Question

please help
A machine with an AMA of 0.650 and an efficiency of 73% produces a force of 9.52 x 10^6 N over a distance of 2.2 x 10^2 m. What is the input distance?

2.87 x 10^9 m

1.46 x 10^7 m

2.09 x 10^9 m

1.96 x 10^2 m

Answer 1

@phi

Answer 2

@ShadowLegendX

Answer 3

what's an AMA ?

Answer 4

actual mechanical advantage

Answer 5

@phi

Answer 6

I have never seen a question like this, and I don't understand it. Do you have more background you can explain?

Answer 7

@johnweldon1993

Answer 8

i dont have that much info on this

Answer 9

If you have notes that show a similar problem with the answer, please post it.

Answer 10

A machine with an AMA of 0.35 and an efficiency of 62% produces a force of 5,798 N over a distance of 17.3 m. What is the input distance?

Answer 11

Ehh, instead of typing everything out...since I see you have another problem which is basically the same...use that as a guideline!

Where are you getting messed up?

Answer 12

well that answers how to do it.

First find the work, Force * distance

Answer 13

it was just one that was provided, and it doesn't make any sense to me

Answer 14

so what numbers would i use? these?
force of 9.52 x 10^6 N over a distance of 2.2 x 10^2 m

Answer 15

yes, that gives the work (or energy)

Answer 16

Got it

Well first, we can agree that a machines efficiency is the ratio of work out to work in

\[\large Efficiency = \frac{W_{out}}{W_{in}} \times 100\text{%}\]

So from that, we know the efficiency...and we can find the output work!

\[\large W_{out} = F\times d\]

So here, what would be the output work?
\[\large W_{out} = 9.52\times 10^6 [N] \times 2.2\times 10^2 [m] = ?\]

Answer 17

2.0944 × 10^9

Answer 18

@johnweldon1993

Answer 19

Looks right to me, and that is in units of Joules [ J ]

Now that is out Output work...so lets plug that into the equation we have so far for efficiency

\[\large Efficiency = \frac{W_{out}}{W_{in}}\times 100\text{%}\]

We can now rearrange this to solve for the unknown "Input work"

\[\large W_{in} = \frac{W_{out}}{Efficiency}\times 100\text{%}\]
\[\large W_{in} = \frac{2.0944\times 10^9 [J]}{73\text{%} }\times 100\text{%} = ?\]

Answer 20

If the percents are throwing you off...this is the same as

\[\large W_{in} = \frac{2.0944\times 10^9 [J]}{0.73}\]

Answer 21

2869041095.89
which is So, the final answer is 2.86904 × 10^9

Answer 22

Right \(\large W_{in} = 2.87\times 10^9\) *Just rounded off a bit to make it look nicer*

Now from here....we recall that the AMA is the ratios of output force to input force...or \(\large AMA = \frac{F_{out}}{F_{in}}\)

So we can, from that, solve for the Input force since we know both of the other terms

\[\large F_{in} = \frac{F_{out}}{AMA} = \frac{9.52\times 10^6 [N]}{0.650} =?\]

Answer 23

1.46462 × 10^7

Answer 24

1.46 X 10^7

Answer 25

Lol glad you rounded, either way correct...so that is out Input force

Now the last step is to solve for what we need

Remember how Work = Force times distance? And we used that to solve for the Output work? We're gonna do the same thing here for the inputs

\[\large W_{in} = F_{in}\times d_{in}\]

We are solving for \(\large d_{in}\) so

\[\large d_{in} = \frac{W_{in}}{F_{in}} = \frac{2.87\times 10^9}{1.46\times 10^7}=?\]

Answer 26

1.96 × 10^2

Answer 27

And we're done!

Answer 28

thank you so much!!!!!!

Answer 29

Not a problem!