Connected particles, pulleys and inclined planesEdexcel International A Level Maths: Revision notes

Two particles are connected by a light inextensible string (or a rod). Light means the tension is the same throughout; inextensible means that, while the string is taut, both particles have the same speed and acceleration (in magnitude, along the string). A smooth fixed pulley or peg changes the direction of the string but not the tension. Method:

• draw a force diagram for each particle,
• write an equation of motion $F=ma$ for each in the direction of its motion,
• solve simultaneously for $a$ and $T$. Alternatively treat the connected system as one body to find $a$ first, then use one particle to find $T$.

Two particles of masses $m_1>m_2$ hanging over a smooth pulley: $a=\frac{(m_1-m_2)g}{m_1+m_2},\qquad T=\frac{2m_1m_2g}{m_1+m_2}.$ Example: $3$ kg and $5$ kg, $g=9.8$: $a=\frac{2(9.8)}{8}=2.45$ m s⁻². For the $3$ kg: $T-3g=3a$ gives $T=36.75$ N. The force on the pulley from the string is the sum of the two vertical tensions, $2T=73.5$ N downwards. A particle on a smooth horizontal table connected over the edge to a hanging particle behaves the same way: the hanging weight drives the total mass.

When one particle hits the ground, or the string goes slack, the forces change from one fixed set to another. Treat the motion in two stages. Stage 1: connected, find $a$ and then the speed $v$ at the end using $v^2=u^2+2as$. Stage 2: the string is slack, so the tension is $0$. A particle left moving upwards decelerates at $g$, so the further rise is $\frac{v^2}{2g}$. The particle that hit the ground stops, assuming it does not rebound. Example: $3$ kg and $2$ kg particles, with $3$ kg $1.2$ m above the ground, give $a=1.96$ and $v^2=4.704$. After impact the $2$ kg particle rises a further $\frac{4.704}{19.6}=0.24$ m. If the other particle is on a slope, find its new acceleration from the forces on it alone.

For a particle of mass $m$ on a plane inclined at $\theta$ to the horizontal, resolve parallel and perpendicular to the plane:

• weight component down the slope: $mg\sin\theta$
• weight component into the slope: $mg\cos\theta$
• normal reaction $R=mg\cos\theta$ (no acceleration perpendicular to the plane if no other force),
• on a smooth plane the acceleration down the slope is $g\sin\theta$. Example: $\theta=30^\circ$, $g=9.8$: $a=4.9$ m s⁻². Projected up a smooth $30^\circ$ slope at $8$ m s⁻¹, the deceleration is $4.9$ so the particle stops after $\frac{64}{9.8}=6.53$ m and returns after $\frac{16}{4.9}=3.27$ s.

On a rough plane friction $F$ acts along the plane, opposing motion. When sliding, $F=\mu R$ where $\mu$ is the coefficient of friction. Equation of motion for a particle sliding down: $mg\sin\theta-\mu R=ma$ with $R=mg\cos\theta$. For a particle moving up, friction acts down the slope: $-mg\sin\theta-\mu mg\cos\theta=ma$, a bigger deceleration than for sliding down. Combined with a pulley, find $R$ first, then write one equation for each particle. Example: $A$ ($2$ kg) on a $30^\circ$ rough plane with $\mu=0.4$ connected to $B$ ($3$ kg) hanging: $R=16.97$ N, $F=6.79$ N, and $3g-2g\sin30^\circ-F=5a$ gives $a=2.56$ m s⁻².

1. Draw a clear diagram for every particle: weight, tension, normal reaction, friction.
2. Resolve perpendicular to a slope first to get $R$ and $F=\mu R$.
3. Write one equation of motion per particle, using the same $a$.
4. Solve, check the sign of $a$ (does the assumed direction make sense?).
5. For later stages use the final velocity of the first stage as the initial velocity. Use $g=9.8$ and give answers to 2 or 3 significant figures. State assumptions in modelling questions: the string is light and inextensible, the pulley is smooth, the particles are points.