Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 4: Mechanics · official syllabus
FORCES AND EQUILIBRIUM LAB
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Four short missions on resolving forces, resultants, equilibrium and static friction. Draw the force balance, predict the rule, then calculate and test your answer. Forces are measured in newtons.
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Mission 1 of 4
Resolve a force
A 10 N force acts at 60 degrees above the horizontal. Replace it with perpendicular components that have the same effect. The white arrow is the 10 N force; teal is your horizontal component. Set that component. The angle is measured from the horizontal, not the vertical.
For a force F at angle theta above the horizontal, its horizontal component is...
White: the 10 N force. Teal: your horizontal component. Dashed: its vertical projection.
Your horizontal component is 2.0 N. Use the side adjacent to the 60-degree angle.
Study notes
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Model the particle and draw its forces 4.1
A particle model ignores an object's size and treats its forces as acting at a single point. Draw a separate force diagram for the particle or system you are analysing.
Weight acts vertically downward. A normal reaction is perpendicular to the contact surface; friction acts along it. Tension pulls along a taut string. Include only forces acting ON the chosen particle, not forces it exerts on another body.
Use newtons for force and kilograms for mass. In these worked examples take g = 10 m/s^2, so a 5 kg particle has weight 50 N. Always use the value of g specified in a question.
Components and resultants 4.1
Force has magnitude and direction. A force F at theta above the horizontal has components F cos theta horizontally and F sin theta vertically, with signs determined by direction.
Worked example: a 10 N force at 60 degrees above the horizontal has horizontal component 5 N and vertical component 5sqrt(3) N, approximately 8.66 N (2 d.p.). Components replace the original force; do not include both the original force and its components in one sum.
Add components in each direction to find a resultant. For 6 N right and 8 N upward, its magnitude is sqrt(36+64)=10 N and its direction is arctan(8/6), approximately 53.13 degrees above the horizontal (2 d.p.). Calculate rather than relying on a scale drawing.
Equilibrium means zero resultant 4.1
For a particle in equilibrium, the sum of force components in each of two perpendicular directions is zero. Choose directions that make the equations simple, often parallel and perpendicular to a surface.
Worked example: a 5 kg particle at rest on a table has an extra upward pull of 10 N. With g = 10, vertical equilibrium gives R+10-50=0, so the normal reaction is 40 N, not 50 N.
To balance a nonzero resultant with one additional force, use the equilibrant: equal in magnitude and opposite in direction. Zero resultant means zero acceleration; the particle need not be at rest if it is already moving at constant velocity.
Smooth contact and normal reaction 4.1
A smooth-contact model neglects friction, leaving only the normal component of contact force. It is a modelling assumption, not a statement that every real surface has no friction.
The normal reaction is not automatically mg. For a particle resting on a plane inclined at alpha with no other forces perpendicular to the plane, perpendicular equilibrium gives R=mg cos alpha; an extra force with a perpendicular component changes that equation.
A surface can push on a particle but cannot pull it towards itself. A negative computed reaction is a warning to reconsider the assumed contact and force diagram.
Friction, limiting equilibrium and mu 4.1
For rough contact, friction opposes relative motion or the tendency to slip. The coefficient mu is the ratio of limiting friction to normal reaction: limiting F=muR. For static contact in general, F<=muR.
Worked example: on a horizontal table with R=50 N and mu=0.4, the maximum available static friction is 20 N. A horizontal pull of 12 N is balanced by actual friction 12 N. The particle is not at its limiting value.
When a question says 'about to slip' or 'limiting equilibrium', set F=muR and choose its direction to oppose the impending motion. Do not assume equality merely because the surface is rough.
Newton's third law: two different bodies 4.1
If body A exerts a force on body B, B exerts an equal and opposite force of the same interaction on A. The two forces act on different bodies, so they do not cancel on a force diagram of either body alone.
The upward force of a table on a particle pairs with the downward force of that particle on the table. The particle's weight and its normal reaction are not a third-law pair: both act on the particle.
Write the two bodies and the interaction when identifying a pair. This keeps a force-balance equation separate from a third-law statement.
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