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 An elevator weight 2500N and is moving vertically downward with a constant acceleration. 

(1) Write the equation for the elevator cable tension. 

(2) Starting from rest it travels a distance of 35m during an interval of 10 sec. Find the cable tension during this time. 

(3) Neglect all other resistance to motion. What are the limits of cable tension. 

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Given data; Weight of elevator WE = 2500N 

Initial velocity u = 0 

Distance traveled s = 35m 

Time t = 10sec 

(1) Since elevator is moving down 

Net acceleration force in the down ward direction 

= WE – T = (2500 – T)N ...(i) 

The net accelerating force produces acceleration ‘a’ in the down ward direction. 

Using the relation, F = ma 2500 – T = (2500/9.81)a 

T = 2500 – (2500/9.81)a

Hence the above equation represents the general equation for the elevator cable tension when the elevator is moving downward. 

(2) Using relation

s = ut + 1/2 at2 = 35 = 0 X 10 + 1/2 X a (10)2 ...(ii)

∴ a = 0.7 m/sec2

Substituting this value of a in the equation of cable tension 

T = 2500 – (2500/9.81) X 0.7T = 2321.61N

(3) T = 2500 – (2500/9.81)a

Limit of cable tension is depends upon the value of a, which varies from 0 to g i.e. 9.81m/sec2 At a = 0, T = 2500

i.e elevator freely down

At a = 9.81, T = 0

i.e elevator is at the top and stationary. 

Hence Limits are 0 to 2500N

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