Light strings & smooth pulleys | CAIE 9709 Mechanics Notes

Mechanics

Mathematics9709

Translate each model word before writing equations

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One taut inextensible string gives one acceleration magnitude. One light string over a smooth pulley gives one tension magnitude.

Tension: the pulling force along a taut string.
Light string: its mass is ignored, so no difference in tension is needed to accelerate the string itself.
Inextensible string: its length does not change.
Taut string: it has no slack, so the connection is active.
Inextensible and taut together: the connected particles have equal acceleration magnitudes along the string.
Smooth pulley: the pulley changes the direction of the string without changing the tension.
Smooth surface: there is no friction from that surface.

Common mistake

“Equal acceleration” means equal magnitude along the string. One particle may move right while the other moves downward.

Write one Newton equation for each particle

Smooth pulley system lab

g = 10 m s⁻²

Mass on table3 kgHanging mass2 kg

a = 4 m s⁻²

T from A = 12 N

T from B = 12 N

The two one-body equations give the same tension. Increasing either mass changes both acceleration and tension; the string constraint still keeps the acceleration magnitudes equal.

Worked example

One particle on a smooth table

Particle A of mass 3 kg lies on a smooth table. It is connected over a smooth pulley to a hanging particle B of mass 2 kg. B moves downward.

For A, take right as positive:

T=3a

For B, take downward as positive:

20-T=2a

Adding eliminates the internal tension:

20=5a

a=4\text{ m s}^{-2}

T=12\text{ N}

Once both signed equations are visible, a calculator simultaneous- equation solver is a fast main method. Enter the two equations in $a$ and $T$ , then record both results. The equations are still essential because they contain the mechanics.

Key idea

Use a positive direction that follows each particle's actual motion. The acceleration symbol can then be positive in both one-body equations even though the particles move in different directions.

Add friction before combining the equations

Particle P of mass 2.5 kg lies on a rough $30^\circ$ plane. It is connected to a hanging 0.5 kg particle Q. The coefficient of friction is 0.2 and P moves down the plane.

R=25\cos30^\circ=12.5\sqrt3

F=0.2R=2.5\sqrt3

For P, down the plane is positive:

12.5-T-2.5\sqrt3=2.5a

For Q, upward is positive:

T-5=0.5a

a=1.06\text{ m s}^{-2}

T=5.53\text{ N}

Examiner note

This follows a current paper's structure. The marks distinguish the normal reaction, friction, the correct Newton equations and the final simultaneous solution. (9709/42/M/J/24 Q7(a))

Your turnIndependent check [5 marks]

A 4 kg particle on a smooth horizontal table is connected over a smooth pulley to a hanging 1.5 kg particle. Find the acceleration and the tension.

Show worked answer

T=4a

15-T=1.5a

15=5.5a

a=\frac{30}{11}\text{ m s}^{-2}

T=4a=\frac{120}{11}\text{ N}

Answer

a=\frac{30}{11}\text{ m s}^{-2}

T=\frac{120}{11}\text{ N}