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Grade: 12th pass

                        

A person of mass 8 kg enters a lift and selects the floor he wants . The lift now accelerates at 1 m/sec^2 for 2 sec , then goes with constant vel . As the lift approaches his floor , it decelerates at the same rate it accelerated . If the lift cable can withstand a tension of 2x 10^4 N and the lift itself has maas of 500 kg , how many persons can it safely carry at once .

3 years ago

Answers : (1)

Eshan
askIITians Faculty
2095 Points
							Dear student,

The stress on the cable will be maximum when the lift is accelerating upwards. From the equation of motion for the body,

T_{max}=(M+m_{max})g+(M+m_{max})a
\implies T_{max}=(M+m_{max})(g+a)
\implies 500kg+m_{max}=\dfrac{2\times 10^4}{10+1}kg=1818kg
\implies m_{max}=1318kg
Hence number of people it can carry=\dfrac{1318}{8}=164
2 years ago
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