3.03k, 3.03o - Connected particles, strings and pulleys

3.03k, 3.03o - Connected particles, strings and pulleys

A string links motions as well as forces. You will use separate particle equations to find acceleration and tension, then adapt the model when a particle reaches the floor or a string runs along an incline.

What a string and pulley model imposes

If both masses on a pulley are doubled, will they accelerate twice as quickly? Start with two equal hanging masses of 2 kg2\,\mathrm{kg}. Each can remain at rest with an upward tension of 19.6 N19.6\,\mathrm N balancing its weight. Unequal masses cannot both balance under that same common tension. The interesting question is how much the whole arrangement accelerates.

You already know

For a fixed mass, resultant force equals mass times acceleration. Weight is mgmg, and equilibrium means zero resultant. Draw the forces acting on each object separately.

A connection adds information that Newton's second law alone does not supply.

Ideal string and pulley

A light string has negligible mass; an inextensible string has fixed length. For a taut light string over a fixed smooth ideal pulley, use a common tension magnitude. In the simple arrangement with one straight segment to each particle, fixed length gives equal displacement, speed and acceleration magnitudes along the two segments. A slack string supplies no tension.

The adjectives do different jobs. Smooth contact removes friction at the pulley; light string and ideal pulley assumptions permit the same tension on its two sides. Inextensibility supplies the motion constraint. A string pulls towards its adjacent segment; it cannot push.

If one hanging particle accelerates downwards and the other upwards, their acceleration vectors differ. You can nevertheless use one signed scalar aa, taking down as positive for one equation and up as positive for the other. A negative answer reverses those assumed acceleration directions. It does not reverse the tension arrows.

For a train translating along one track, all its vehicles have the same acceleration vector. For a pulley system, adding compatible equations along different string segments is an algebraic elimination of tension. It is not a vector claim that two particles travelling in different directions form one translating particle. If you include the string in a vector system diagram, the support force from the fixed pulley is external too.

Key idea

The force model gives common tension; the taut fixed-length constraint gives common acceleration magnitude. These are separate assumptions, and neither says tension equals weight.

These connection rules are not in the formula sheet. Before using them, check the geometry rather than relying on the word “connected”.

A whole train and its individual vehicles

Use the whole train to find acceleration, then one vehicle to recover the force in the coupling.

A coupling force is internal to the engine-plus-trailer system but external to either vehicle separately. Resistances on both vehicles remain external. With forwards positive, an engine's drive DD, resistances RE,RTR_E,R_T, and masses mE,mTm_E,m_T give the following translating-system equation.

Whole train and trailer equations

For a taut light inextensible coupling on a horizontal track,

D−RE−RT=(mE+mT)aT−RT=mTa\begin{aligned} D-R_E-R_T&=(m_E+m_T)a \\ T-R_T&=m_Ta \end{aligned}

The first equation eliminates the internal tension; the second concerns the trailer alone. These are applications of Newton's second law, not additional laws.

Not in the formula sheet: learn it.

The separate equation is essential: the whole-system equation contains no tension to solve for.

Worked example 1: drive and two resistances

Question. A 600 kg600\,\mathrm{kg} engine pulls a 400 kg400\,\mathrm{kg} trailer through a taut light inextensible coupling on a horizontal track. Drive is 2200 N2200\,\mathrm N; resistances are 400 N400\,\mathrm N on the engine and 200 N200\,\mathrm N on the trailer. Find acceleration and tension.

Plan. Apply the second law to both vehicles together, then to the trailer.

2200−400−200=1000awhole traina=1.6divide by total massT−200=400(1.6)trailer aloneT=840solve for tension\begin{aligned} 2200-400-200&=1000a &&\text{whole train} \\ a&=1.6 &&\text{divide by total mass} \\ T-200&=400(1.6) &&\text{trailer alone} \\ T&=840 &&\text{solve for tension} \end{aligned}

Why this step? The coupling forces cancel for the whole train but must appear when the trailer is isolated.

Answer. Acceleration is 1.6 m s−21.6\,\mathrm{m\,s^{-2}} forwards; tension is 840 N840\,\mathrm N.

Check. The engine equation gives 2200−400−840=960=600(1.6)2200-400-840=960=600(1.6).

Change only the resistance on the trailer. It changes both the acceleration of the whole train and the force needed to pull that trailer.

Worked example 2: increased trailer resistance

Question. Keep the same engine, trailer, drive and engine resistance, but increase the trailer resistance to 500 N500\,\mathrm N. Find acceleration and tension.

Plan. Recalculate the whole-system resultant, then use the new trailer equation.

2200−400−500=1000ainclude both resistancesa=1.3whole-system accelerationT−500=400(1.3)trailer aloneT=1020recover coupling force\begin{aligned} 2200-400-500&=1000a &&\text{include both resistances} \\ a&=1.3 &&\text{whole-system acceleration} \\ T-500&=400(1.3) &&\text{trailer alone} \\ T&=1020 &&\text{recover coupling force} \end{aligned}

Why this step? A larger resistance can require more tension even though the acceleration is smaller.

Answer. Acceleration is 1.3 m s−21.3\,\mathrm{m\,s^{-2}} forwards; tension is 1020 N1020\,\mathrm N.

Check. 2200−400−1020=780=600(1.3)2200-400-1020=780=600(1.3). Both vehicle equations agree.

A flexible coupling must remain in tension. A negative calculated tension would require a push and would invalidate that assumed connection. A rigid towbar can transmit compression, but then identify it as a push.

Watch out

Wrong: D−RE=(mE+mT)aD-R_E=(m_E+m_T)a. This omits the trailer's resistance. Correct: include every external horizontal force on the chosen system, then use a separate vehicle equation for tension.

Try the same two-system choice for a different vehicle pair.

Two hanging particles and compatible signs

For two hanging particles, choose a positive direction for each particle that respects the string constraint.

Take the heavier mass MM downwards and lighter mass mm upwards. Each particle has two forces: upward tension and downward weight. Keep the same tension magnitude in both equations.

[DIAGRAM: asset_name: Separate equations for two hanging particles; asset_slug: hanging_particle_forces; description: Five-kg and three-kg particles on opposite sides of an ideal pulley, separate upward tension and downward weight arrows, acceleration 5 kg down and 3 kg up distinguished from force arrows; caption: The particles have opposite acceleration vectors and one common tension]
The particles have opposite acceleration vectors and one common tension

The equations also explain a useful surprise: doubling both masses doubles the driving weight difference and the total inertia. The acceleration stays the same.

Two hanging particles

For a taut light inextensible string over a fixed smooth ideal pulley, with M>mM>m,

Mg−T=MaT−mg=maa=(M−m)gM+m\begin{aligned} Mg-T&=Ma \\ T-mg&=ma \\ a&=\frac{(M-m)g}{M+m} \end{aligned}

Recover tension from either separate equation. The acceleration formula is derived by adding the first two equations; it is not a substitute for setting up a new arrangement.

Not in the formula sheet: learn it.

Keep unrounded values while checking both equations.

Worked example 3: unequal hanging masses

Question. Masses 5 kg5\,\mathrm{kg} and 3 kg3\,\mathrm{kg} hang from a taut light inextensible string over a fixed smooth ideal pulley. Released from rest, find acceleration and tension, and identify the descending mass. Use g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}; give non-exact answers to 33 significant figures.

Plan. Write equations in the directions 5 kg5\,\mathrm{kg} down and 3 kg3\,\mathrm{kg} up, then add.

49−T=5aheavier particleT−29.4=3alighter particle19.6=8aeliminate tensiona=2.45solveT=29.4+3(2.45)=36.75lighter particle\begin{aligned} 49-T&=5a &&\text{heavier particle} \\ T-29.4&=3a &&\text{lighter particle} \\ 19.6&=8a &&\text{eliminate tension} \\ a&=2.45 &&\text{solve} \\ T&=29.4+3(2.45)=36.75 &&\text{lighter particle} \end{aligned}

Why these signs? The positive directions are opposite in space but compatible with one taut string.

Answer. Acceleration is 2.45 m s−22.45\,\mathrm{m\,s^{-2}}; tension is 36.8 N36.8\,\mathrm N. The 5 kg5\,\mathrm{kg} particle descends.

Check. 49−36.75=12.25=5(2.45)49-36.75=12.25=5(2.45); tension lies between the two weights.

Now change only the common mass scale: double both masses and leave the geometry and gravity unchanged.

Worked example 4: doubling the masses

Question. Replace those masses by 10 kg10\,\mathrm{kg} and 6 kg6\,\mathrm{kg}. Under the same conditions, find acceleration and tension and identify the descending mass. Give non-exact answers to 33 significant figures.

Plan. Form the two new second-law equations and compare their sum with the previous sum.

98−T=10aheavier particle downT−58.8=6alighter particle up39.2=16aadd equationsa=2.45same force-to-mass ratioT=58.8+6(2.45)=73.5recover tension\begin{aligned} 98-T&=10a &&\text{heavier particle down} \\ T-58.8&=6a &&\text{lighter particle up} \\ 39.2&=16a &&\text{add equations} \\ a&=2.45 &&\text{same force-to-mass ratio} \\ T&=58.8+6(2.45)=73.5 &&\text{recover tension} \end{aligned}

Why this step? Doubling both sides' masses does not double their acceleration: it doubles the net drive and the total mass together.

Answer. Acceleration is 2.45 m s−22.45\,\mathrm{m\,s^{-2}}; tension is 73.5 N73.5\,\mathrm N. The 10 kg10\,\mathrm{kg} particle descends.

Check. 98−73.5=24.5=10(2.45)98-73.5=24.5=10(2.45); tension has doubled, as every force in the equations has doubled.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account