0.4\sin30^\circ)=0.4(5g)$, giving $P=18.4$ N to three significant figures. The angled pull reduces $R$, so using $R=5g$ would overestimate the limiting friction.

Questions asking for the smallest or largest applied force that preserves equilibrium usually have two limiting boundaries. At one boundary the particle is about to move in one direction; at the other it is about to move in the opposite direction, so friction reverses. Set up and solve the two cases separately, then state the permitted interval.

The moment of a force about a point measures its turning effect. It equals the force magnitude multiplied by the perpendicular distance from the point to the force's line of action. A force whose line of action passes through the point has zero moment about that point.

Choose one rotational sense as positive and use it consistently; for example, anticlockwise moments positive and clockwise moments negative. For parallel vertical forces on a horizontal body, $d_\perp$ is the horizontal separation. The distance is measured to the line of action, not along the body unless that length is perpendicular to the force.

First draw every parallel force at its correct position. Resolve vertically to relate the reactions, weights and tensions. Then take moments about a point that removes an unwanted force—usually a support whose reaction is unknown. Keep each term dimensionally as force times distance, solve, and check that all required contact reactions are non-negative.

A uniform $4$ m beam of mass $10$ kg is supported at its ends $A$ and $B$. A $6$ kg particle is attached $1$ m from $A$. If the upward reactions are $R_A$ and $R_B$, vertical equilibrium gives$$R_A+R_B=16g.$$Taking moments about $A$ gives$$4R_B=(10g)(2)+(6g)(1),$$so $R_B=6.5g$ N and $R_A=9.5g$ N. Both are positive, so both supports can remain in contact.