• LAVA Moderator: Mysterier

⫸STICKY⫷ 😃 LAVA Chit-Chat Thread! 🤪

Buying dress clothes...and further, being subjected to wearing them too...is something I don't wish on even my worst of enemies.

Do you wish them on the floor, next to your enemies' beds? Maybe with some smooth jazz playing and dimly lit red lights?
 
^Now that you have quite a bit of lifting under your belt, I can't imagine you're too comfortable in dress clothes either
 
^Now that you have quite a bit of lifting under your belt, I can't imagine you're too comfortable in dress clothes either

I just walk around the local mall until I find an attractive woman working in a tailor shop who can help measure me, tailor my clothes to fit my nonstandard frame, and then let me wine, dine and show her the divine...
 
Anyway...I wonder if I'm still on good enough terms with some of my old profs to take the Putnam under some other student's name.
 
Ugh...the LSAT.


I dunno. Maybe if I spend a few hours a day practicing for the next month or two I could crack 140
 
I'm really off my game again this morning.

Between our social-to-social whoring, I mistook Leftwink for you. Why did you have to go and change your avatar?
 
I think I'm doing something wrong:

"An old oaken bucket of mass 6.75 kg hangs in a well at the end of a rope. The rope passes over a frictionless pulley at the top of the well, and you pull horizontally on the end of the rope to raise the bucket slowly a distance of 4.00m"

How much work do you do on the bucket?

Shouldn't it be W=Fd=mgd=6.75(9.8)(4)=264.6
The book says it should be 3.60J
 
Fire away! I have 4 hard ciders in my belly, though. So we'll see how it goes. "lol."
 
A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.25.

What magnitude of force must the worker apply?

if W=Fd
d=4.5m

where would I get W from to solve for F?
 
Horizontal force, so:

pushing force [work] = friction

friction = uN = u(mg)

So we have

friction = 0.25x30kgx9.8m/s/s = 73.5N

work done = (force)(distance)

so

73.5N x 4.5m=330.75J
 
Mkay, now that I see it done it all makes sense. For some reason when I read the problem my mind goes blank and I have no idea how to apply what I just learned. And thank you again. I really appreciate it.
 
No problem. Again, and I didn't really appreciate this freshman year either, but free body diagrams can save your life :)

Keep asking if you want. I don't know how much more I got in me, but if not tonight, I can answer tomorrow morning!
 
Free body diagrams are a fantastic way for me to figure out what's going on in a problem otherwise I'm completely lost.
If I have any more questions I'll post them :)
 
BA, did you ever watch the Howard Stern show when he tormented the PETA women? I felt HORRIBLE enjoying that. Soul = black.
 
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