4. Mechanics

Force has magnitude and direction. Resolve it into perpendicular components, add components to find the resultant, and use Newton’s law or equilibrium equations along each axis.

Draw a force diagram, choose positive directions, and keep units consistent. Equal and opposite forces act on different bodies and should not be cancelled across a single free-body diagram.

Forces (3,4) N and (−1,2) N have resultant (2,6) N with magnitude √40 N.

A scalar magnitude alone cannot determine a resultant; direction is part of the force.

Forces combine by vector addition, so their lines of action and directions matter. A force can be resolved into components along convenient axes without changing its physical effect.

Use a consistent axis system, resolve oblique forces with sine/cosine, and interpret a negative component as opposite to the chosen positive direction.

Two 10 N forces at right angles have resultant 10√2 N, not 20 N; two opposite 10 N forces have resultant zero.

Adding magnitudes ignores angle and can overestimate the resultant.

A particle is in equilibrium when ΣF=0. Resolve horizontally and vertically (or along chosen axes), giving one scalar equation per independent 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.

Equilibrium does not mean no forces act; it means their vector sum is zero.

Friction acts along the contact surface and opposes relative motion or its tendency. In limiting equilibrium F=μR; otherwise F≤μR.

Draw the normal reaction and friction separately, identify the possible direction of motion, and solve force/moment equations before checking the friction limit.

A block on a rough incline has friction up the slope if it would otherwise slide down; the limiting value μR determines the largest angle for rest.

Friction is not always μR: that equality applies at limiting friction, not every static situation.

“Smooth” means the contact force is perpendicular to the surface. There is no friction component along the surface, so the unknown contact force is a single normal reaction.

Use the geometry of the surface to choose the normal direction, and do not add a friction force to a smooth-contact free-body diagram.

A particle on a smooth plane has a reaction normal to the plane and weight vertically downward; resolving these gives the motion along the plane.

Smooth does not mean weightless or force-free; it removes friction only.

A particle is in equilibrium when the resultant force is zero, so the sums of components along two independent axes both vanish.

Choose axes that simplify the geometry, include all applied forces and reactions, then solve simultaneously. A negative unknown indicates the assumed direction was opposite.

For a ring held by two strings, horizontal components cancel and vertical components add to the load; the ring remains at rest.

Equal forces are not required—different magnitudes can balance when their directions differ.

Newton’s second law is ΣF=ma for a chosen body and inertial frame. First-law equilibrium is the special case a=0; the third law pairs equal opposite forces on different bodies.

Draw one free-body diagram at a time, choose axes, and distinguish action–reaction pairs from forces that act on the same object.

A 4 kg block with resultant horizontal force 12 N has acceleration 3 m s⁻²; the reaction and weight are separate vertical forces.

The third-law partner of a normal reaction acts on the other body, so it cannot cancel the weight of the same body.

|z−a|=r describes a circle, arg(z−a)=θ a ray, and |z−a|=|z−b| the perpendicular bisector of AB. Each condition is a set of allowed points.

Sketch reference points first, then apply equality or inequality signs. For multiple conditions, keep only their intersection and note excluded endpoints or rays.

|z−(1+i)|=2 is a circle centred at (1,1) with radius 2.

The modulus condition gives distance, while an argument condition gives direction; confusing them changes the locus completely.

|z−a|<r selects the interior of a circle and |z−a|>r its exterior. An argument inequality selects a wedge, with boundary rays included or excluded according to the sign.

Test one convenient point to decide which side of each boundary is wanted, then combine conditions. State whether boundaries are included.

|z|≤2 and arg z between 0 and π/2 gives a closed quarter-disk in the first quadrant.

Sketching only the boundary is incomplete when an inequality describes a region.

A condition such as |z−a|=|z−b| gives a perpendicular bisector, while |z−a|=r gives a circle and arg(z−a)=θ gives a ray. Multiple conditions mean an intersection of sets.

Translate each condition separately, sketch boundaries, and test a point to select the correct side for inequalities.

|z−1|=|z+1| is the imaginary axis: points equidistant from 1 and −1 have real part zero.

The equal-distance locus is not the line joining the two points; it is perpendicular to that line.

To find where a line meets a locus, parameterise the line, substitute its coordinates into the modulus or argument condition, and solve for permitted parameter values.

Check the parameter interval and retain only points satisfying every original condition. A quadratic parameter may represent two, one or no intersections.

Substituting z=1+iλ into |z|=√2 gives 1+λ²=2, so λ=±1 and two points on the circle.

Solving the squared modulus equation can introduce a sign or domain issue; verify in the unsquared condition.

Total momentum p=mv is a vector. If external impulse is negligible, Σm u=Σm v before and after a collision or separation.

Choose a positive direction, keep signs on velocities, and define the system before writing the conservation equation.

A 2 kg cart at 4 m s⁻¹ and a 1 kg cart at −1 m s⁻¹ have total momentum 7 kg m s⁻¹ before impact.

Momentum conservation does not imply kinetic-energy conservation; inelastic impacts can dissipate energy.

Impulse J=∫Fdt equals change in momentum Δp. For a constant force, J=FΔt; for a varying force, use the signed area under the force–time graph.

Choose the positive direction and distinguish peak force from average force. The impulse includes direction and can be negative.

A triangular force pulse of base 0.20 s and height 50 N gives impulse 5 N s.

A large peak force over a very short time may give less impulse than a smaller sustained force; peak force is not momentum change.

For a body of mass m, ΣF=ma. The acceleration points in the direction of the resultant, not necessarily in the direction of the largest individual force.

Draw a free-body diagram, resolve along convenient axes and include friction, tension, weight and reactions only when they act on the chosen body.

A 5 kg block with 20 N right and 8 N left has resultant 12 N right and acceleration 2.4 m s⁻² right.

Action–reaction pairs belong to different bodies and cannot be cancelled inside one body’s force equation.

Mass m is measured in kilograms and remains the same in a given object; weight W=mg is a force in newtons and depends on local gravitational field strength g.

Use mass in F=ma and weight as a force in vertical equations. A scale reading is a contact force and may differ from mg during acceleration.

A 3 kg object weighs about 29.4 N where g=9.8 m s⁻²; its mass is still 3 kg on the Moon.

Mass and weight are not interchangeable, and “weighing zero” in free fall does not mean mass disappears.

For constant acceleration, v=u+at, s=ut+½at², v²=u²+2as and s=½(u+v)t. Choose equations containing the known quantities.

State the positive direction and use signed displacement, velocity and acceleration consistently. If acceleration changes, split the motion or use a different method.

From rest with a=3 m s⁻² for 4 s, v=12 m s⁻¹ and s=24 m.

These equations are not valid for variable acceleration, and distance is not always equal to displacement.

For linked particles, an inextensible string over a smooth pulley gives equal magnitude accelerations and tension throughout the string. Newton’s second law is written separately for each mass.

Draw one diagram per particle, choose a common positive direction and use the string constraint only after writing the force equations.

A hanging mass and a block on a smooth table have the same acceleration; their combined equations can eliminate the internal tension.

Equal tension does not mean equal net force or equal mass; the particles can accelerate together with different forces.

a·b=a₁b₁+a₂b₂=|a||b|cosθ. It is zero for perpendicular vectors and gives the component of one vector along another after division by the reference magnitude.

Check both vectors are non-zero, use the principal angle and preserve units when a physical projection is requested.

For a=(2,1) and b=(1,2), a·b=4 and cosθ=4/5.

The dot product is not |a||b| without the cosine factor and is not a vector direction.

Near Earth’s surface, gravitational potential energy change is ΔE=mgh relative to a chosen reference level. Only differences matter, so the zero level can be selected for convenience.

Use energy conservation when no non-conservative work acts, and include kinetic and elastic terms as needed. Far from Earth, the uniform-field approximation may fail.

Raising 2 kg by 3 m where g=9.8 gives an increase of 58.8 J.

Potential energy itself is reference-dependent; a negative value is not automatically an error.

Forces arise from interactions such as weight, contact, tension, friction or thrust. A complete free-body diagram lets ΣF=ma be applied without double-counting.

Name the body, draw force directions and distinguish applied forces from resultant shorthand. Resolve only after the physical forces are identified.

For a block pulled by a rope on a rough plane, include weight, normal reaction, tension and friction before resolving along the plane.

Centripetal force is not an extra force; it is the name for the resultant inward force in circular motion.

Write z=x+iy and translate modulus or argument conditions into equations or inequalities in x and y. The resulting curve should match the geometric interpretation.

For |z−a|=r, expand to a circle; for equal distances, subtract squared distances to obtain a line. Check restrictions introduced by squaring or arguments.

|z−(1+i)|=|z−(−1+i)| simplifies to x=0, the vertical bisector of the two centres.

Squaring distances is safe for non-negative moduli, but argument equations still need branch and quadrant checks.

Acceleration is dv/dt, the instantaneous rate of change of velocity. Since velocity is ds/dt, acceleration is d²s/dt².

Keep signs and units consistent, and distinguish average acceleration Δv/Δt from the derivative at one instant.

If v(t)=t²−4t, then a(t)=2t−4; at t=3 the acceleration is 2 m s⁻².

Zero velocity does not imply zero acceleration: an object can be momentarily at rest while its velocity changes.