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Explainer Video

Tutor: Daniel

Summary

Pulleys

In a nutshell

Pulleys are systems of connected particles joined by a string that passes over a wheel. Problems involving pulleys involve weight, tension, friction, acceleration, speed, time and distance. In this lesson, the pulleys are assumed to cause no friction and be weightless.


Equations


description

equation

Newton's second law
​F=maF=maF=ma​​

Weight
​W=mgW=mgW=mg​​

Second equation of motion
​s=ut+12at2s=ut+\dfrac12at^2s=ut+21​at2​​

​

Variable definitions


quantity

symbol

unit name

unit

​ForceForceForce​​
​FFF​​
​NewtonNewtonNewton​​
​NNN​​
​WeightWeightWeight​​
​WWW​​
​NewtonNewtonNewton​​
​NNN​​
​TensionTensionTension​​
​TTT​​
​NewtonNewtonNewton​​
​NNN​​
​DisplacementDisplacementDisplacement​​
​sss​​
​MetreMetreMetre​​
​mmm​​
​TimeTimeTime​​
​ttt​​
​SecondSecondSecond​​
​sss​​
​Initial speedInitial\space speedInitial speed​​
​uuu​​
​Metres per secondMetres \space per \space secondMetres per second​​
​m/sm/sm/s​​
​AccelerationAccelerationAcceleration​​
​aaa​​
​Metres per second squaredMetres \space per \space second\space squaredMetres per second squared​​
​ms−2ms^{-2}ms−2​​
​Acceleration due to gravityAcceleration\space due\space to \space gravityAcceleration due to gravity​​
​ggg​​
​Metres per second squaredMetres \space per \space second\space squaredMetres per second squared​​
​ms−2ms^{-2}ms−2​​


​

A pulley system

Below is a common pulley system involving two objects xxx and yyy, where yyy has twice the mass of xxx.


In pulley systems, the same forces affect the objects involved differently. For example, if you look at the forces affecting xxx, you receive this equation:

​​F=maT−mg=ma\begin{aligned}F&=ma\\T-mg&=ma\end{aligned}FT−mg​=ma=ma​


Now, consider the forces acting on yyy. As yyy is heavier, it begins accelerating downwards. Now, TTT is against the direction of motion and is negative.​

2mg−T=2ma\begin{aligned}2mg-T=2ma\end{aligned}2mg−T=2ma​​​​​


This is why you can't consider connected particles on a pulley as one single object like in other cases. The particles move in different directions and so do not move singularly. 

Finally, the force to hold up the pulley, PPP, is equal to the tension in the string on both sides of the pulley, 2T2T2T. 

P=2TP=2TP=2T​​​



Speed and distance

A problem involving pulleys will often ask for the distance the object travels in a certain time, or vice versa. The second equation of motion, s=ut+12at2s=ut+\dfrac12at^2s=ut+21​at2, is necessary to solve these problems. Most (but not all) pulley system problems begin the moment the system is released from rest, making the initial speed u=0u=0u=0. As such, the equation can be simplified:

​s=(0)t+12at2s=(0)t+\dfrac12 at^2s=(0)t+21​at2​​


This simplifies to give:

s=12at2\boxed{s=\dfrac 12at^2}s=21​at2​​​



Once one object stops moving, the string becomes 'slack', and the tension force affecting the objects ceases. 


Example 1

A 444 kg\ kg kg​ object and a 6 kg6\ kg6 kg object connected in a pulley system are released from rest. The 6 kg6\ kg6 kg falls for 333 seconds before hitting the floor. Find out the following information, giving all answers in 333 s.f:

(i) The acceleration felt by the objects, aaa.

(ii) The tension in the string, TTT.

(iii) The distance travelled by the 6 kg6\ kg6 kg block.

Maths; Forces and motion; KS5 Year 12; Pulleys


(i) First, construct force equations for 4kg4kg4kg​ and 6kg6kg6kg. 

​4 kg:F=maF=T−4gm=4T−4g=4a6 kg:F=maF=6g−Tm=66g−T=6a\begin{aligned}4\ kg:&\\F&=ma\\F&=T-4g\\m&=4\\T-4g&=4a\\\\6\ kg:&\\F&=ma\\F&=6g-T\\m&=6\\6g-T&=6a\end{aligned}4 kg:FFmT−4g6 kg:FFm6g−T​=ma=T−4g=4=4a=ma=6g−T=6=6a​


Add the two equations to cancel out the TTT and find the acceleration.

​6g−T+T+4g=6a+4a2g=10aa=15g\begin{aligned}6g-\cancel{T}+\cancel{T}+4g&=6a+4a\\2g&=10a\\a&=\dfrac15g\end{aligned}6g−T​+T​+4g2ga​=6a+4a=10a=51​g​​​


​a=1.96 ms−2‾\underline{a=1.96\ ms^{-2}}a=1.96 ms−2​​​

​​

(ii) Use the acceleration to find the tension.

T−4g=4g×15gg=9.8T=4(9.8)×15(9.8)+4(9.8)T=116.032\begin{aligned}T-4g&=4g\times \dfrac15g\\g&=9.8\\T&=4(9.8)\times \dfrac15(9.8)+4(9.8)\\T&=116.032\end{aligned}T−4ggTT​=4g×51​g=9.8=4(9.8)×51​(9.8)+4(9.8)=116.032​


T=116 N‾\underline{T=116\ N}T=116 N​​​


(iii) Use the second equation of motion to find the distance sss.

s=12at2a=1.96t=3s=12(1.96)(3)2\begin{aligned}s&=\dfrac12at^2\\a&=1.96\\t&=3\\s&=\dfrac12(1.96)(3)^2\end{aligned}sats​=21​at2=1.96=3=21​(1.96)(3)2​​


s=8.82 m‾\underline{s=8.82\ m}s=8.82 m​​​


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FAQs - Frequently Asked Questions

What happens in a pulley when one object stops and the other keeps going?

Once one object stops moving, the string between the objects becomes 'slack', and the tension force affecting the objects ceases.

What is the second equation of motion?

s=ut+1/2at^2, where s is displacement, u is initial velocity, t is time and a is acceleration.

Why can't you consider connected objects as one object in a pulley system?

In pulley systems, the same forces affect the objects involved differently. As such, you can't consider them as one object.

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