Any Math Whizzers Out There?

[spedxprt]

I got the answer like this: Mikes age=3x Mary's age=4x

4x+5=9y
3x+5=7y
subtracting,
x=2y
plugging back in,
8y+5=9y
y=5
x=10
Mary 40
Mike is 30

Nice job Danno, we are currently doing linear systems in my classroom. I wish that I could access our site at school tomorrow to show my students this thread!

[Danno]

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... original post from the Bike lift ... DFW crew thread ...

Too..... Much...... Information......

Ok Grasshopper .... We'll start with the basics .......

Ok class ... welcome to .....
" See What Happens When You Open Your Big Mouth 101" ...... =^>

Attachment 3667
Oops ... Might help if you knew that for this exercise the 500 pound weight is pushing down on a pivot that is 36" long and the pull point is two thirds the way down .........

What is the value of "?"
.. and .. does the specifications of the winch match "?" .

This shows a winch ... It could be the lift point for a floor jack or the push point of a hydraulic ram.
Winch doesn't match "?" ? ... under powered? .. What would a single block .. as in block and tackle .. do to the equation?
I like the winch ideal ... don't like having to jack with a battery ...

Ok Maestro ... your move.

I printed the image but I have increased the weight to 700 pounds since the Zed is 605 w/ fuel I wanted to be good on my figures. Of course I never took a physics course so this will take me a while (envision Jethro Bodine saying: one knot knot plus 2 knot knot equals........)

The 500 Lb. is just an easy value to work with and the example is just a generic single leg ... One end/ leg does all the lifting where as a scissor there would be equal weight at both ends , On a single leg the other leg is just there pretty much to keep the platform parallel to the base ... There is some weight transferred to the second leg but in my opinion it's negligible.
change the weight to 1000 Lb. if it helps ... All we're really looking for is the formula.

Thanks for the "formula" Dan ......

Point well taken Reid ...
tensile strength of quarter inch aircraft cable shows to be only 6.4K to 7K .... three inch polyester strap 3.5K ..
and yes you can reduce the pull by adding pulleys but the ends will still be at static weight .. or do have this wrong?

If the added pulleys double the length of cable required to lift the load, i.e. it takes 4 feet of cable to lift the load 2 feet, the force required is cut in half. The load will be the same on the lifted end but only half at the lifting end. That's why a chain fall (comealong) has a loop at the lifting end- it would take too many feet of cable (or chain) to cut the lifted load down to a manageable level.

[Danno]

Oh, yeah, forgot my manners. You're welcome.

[Maximus]

Thanks Danno! You make it look easy.


I should be on my bike. Instead, I'm on my iPhone.

[DWC]

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.... Arkansas math ...
Divide load by number of pulleys used = pull load.
Just make sure that rafter you tied off to can support load + pull.

[Danno]

Thanks Danno! You make it look easy.


I should be on my bike. Instead, I'm on my iPhone.

Numbers don't lie, my friend. Math is one thing that can't be politicized, nuanced or denied by non-believers. It is what it is.