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Pulleys

Theoretical Mechanical Advantage

The theoretical mechanical advantage (TMA) is the ratio of the useful load force to the applied effort, assuming no friction:

\[ \text{TMA} = \frac{\text{Load}}{\text{Effort}} \]

Velocity Ratio

Velocity ratio (VR) is the ratio of the distance moved by the effort to the distance moved by the load:

\[ \text{VR} = \frac{\text{Distance moved by effort}}{\text{Distance moved by load}} \]

Work Done

The work done by a constant force \(F\) acting through a displacement \(s\) is:

\[ W = Fs \]

One Wheel

Single fixed pulley diagram

A single fixed pulley changes the direction of the force but does not reduce the effort.

Single pulley force direction change

Pulling down 1 m of rope raises the load by 1 m.

The velocity ratio is:

\[ \text{VR} = 1 \]

Single pulley VR demonstration

Loaded Pulley

Loaded pulley diagram

The tension \(T\) is the same throughout the string.

Tension in pulley rope

What happens next depends on the values of M1 and M2

Case 1: \(M_1 > M_2\)

Pulley with M1 heavier than M2

Mass \(M_1\) moves down, pulling \(M_2\) up.

This causes downwards acceleration of load \(M_1\) and upwards acceleration of load \(M_2\)

\[ F = ma \] \[ M_1 g - T = M_1 a \] \[ T - M_2 g = M_2 a \]
\[ \text{Adding} \] \[ M_1 g - M_2 g = M_1 a + M_2 a \] \[ a = \frac{M_1 g - M_2 g}{M_1 + M_2} \] \[ a = \frac{g\,(M_1 - M_2)}{M_1 + M_2} \]

Case 2: \(M_2 > M_1\)

Pulley with M2 heavier than M1

Mass \(M_2\) moves down, pulling \(M_1\) up.

This causes downwards acceleration \( a_2 \) and upwards acceleration \( a_1 \)

\[ T - M_1 g = M_1 a \] \[ M_2 g - T = M_2 a \] \[ \text{Adding} \] \[ M_2 g - M_1 g = M_1 a + M_2 a \] \[ a = \frac{M_2 g - M_1 g}{M_1 + M_2} \] \[ a = \frac{g\,(M_2 - M_1)}{M_1 + M_2} \]

Example

Find the downward acceleration of the block:

Example pulley problem diagram


\[ a = \frac{(M_1 - M_2)g}{M_1 + M_2} \]
\[ a = \frac{g(3 - 1)}{3 + 1} \] \[ = \frac{2g}{4} \] \[ = \frac{g}{2} \]

Alternatively, from the diagram:

\[ 3g - T = 3a \] \[ T - g = a \] \[ \text{Adding} \] \[ 3g - g = 3a + a \] \[ a = \frac{3g - g}{3 + 1} \] \[ a = \frac{2g}{4} = \frac{g}{2} \]

Two Wheels

Two-wheel pulley system

Pulling down 1 m of rope raises the load by 0.5 m.

Two-wheel pulley VR demonstration

\[ \text{VR} = 2 \] \[ \text{Effort} = \frac{1}{2}\text{Load} \]

Effort reduction in two-wheel pulley

Youtube example

If a pulley system is perfectly efficient, the mechanical advantage and velocity ratio equal the number of pulleys.

Example

Calculate the acceleration of the load and the tension in the system.

Four-pulley example diagram


\[ 3g - T = 3a \] \[ 2T - 2g = 2a \] \[ \text{So} \] \[ 6g - 2T = 6a \] \[ 2T - 2g = 2a \] \[ \text{Adding } \] \[ 4g = 8a \] \[ a = \frac{4g}{8} \] \[ a = \frac{g}{2}\ \text{ms}^{-2} \]
\[ 3g - T = 3a \] \[ T = 3g - 3a \] \[ \text{substitute } a = \frac{g}{2} \] \[ T = 3g - 3\left(\frac{g}{2}\right) \] \[ = \frac{6g - 3g}{2} \] \[ = \frac{3g}{2}\,N \]

Three Wheels

Three-wheel pulley system

\[ \text{VR} = 3 \]
\[ M_1 g - T = M_1 a_1 \] \[ 2T - xg = x\left(\frac{a_1 + a_2}{2}\right) \] \[ M_2 g - T = M_2 a_2 \]
Example

Calculate the acceleration of the moveable pulley and the tension in the rope.

Three-wheel example diagram


\[ 3g - T = 3a_1 \] \[ 2T - 4g = 4\left(\frac{a_1 + a_2}{2}\right) \] \[ 3g - T = 3a_2 \]

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\[ 2T - 4g = 2a_1 + 2a_2 \] \[ \text{Multiply first equation by 2} \] \[ 6g - 2T = 6a_1 \] \[ \text{Adding} \] \[ 2g = 8a_1 + 2a_2 \]
\[ 2T - 4g = 2a_1 + 2a_2 \] \[ 3g - T = 3a_2 \] \[ \text{Multiply second equation by 2} \] \[ 6g - 2T = 6a_2 \] \[ \text{Adding} \] \[ 2g = 2a_1 + 8a_2 \]
\[ 2g = 8a_1 + 2a_2 \] \[ 2g = 2a_1 + 8a_2 \] \[ \text{subtracting} \] \[ -6g = -30a_2 \] \[ a_2 = \frac{1}{5}g \]
\[ \text{substitute } a_2 = \frac{1}{5}g \] \[ 2g = 8a_1 + \frac{2}{5}g \] \[ \frac{8}{5}g = 8a_1 \] \[ a_1 = \frac{1}{5}g \]
\[ a_{\text{pulley}} = \frac{a_1 + a_2}{2} \] \[ = \frac{\frac{1}{5}g + \frac{1}{5}g}{2} \] \[ = \frac{1}{5}g \]
\[ 3g - T = 3a_2 \] \[ T = 3g - 3a_2 \] \[ = 3g - \frac{3}{5}g \] \[ T = \frac{12}{5}g \]
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