PowerIce
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Posts posted by PowerIce
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I really wanted to reply to this thread but it got locked.
http://forum.canucks.com/topic/289553-grade-two-teachers-name-one-good-memory/
Anyways, my grade 1 teacher was called Mrs. Willow
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Toooonight amma fight till we see the suuunlight
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Got kicked out of Riverrock again tonight.
The sucky thing is that they waited until I lost my last dollar to do it.
When you get kicked out, is that permanently or just for the night?
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Turning 38!
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On the bright side of this night.
I've gained 7x my original chip count
I started with 140k. I'm over a million right now. I wish this was real life
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Let's talk about the Cleveland Browns here.
New QB? Fire the front office? What can be done to help this 1-4 team?
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.5 feet represents the 6 inches that is the distance on the other end of the fulcrum. the 4.833 is the 58 inches on one end of the rod the .5 is the 6 inches on the other end. the end with 6 inches is the end that is doing the lifting
each side of the lever has it's own Force and distance. you really dont need to add in the torque the equation could be simplified to
F1d1 = F2d2
220(4.833) = F(.5)
you can keep it in inches too
220(58) = F(6)
probably easier to work with that way
Ohh okay, that's what I was wondering. On the next question it is 4.5 inches from the end. So I would divide my F by .375?
220(4.833) = 1063.26
1063.26/.5 = 2126.52
Ah makes sense, except you had 1063.33 instead of 1063.26
If you multiplied 220 by 4.3331818 you would end up with your answer, 1063.33
And if you multiplied 220 by just 4.833 that's how you would end up with my answer in 1063.26
Which one would be the correct choice?
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the equation for torque i think is t=Fd, where t is the torque. f is the force exerted, which is 220 lbs in the first question. d is the distrance from the fulcrum, which would be 58 inches. which you could switch to 4.833 feet since torque is more often measured in foot pounds than inch pounds
t=220(4.833)
t=1063.33 foot pounds of torque
i have no idea what the lift force your talking about is though
if we reverse the equation though we can find the force exerted on the other end of the rod
1063.33=F(.5ft)
F =1063.33/.5
F= 2126.66
When I multiplied 220 by 4.833 I didn't get 1063.33 but 1063.26
Of course, you must have multiplied it by 4.83331818 and not 4.833. Why do I divide by .5ft? Wouldn't it be .483331818 instead?
58/6 x 220 or
6/58 x 220 depending on the location of the force in relation to the fulcrum. Unfortunately this question leaves out that little tidbit so give both answers and tell the instructor to get some better problems
Those are the ways I thought it would be done. Except I wouldn't take away the 6" from the 64" straight bar. So the equation would look like 64/6 x 220 or 6/64 x 220
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This is all the information included in the question.
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Thanks a ton guys.
Somehow the 6 inches needs to come in play though. I have to divide or multiply something. I don't think it's a trick question, their hasn't been any all year. Yep, we do. There's 2 questions like this actually, the next one is: If a 180 pound for is applied to a 60 inch straight bar that is pivoted 4.5 inches from the end, what life force is exerted?
you still never answered my question.
you dont have to do stupid crap like that in carpentry school do you?
That was me answering your question.
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If the force being applied is closer to the fulcrum there will be a force reduction and an increase in distance/speed. If it is further away the opposite is true.
I see. I just don't get how I format my equation. Do I divide the force by the amount pivoted from the end or divide the force by the length of the straight bar?
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you dont do stupid crap like that in carpentry school do you
im pretty sure though its a trick question 220 lbs is the force that is used to lift. the forces on either side of the pivot point need to be equal.
and i thought lift force reffered to airplanes and such
Somehow the 6 inches needs to come in play though. I have to divide or multiply something. I don't think it's a trick question, their hasn't been any all year. Yep, we do. There's 2 questions like this actually, the next one is: If a 180 pound for is applied to a 60 inch straight bar that is pivoted 4.5 inches from the end, what life force is exerted?
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yes
what question
If a 220 pound force is applied to a 64 inch straight bar that is pivoted 6 inches from the end, what lift force is exerted? It's helpful to draw a diagram to help solve.
I need the equation to solve this question.
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To answer my question?
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Anybody up for some math questions?
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We need a separate website for GDT's.
Get on it Gilli... Stealth!
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Of course, you must have multiplied it by 4.83331818 and not 4.833.
CDC Latenight Lounge
in White Noise
Posted
Was that your best memory of grade 2?
You may elaborate on your story here. Please, go ahead.