An elevator (mass 4100 kg) is to it is in designed so the the maximum acceleration is 0.0300g.What is the maximum pressure the motor have to exert on the sustaining cable?\sum F=m.03gT=mg(1+.03)41385 N What is the minimum pressure the motor need to exert on the supporting cable? I assumed this would simply be weight?? but that is no correctNI am plainly misunderstanding what the inquiry is askingCasey

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What is the minimum pressure the motor have to exert on the supporting cable? I assumed this would just be weight??
No. If the motor detailed a pressure equal come the weight, climate the elevator would certainly not move. The force of the motor would certainly be well balanced by the weight.When one feel an acceleration that is a differential acceleration, i.e. The is in enhancement to the acceleration that gravity.So T = m(g+a) = mg (1 + 0.03) = 4100 kg * 9.8 m/s2 * (1.03) = 41385.4 N
No. If the motor detailed a pressure equal come the weight, climate the elevator would not move. The force of the motor would be balanced by the weight.When one feel an acceleration that is a differential acceleration, i.e. That is in addition to the acceleration of gravity.So T = m(g+a) = mg (1 + 0.03) = 4100 kg * 9.8 m/s2 * (1.03) = 41385.4 N
But the is what i did use (exactly^^^) because that the best force. The is the minimum that i am involved with.I acquired 41385.4 N because that MAX force. And it to be correct.Casey
I don"t know. It is just one of those net assign problems. That is for a friend of mine that i am helping. I never ever used webassign in school. But I have the right to say that i don"t choose it. Fifty percent the time, ns cannot phone call what they space asking.Why? Where room you getting the 38975 from? What is her stream of assumed here?Casey
I the elevator was dropping v an acceleration, then the stress would bg m (g-a) or mg (1-0.03) = mg (0.97).I to be trying to understand the question you to be being asked.There space two means to feeling the enhanced weight as soon as standing in one elevator: 1) when it start upward v some acceleration a indigenous rest, or 2) as soon as it slows down to a stop during a descent. Once the elevator is falling (with some acceleration) the anxiety is less than the weight when it is stopped.I"m simply trying to know the recommendation to minimum in this problem.
I the elevator was dropping v an acceleration, then the tension would bg m (g-a) or mg (1-0.03) = mg (0.97).I to be trying to recognize the question you to be being asked.I"m simply trying to recognize the reference to minimum in this problem.

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Well. The was the correct answer Astronuc!!! I never ever would have thought that THAT was what was implied by the question. Therefore let me gain this straight. The REAL question implied is what is the minimum pressure that the motor have the right to exert and still have the elevator speeding up at .03g?Does the make sense?Casey