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tfguss
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I'm not sure where to start. The drawing shows two crates that are connected by a steel wire that passes over a pulley. The unstretched length of the wire is 2.0 m, and its crosssectional area is 1.6 *10^5 [m^2]. The pulley is frictionless and massless. When the crates are accelerating, determine the change in length of the wire. Ignore the mass of the wire.
 2 years ago
 2 years ago
tfguss Group Title
I'm not sure where to start. The drawing shows two crates that are connected by a steel wire that passes over a pulley. The unstretched length of the wire is 2.0 m, and its crosssectional area is 1.6 *10^5 [m^2]. The pulley is frictionless and massless. When the crates are accelerating, determine the change in length of the wire. Ignore the mass of the wire.
 2 years ago
 2 years ago

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tfguss Group TitleBest ResponseYou've already chosen the best response.0
Attached is the image. Please help!
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.1
your goal is tension ?
 2 years ago

tfguss Group TitleBest ResponseYou've already chosen the best response.0
Young's modulus of steel is 200gPa I think how do I factor in the weight? I tried the usual deformation formula using 9.81*(m1+m2) as the force in newtons, but this does not yield the answer. I'm confused.
 2 years ago

banditelol Group TitleBest ResponseYou've already chosen the best response.0
what is the cross sectional area means?
 2 years ago

tfguss Group TitleBest ResponseYou've already chosen the best response.0
That is the area across the end of the wire, aka you cut the wire in half and its the area of the circle at the end of the wire.
 2 years ago

banditelol Group TitleBest ResponseYou've already chosen the best response.0
oooo.. I think you should just search the tension by using the (sigma)F=ma for both crates, and just use the modulus young equation to search for the change of length
 2 years ago

banditelol Group TitleBest ResponseYou've already chosen the best response.0
dw:1329064147162:dw in which, Y=modulus young F= the tension(in this case I think 2 times the rope's tension) A = cross sectional area l = the original length (delta) l = the change of length
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.1
do you know relation of atwood acceleration & tesion ? use from that
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.1
\[T=2m _{1}m _{2}g/m _{1}+m _{2}\] for more see this: http://en.wikipedia.org/wiki/Atwood_machine
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.1
did you get it?
 2 years ago

tfguss Group TitleBest ResponseYou've already chosen the best response.0
Yes! Thank you!
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.1
your wellcome
 2 years ago
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