An elevator in a mine shaft is supported by a steel cable ofdiameter 2.5 cm. The

Question

An elevator in a mine shaft is supported by a steel cable ofdiameter 2.5 cm. The total mass of the elevator and passengers is670 kg. By how much does the cable stretch when the elevatoris a) 12m below the attachment point. b) 500m below. Note: Young's modulus for steel: Y = 2.0 X 10^11 N/m^2 Thanks. An elevator in a mine shaft is supported by a steel cable ofdiameter 2.5 cm. The total mass of the elevator and passengers is670 kg. By how much does the cable stretch when the elevatoris a) 12m below the attachment point. b) 500m below. Note: Young's modulus for steel: Y = 2.0 X 10^11 N/m^2 Thanks.

Explanation / Answer

THE FIRST POST, ABOVE, IS INCORRECT. HE ONLY CONSIDERED THEWEIGHT OF THE ELEVATOR, BUT THE CABLE ITSELF IS VERY HEAVY. THIS ISWHY YOU ARE ASKED TO CONSIDER 12 meters AND 500meters. YOU MUST ALSO INCLUDE THE EFFECT OF THE WEIGHT OF THE CABLEON THE CABLE ITSELF. See below... . This problem is a pain. First you have to consider that 500 meters of this cable has a mass of 1929 kg. So the weightof the cable itself is far greater than the weight of the elevator.Even just 12 meters of the cable has a mass of 46.3 kg, soits weight is not insignificant. . Technically, you have to consider the tension in the cable forevery point in the cable. Then you have to apply the stress /strain equation: .                tension / A = Y L /L     . Since you will consider every infinitesimal bit of the steelcable separately, you can call each little bit length "dy", anddefine the y direction to be vertical (so the cable runs along they axis). Then you can replace L with the differential dL and you get: .             tension / A =   Y dL / dy   . tension is a function of y...   the further you godown, the less the tension on the cable (because the tension is theweight of the cable below plus elevator). So you can separatevariables: .             tension dy = A Y dL . Or...    total stretch = integralof dL =   ( 1 / AY) integral of (tensiondy)      where  tension   is a function of y. . Also,   A is the cross sectional area of thecable (given by the diameter) and Y is youngs modulus(also given). . This problem is kind of a pain but... awww hell, I'll finishit. Hold on... .    So we define y = 0 to be at thebottom of the cable where the weight is just   mg = 670 *9.8 = 6566 newtons . And y = c     where  c   is the length of the cable,   at the top.At any point y then the tension is .    tension = weight of cable below pointy + 6566 =    .             = g * density of steel * area of cable * length of cable belowy + 6566 .           =  9.8 * 7860 * * 0.01252 *y   + 6566 =   37.811 y + 6566 . so now... . total stretch = ( 1 / AY) integral of(tension dy)   = ( 1 / *0.01252 * 2 x 1011 ) * (37.811 y+ 6566) dy .            =    ( 1.0186 x 10-8 ) (18.9055 y2   + 6566 y)         where we evaluatefrom 0 to c .    total stretch = (1.0186 x10-8 ) (18.9055 c2 + 6566c)   . So when the cable is 12meters, the total stretch is .              (1.0186 x 10-8 ) (18.9055 * 122 + 6566 * 12) =   0.830 mm . And when the cable is500 meters... .        (1.0186 x 10-8 ) (18.9055 * 122 + 6566 * 12) =   8.16 cm . Notice... the personabove got    0.803 mm for the first answer... mineis just a little more, because the cable is only 12 meters long andstretches just a little bit more because of its weight. But thesecond answer... he got 3.345 cm, while mine is 8.16 cm. Thecable stretches much more when it is 500 meters long because of itsown weight. . NOTE: if someone tells you my answers are wrong, it isprobably because they are not including the weight of the cable!!Ask them "Are you including the weight of the cable?" and you canask if they did an integral to get their answer. The cable is veryheavy, so it stretches itself. . Let me know if there's a problem. I'm always willing tocorrect my work if there's an error.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.