4. Mechanics

Choose the particle/body, replace it by a point or simple outline, inspect every external interaction, draw one labelled arrow per force from the body, and include angles needed for later resolution.

A particle resting on a rough inclined plane may have weight $mg$ downward, normal reaction perpendicular to the plane and friction along the plane. Motion or acceleration is not an extra force.

Draw only forces acting on the selected body. A force that this body exerts on something else belongs on the other body’s diagram.

Resolve every force along the same perpendicular axes, add signed components to $(R_x,R_y)$, then calculate $R=\sqrt{R_x^2+R_y^2}$ and determine its quadrant-correct direction.

Choose axes along an incline or along a force when that reduces unknown components. A negative component means opposite to the chosen positive direction.

Include all external forces, use geometry to express angles, and solve the component equations together. A zero horizontal resultant alone does not ensure equilibrium.

A weight supported by two symmetric strings has equal tensions; horizontal components cancel and vertical components sum to the weight.

The normal component pushes away from the surface. Friction opposes actual relative motion or the tendency/impending relative motion between the contacting surfaces.

On an incline, draw $R$ perpendicular to the plane and $F$ along it. Weight remains vertical; it is not one of the contact components.

Remove friction from the free-body diagram, keep weight and all other forces, choose axes along/perpendicular to the surface, and solve with the remaining reaction.

For a particle on a smooth incline, $R$ is perpendicular to the plane and the component $mg\sin\theta$ acts down the plane; the model predicts no contact resistance along the plane.

Real surfaces generally have friction. The smooth model is unsuitable when tangential resistance, sticking, impending slip or energy loss materially affects the result.

Identify the impending motion, draw friction opposite that tendency, resolve equilibrium to find $F$ and $R$, then use equality only if the wording indicates limiting equilibrium; otherwise verify $F\le\mu R$.

“About to slip”, “on the point of moving” and “limiting equilibrium” all signal maximum static friction. If the tendency reverses, friction direction reverses.

The ground pushes upward on a particle with normal reaction $R$; the particle pushes downward on the ground with force $R$. These arrows belong on different free-body diagrams.

Weight is Earth pulling the particle; its partner is the particle pulling Earth. The normal reaction is therefore not the third-law partner of weight.

Choose one positive direction before calculating. Negative displacement/velocity/acceleration means opposite to that direction, not an invalid value.

A particle moves from $x=0$ to $x=5$ m then to $x=2$ m. Distance $=5+3=8$ m; displacement $=+2$ m. If the return takes $1$ s, its velocity then is negative while speed is positive.

A secant gradient gives average velocity/acceleration over an interval; a tangent gives instantaneous value. Horizontal $s$–$t$ means rest; horizontal $v$–$t$ means zero acceleration.

Signed area above minus area below the time axis gives displacement. Total distance is the sum of absolute areas, splitting at every $v=0$ direction change.

A positive/negative $s$–$t$ slope shows positive/negative velocity. On a $v$–$t$ graph, crossing the axis may reverse direction; increasing/decreasing height alone must be interpreted with sign.

Differentiate when moving right in the chain $s\to v\to a$; integrate when moving left. Use stated values such as $v(t_0)$ or $s(t_0)$ to determine each constant, preserving one sign convention.

Solve $v=0$ for rest/possible direction changes; test the sign of $v$ around each root. For total distance, evaluate displacement changes separately on intervals of constant velocity sign.

Choose a positive direction, assign signed $u,v,a,s$, list known/required quantities, select a formula with no extra unknown, solve and check time/position against the event wording.

From $u=5$ m s$^{-1}$ to $v=17$ m s$^{-1}$ in $4$ s, $a=3$ m s$^{-2}$ and $s=\tfrac12(5+17)4=44$ m.

For different particles, write a separate equation for each using the same time origin and coordinate line. At a meeting, their positions are equal—not necessarily their displacements from different starting points. Other events may share time but not velocity.

Choose a positive direction once. Momentum is positive or negative according to velocity; reversing motion reverses momentum without changing mass.

Momentum is a vector quantity restricted here to one dimension, so it is represented by a signed scalar. Its magnitude is $m$ times speed.

Define the two-body system and positive direction, label velocities immediately before/after, write one signed momentum equation, include any stated relation between final velocities, solve and interpret a negative result as opposite to the assumed direction.

Isolate one particle, draw all forces acting on it, choose a positive line/axes, resolve signed forces, write one $\sum F=ma$ equation per required direction, then solve and interpret any negative acceleration.

Include friction, string tension, rod thrust and weight/reaction when they act. Other resistance such as air resistance is included only when the question states it.

Use $m$ on the right of $F=ma$ and include $mg$ as one force in the free-body equation. A normal reaction or scale reading is a separate contact force and need not equal $mg$ during acceleration or on an incline.

For each phase, draw forces, choose a positive direction, resolve $\sum F=ma$ to obtain a constant $a$, then use constant-acceleration formulae only within that phase. Start a new phase when motion reverses or a force changes.

For vertical free motion with negligible resistance, acceleration is $g$ downward. If upward is positive, $a=-g$ during both ascent and descent; velocity changes sign at the highest point.

Draw a separate free-body diagram and $\sum F=ma$ equation for each particle. Choose compatible positive directions, apply the kinematic connector constraint, then solve simultaneous equations for acceleration and internal force.

A whole-system equation may eliminate internal tension/thrust, but a separate-body equation is still needed when the connector force is required.

Use displacement of the force’s point of application and the angle between the two directions. Resolve first if that makes the along-motion component clearer.

Potential-energy zero is chosen conveniently; only differences matter. Keep joules and use vertical height, including on a curved path.

Work by resistance/friction is negative and must be included, or represented as energy dissipated on the loss side. Driving work is positive when along motion.

Energy depends on initial/final speed and height, not path shape. A child moving on a smooth curved slide can be solved from overall height change without resolving forces along every segment.

If force is not parallel to motion, only its along-velocity component contributes: $P=Fv\cos\theta$. The syllabus $P=Fv$ form assumes force acts in the direction of motion.

At non-zero speed, driving force from known power is $F=P/v$; this feeds the next objective’s force equation.

Convert power units to watts, identify the instantaneous speed, compute $D=P/v$, draw/resolve slope forces with one positive direction, then calculate $a$ and interpret its sign as velocity change—not automatically speed change.

If motion is downhill, reassign signs: the component of weight along the slope aids motion while resistance opposes it. Re-derive rather than memorising one sign pattern.