CDC Latenight Lounge

[Canuckette_84]

i hate math.

[Super19]

i hate math.

[JoeyJoeJoeJr. Shabadoo]

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

This works too. The force will be 2126.66lbs/ft if the force is applied to the long side of the lever. 22.76 lbs/ft if is on the closer side

[-Goose-]

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

That's what I'm wondering about too. I have no knowledge on carpenters math but I'm going to add my own physics two cents. I'd say the 220lbs of force is synonymous with the lift force assuming we have an object (in this case the fulcrum) in contact with the ground. That in itself has a whole different set of rules because now only a part of the system is touching the ground.

[PowerIce]

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.

[-Goose-]

i hate math.

I hate literature

[JoeyJoeJoeJr. Shabadoo]

i hate math.

Math speaks highly of you. It physics that cant stand ya (:

[PowerIce]

This is all the information included in the question.

[JoeyJoeJoeJr. Shabadoo]

That's what I'm wondering about too. I have no knowledge on carpenters math but I'm going to add my own physics two cents. I'd say the 220lbs of force is synonymous with the lift force assuming we have an object (in this case the fulcrum) in contact with the ground. That in itself has a whole different set of rules because now only a part of the system is touching the ground.

its very common for trades training to include a math component that includes simple machines in this case being the lever. I wouldnt try to over complicate this one i think its pretty straight forward.

[-Goose-]

Yea, I'm going to leave this with Joey and >you as I really don't know what's goin on. I'll take a look at it though and see if I can come up with anything.

[Barney Stinson]

You rang?

[JoeyJoeJoeJr. Shabadoo]

This is all the information included in the question.

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 or youll go hacksaw jim duggan on him.

Edited October 12, 2010 by JoeyJoeJoeJr. Shabadoo

[>you]

This is all the information included in the question.

im assuming the lift force is the answer i came up with the 2126.66 it makes sense to me, you apply the down force on the long end of the lever and the short end is the one that is lifted

[PowerIce]

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

Edited October 12, 2010 by SmirnoffIce

[>you]

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?

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

.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

[PowerIce]

.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?

Edited October 12, 2010 by SmirnoffIce

[>you]

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?

1026.3 round up the .26

i dont know where you get the 4.83331818 from but my calculator has it as 4.833 repeating. however many decimal places you use for the final equation determines where the .26 goes, if you use 4.8333 you get 1063.326 which rounds up to 1063.33.

and yes .375 is the decimal equivalent for 4 and a half inches

make sure you put the units on the question because you switched from inches to feet. i did that because the inches creates an unnecessarily long answer.

[d^/ > _ < \^b]

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[-Goose-]

Anon is a mysterious person.

Good night.

[>you]

im anonymous