Why does a simple machine with Mechanical Advantage (MA) less than 1 acts as speed multiplier? My book mentions that A simple machine with Mechanical Advantage (MA) less than 1 acts as a speed multiplier. But I could not understand why is it so and I searched on the internet and some books but I found no Useful insights for  why a simple machine with MA less than 1 acts as a speed multiplier.
 A: It would be more accurate to say that a machine with a mechanical advantage less than $1$ can act as a speed multiplier - whether it does or not depends on the efficiency of the machine.
If the input to the machine is force $F_{in}$ at speed $v_{in}$ then the input power is $P_{in}=F_{in}v_{in}$. If the output is force $F_{out}$ at speed $v_{out}$ then the output power is $P_{out}=F_{out}v_{out}$. If the efficiency of the machine is $\eta$ then
$P_{out} = \eta P_{in}
\\ \displaystyle \Rightarrow v_{out} = \left ( \frac {\eta F_{in}}{F_{out}}\right ) v_{in}$
But $\frac {F_{out}}{F_{in}}$ is the mechanical advantage of the machine $MA$, so
$v_{out} = \frac \eta {MA} v_{in}$
So if $MA < \eta$ then $v_{out} > v_{in}$.
A: Think of a simple machine like a see-saw lever for instance.
If the fulcrum is exactly in the middle, then moving one end by 1 meter will move the other end exactly by 1 meter. This is the the machine with Mechanical advantage exactly equal to 1.
Now, if you change the position of the fulcrum such that the distance from fulcrum to you is less than the distance to the other end, then if you move your end by 1 meter, the other end will move by MORE than one meter. So, this is the case of mechanical advantage less than 1. Since, displacement of other end is more than on your end, hence we can say it is a speed multiplier.
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A: The mechanical advantage refers to (force of the load)/(force of effort to move the load) . If that's less than 1, you have the top lower than the bottom (in the fraction), so you are putting much effort for the amount of load. This is like pushing the shorter end of the lever: the other end moves faster, and you'll have to push hard to move the load. In the lever analogy, the load is the longer part of the lever. Any time you get a mechanical advantage less than 1, the output force is lower (than input), but the output speed is higher (than input). This all flips around appropriately. In the opposite case, that would be using a lever to lift a heavy object. A gearbox transmission is a great example of using this tradeoff to improve daily life.
If you understand all of that and are still wondering "why" , then the last bit I can share is that the total work done must be equal (Conservation of energy) on the input and output in these ideal cases. So, if you integrate the forcexdistance over the travel path for each the input and output side, they will equal, so if force is lower, then distance must be higher to make up for that. Hope this helps.
