An elevator (mass 4850 kg) is to be designed so that the maximum acceleration is 0.0680 g. What are the maximum and minimum forces the motor should exert on the supporting cable?
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This is Giancoli Answers with Mr. Dychko. The supporting cable exerts a tension force upwards and then there is gravity downwards. And the acceleration of the elevator is a maximum of 0.068 g and that's either upwards or it's downwards. And so, will the case where the acceleration is upwards is when this tension force will be at its maximum. And the case where acceleration is downwards is when this tension will be at its minimum. And so, we'll solve for those two cases here a greater than zero, less than zero. So, we will turn this acceleration into m/s squared by multiplying 0.0680 g by... Well, I could probably write that a bit more clear. I can say, 0.0680 g times 9.8 m/s squared for every g. And you can see the g's cancel, and so we have 0.0680 times 9.8. That gives 0.666 m/s squared either positive or negative. So, the net force on this elevator is the tension force up minus the gravity down, and that total has to equal mass times acceleration. And we can solve for tension force by adding Fg to both sides. So, we have ma plus Fg. So, that's tension force equals ma plus mg. And we can factor out the m's just to make it look a little more clean. And we have the tension force maximum is when the elevator is accelerating upwards and that's the elevators mass, 4850 kilograms times 9.8 m/s squared g plus the acceleration up, 0.664 m/s squared and that gives 5.08x10^4 Newtons. And when acceleration is downwards, the tension force will be the mass times 9.8 minus 0.6664 m/s squared which gives 4.343x10^4 Newtons of tension force.
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