You are designing a simple elevator systemfor an old warehouse You are designing

Question

You are designing a simple elevator systemfor an old warehouse

You are designing a simple elevator system for an old warehouse that is being converted to loft apartments. A 22,500-N elevator is to be accelerated upward by connecting it to a counterweight by means of a light (but strong!) cable passing over a solid uniform disk-shaped pulley. The cable does not slip where it is in contact with the surface of the pulley. There is no appreciable friction at the axle of the pulley, but its mass is 819 kg and it is 1.70 m in diameter. What mass should the counterweight have so that it will accelerate the elevator upward through 6.75 m in the first 3.00 s, starting from rest? Express your answer with the appropriate units. Under these conditions, what is the tension in the cable on the elevator side of the pulley? Express your answer with the appropriate units. Under these conditions, what is the tension in the cable on the counterweight side of the pulley? Express your answer with the appropriate units.

Explanation / Answer

Applying Newton’s second law to mass m1,

m1g – T1 = m1a

T1= m1g – m1a ----------(1)

Applying Newton’s second law to mass m2,

T2 – m2g = m2a

T2 = m2g + m2a ----------(2)

T2r- T1r = I = I(a/r) ----------(3)

From (1), (2) and (3)

m2g + m2a - m1g + m1a = Ia/r^2

a = [m1g – m2g)/(m1+m2+ Ia/r^2)

I=1/2Mr^2 = 1/2*819*1.70^2 = 1183.5 kg*m^2

M= mass of the pulley,

m1 = counterweight

m2 = mass of the elevator = 22500N/9.8m/s^2 = 2295.9 kg

A) To calculate a use equation,

d=vi*t + 1/2at^2

6.75 = 0*3.0 +1/2*a*3.0^2 => a = 1.5 m/s^2

Plugging values,

a = [m1g – m2g)/(m1+m2+ Ia/r^2)

1.5 = (m1*9.8 – 2295.9*9.8)/(m1+2295.9+1183.5)   => m1 = 3339.6 kg

B)

From (1)

T1= m1g – m1a = 3339.6(9.8 – 1.5) = 27718.7 N

C) From (2),

T2 = m2g + m2a = 3339.6(9.8 + 1.5) = 37737.7 N

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