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Don't they both deal with input and output forces?

2006-12-15 10:43:02 · 4 answers · asked by ZZ 4 in Science & Mathematics Physics

4 answers

Ideal mechanical advantage is calculated based on geometry. Actual mechanical advantage is always less because of friction.

2006-12-15 10:46:19 · answer #1 · answered by hevans1944 5 · 1 0

In an ideal machine the work energy given to the machine (input energy) is equal to the work energy done by the machine (out put energy).
There is no loss of energy due to friction etc.
Therefore the efficiency of the machine will be 1 or (100 %).
The mechanical advantage of this ideal machine is W/P.
W: Load (Force lifted or given by the machine).
P: Effort (Force applied on the machine by us)

But in practice in all machines, there is some loss of energy due to friction etc.
To compensate this loss, we do some extra energy or work on the machine.
That is, for the same out put energy, we have to give more input energy than needed for an ideal machine.

The efficiency of this practical machine will always be less than 1 (less than 100%).

To do that extra work, usually the effort (the force applied on the machine by us will be more than for an ideal machine.)
Since M.A = Load / effort and since the effort is more for practical machines, the M.A will be less than that for an ideal machine.

2006-12-15 12:37:16 · answer #2 · answered by Pearlsawme 7 · 0 0

The ideal MA doesn't take friction into account, so it is always a bit more than the actual MA.

2016-05-22 22:03:03 · answer #3 · answered by ? 4 · 0 0

IMA=Ideal Mechanical Advantage

AMA=Actual Mechanical Advantage

In your system,
Wheel, angle, pulley system,
Take Distance

Force actually pulled
----------------- (Friction adds to the top one)=AMA
Force ideally applied

Distance you pulled
-------------------- =IMAexp. 4:1 if you had a 4 string pulley system.
Distance machine moved

% Effeciency = AMA/IMA *100

2006-12-15 13:36:17 · answer #4 · answered by ncsbee 2 · 0 0

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