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March 26, 2017

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An elevator (mass 4700 kg) is to be designed so that the maximum acceleration is 6.00×10^-2 g

What is the maximum force the motor should exert on the supporting cable?

What is the minimum force the motor should exert on the supporting cable?

  • Physics - ,

    forcemax=mass*amax=4700(.06+1)g

    force min=mass*g that is the force to hold the elevator still.

  • Physics - ,

    I've tried both of those. Does not work.

  • Physics - ,

    ma = tension - weight so ma + weight = tension so 4700(.06)(9.8) + 4700(9.8)

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