Hi all, Here\'s my question. It seemed simple atfirst. An elevator with passenge

Question

Hi all, Here's my question. It seemed simple atfirst. An elevator with passengers has a total mass of 1700 kg.What is the net force on the cable needed to give the elevator avelocity of 4.5 m/s [up] in 8.8 s if it is starting fromrest? Iinitially did this as a simple momentum impulse question F^t =m^v and got an answer of 870N [up] which is the answer in theback of the text. But now I'm wondering about the mass of thehanging elevator. The 870N would be the answer for ahorizontal momentum change but the elevator is hanging.Shouldn't it be + mg? I also redid the question as a netforce = sum of forces: Fnet= T + mg ma =T + mg (1700)(4.5/8.8) = T + (-9.8)(1700) T =17500N Isit 870N or 17500N?? Thanks in advance for your time Hi all, Here's my question. It seemed simple atfirst. An elevator with passengers has a total mass of 1700 kg.What is the net force on the cable needed to give the elevator avelocity of 4.5 m/s [up] in 8.8 s if it is starting fromrest? Iinitially did this as a simple momentum impulse question F^t =m^v and got an answer of 870N [up] which is the answer in theback of the text. But now I'm wondering about the mass of thehanging elevator. The 870N would be the answer for ahorizontal momentum change but the elevator is hanging.Shouldn't it be + mg? I also redid the question as a netforce = sum of forces: Fnet= T + mg ma =T + mg (1700)(4.5/8.8) = T + (-9.8)(1700) T =17500N Isit 870N or 17500N?? Thanks in advance for your time

Explanation / Answer

Seems you are correct T= m( g +a) a= 4.5/8.8= 0.511m/s2 T= 1700( 9.81 +0.511) T=Fnet= 17,500N (force on the cables) and as you have said F= ma= 1700 x 0.511 = 870N (I just do'nt know how you got T =48000N ) (I just do'nt know how you got T =48000N )

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