Question #8a35a

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Question #8a35a

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Answer 1 · 2015-06-25T22:06:06

For a simple machine mechanical advantage (MA) = $\frac{load}{effort}$ $"load"/"effort"$

Efficiency = $\frac{Mechanical advantage}{Velocity ratio} \times 100$ $"Mechanical advantage"/"Velocity ratio"xx100$

Explanation:

Mechanical advantage = $\frac{Load}{Effort}$ $"Load"/"Effort"$

So if we look at a simple 1 pulley system:

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The effort = the load so $M A = 1$ $MA=1$ . In this case there is no mechanical advantage gained from such a system.

You can increase the number of pulleys to increase the $M A$ $MA$ .

Gears work like this.

Velocity Ratio ( $V R$ $VR$ ) = $\frac{distance moved by effort}{distance moved by load}$ $"distance moved by effort"/"distance moved by load"$ .

Again, in the 1 pulley system $V R = 1$ $VR=1$ .

Work done = Force x distance moved in the direction of that force.

If we divide $M A$ $MA$ by $V R$ $"VR$ we get:

$\frac{MA}{VR} = \frac{load \times distance moved by load}{effort \times distance moved by effort}$ $"MA"/"VR"=("load"xx"distance moved by load")/("effort"xx"distance moved by effort")$

So this ratio is a measure of the work done on the load compared with the work done by the effort.

In an ideal world the work put in by the effort would equal the work done on the load. However, we don't live in an ideal world.

There are energy losses due to friction and heat etc which means that machines are rarely 100% efficient.

We usually express this as a percentage:

$Efficiency= M A / V R \times 100$ $"Efficiency="MA"/"VR"xx100$

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Question #8a35a

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