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

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Variable definitions

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A pulley system

Below is a common pulley system involving two objects $$x$$ and $$y$$, where $$y$$ has twice the mass of $$x$$.

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

​​$$\begin{aligned}F&=ma\\T-mg&=ma\end{aligned}$$

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

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

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, $$P$$, is equal to the tension in the string on both sides of the pulley, $$2T$$. 

$$P=2T$$​​​

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, $$s=ut+\dfrac12at^2$$, 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 $$u=0$$. As such, the equation can be simplified:

​$$s=(0)t+\dfrac12 at^2$$​​

This simplifies to give:

$$\boxed{s=\dfrac 12at^2}$$​​

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

Example 1

A $$4$$$$\ kg$$​ object and a $$6\ kg$$ object connected in a pulley system are released from rest. The $$6\ kg$$ falls for $$3$$ seconds before hitting the floor. Find out the following information, giving all answers in $$3$$ s.f:

(i) The acceleration felt by the objects, $$a$$.

(ii) The tension in the string, $$T$$.

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

(i) First, construct force equations for $$4kg$$​ and $$6kg$$. 

​$$\begin{aligned}4\ kg:&\\F&=ma\\F&=T-4g\\m&=4\\T-4g&=4a\\\\6\ kg:&\\F&=ma\\F&=6g-T\\m&=6\\6g-T&=6a\end{aligned}$$

Add the two equations to cancel out the $$T$$ and find the acceleration.

​$$\begin{aligned}6g-\cancel{T}+\cancel{T}+4g&=6a+4a\\2g&=10a\\a&=\dfrac15g\end{aligned}$$​​

​$$\underline{a=1.96\ ms^{-2}}$$​​

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(ii) Use the acceleration to find the tension.

$$\begin{aligned}T-4g&=4g\times \dfrac15g\\g&=9.8\\T&=4(9.8)\times \dfrac15(9.8)+4(9.8)\\T&=116.032\end{aligned}$$

$$\underline{T=116\ N}$$​​

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

$$\begin{aligned}s&=\dfrac12at^2\\a&=1.96\\t&=3\\s&=\dfrac12(1.96)(3)^2\end{aligned}$$​

$$\underline{s=8.82\ m}$$​​