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4.4 Newton's laws of motion

Syllabus
9709–2028–2029
Topic
4.4
Level
A2

Learning objectives

4.4.1Forces• apply Newton's laws of motion to the linear motion of a particle of constant mass moving under the action of constant forces, which may include friction, tension in an inextensible string and thrust in a connecting rod If any other forces resisting motion are to be considered (e.g. air resistance) this will be indicated in the question.4.4.2Mass and weight• use the relationship between mass and weight W = mg. In this component, questions are mainly numerical, and use of the approximate numerical value 10 (m s-2) for g is expected.4.4.3Constant acceleration• solve simple problems which may be modelled as the motion of a particle moving vertically or on an inclined plane with constant acceleration Including, for example, motion of a particle on a rough plane where the acceleration while moving up the plane is different from the acceleration while moving down the plane.4.4.4Connected-particle models• solve simple problems which may be modelled as the motion of connected particles. e.g. particles connected by a light inextensible string passing over a smooth pulley, or a car towing a trailer by means of either a light rope or a light rigid tow-bar.

Apply Newton’s second law to the resultant on one particle

Foraconstant−massparticleinaninertialframe:For a constant-mass particle in an inertial frame:\sum F=maalongeachchosendirection.Theaccelerationfollowstheresultantforce.along each chosen direction. The acceleration follows the resultant force.

Isolate one particle, draw all forces acting on it, choose a positive line/axes, resolve signed forces, write one ∑F=ma\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.

A $5$ kg block with $20$ N right and $8$ N left has20-8=5a\Rightarrow a=2.4\text{ m s}^{-2}totheright.to the right.

Use the resultant, not the largest force. Third-law partners act on other bodies and do not enter this particle’s equation.

Convert mass to downward weight using the Mechanics value of $g$

W=mg,where mass $m$ is in kg, gravitational acceleration $g$ is in m s$^{-2}$ and weight $W$ is a downward force in N.

In this Mechanics component, use the expected approximation g=10g=10 m s−2^{-2} unless the question states another value.

A $3$ kg particle has weightW=3\times10=30\text{ N}verticallydownward.vertically downward.

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

Mass is not measured in newtons and weight is not measured in kilograms. Do not default to 9.89.8 when this component expects 1010.

Derive acceleration separately for each vertical or incline phase

For each phase, draw forces, choose a positive direction, resolve ∑F=ma\sum F=ma to obtain a constant aa, 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 gg downward. If upward is positive, a=−ga=-g during both ascent and descent; velocity changes sign at the highest point.

For a rough incline, friction opposes motion:

Motion Friction direction Typical down-slope resultant
moving up down slope mgsin⁡θ+Fmg\sin\theta+F
moving down up slope mgsin⁡θ−Fmg\sin\theta-F

If a particle slides on a rough plane with $F=\mu mg\cos\theta$, then down-slope acceleration while moving down isa=g(\sin\theta-\mu\cos\theta),providedthemodelgivesmotiondowntheplane.provided the model gives motion down the plane.

Do not carry the same friction direction or acceleration through a reversal. SUVAT is valid only while acceleration is constant.

Combine connector constraints with one force equation per particle

Connector model Shared/connector consequence
light inextensible string over smooth pulley equal acceleration magnitudes; same tension throughout
light rope towing rope carries tension (pull)
light rigid tow-bar may carry tension or thrust/compression

Draw a separate free-body diagram and ∑F=ma\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.

For masses $m_1,m_2$ hanging on a smooth pulley with $m_2>m_1$:m_2g-T=m_2a,\qquad T-m_1g=m_1a.Adding eliminates $T$ and gives $a=(m_2-m_1)g/(m_1+m_2)$.

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

Equal acceleration does not imply equal resultant force when masses differ. A rigid tow-bar force is not automatically tension.