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A 5000kg elevator is to be designed so that the maximum acceleration is 10% og gravitational acceleration. What are the maximum and minimum forces the motor should exert on the supporting cable? |
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Answer» imum FORCE is 55000 N,The minimum force is 50000 N EXPLANATION:Given as :The MASS of the elevator = m = 5000 kgThe gravitational acceleration = = 10 m/s²The maximum acceleration = = 10% of Let The maximum force of motor = NLet The minimum force of motor = NACCORDING to question∵ Force = Mass × accelerationFor maximum force∵ = 10% of i.e = 10% × 10 m/s²So, = 1 m/s² Now, Maximum force = mass × maximum acceleration So, = 5000 KG × ( + 10)Or, = 5000 kg × 11 m/s² ∴ = 55000 NAgainMinimum force = mass × minimum acceleration So, = 5000 kg × Or, = 5000 kg × 10 m/s² ∴ = 50000 NHence, The maximum force is 55000 N,And The minimum force is 50000 N Answer |
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