Question

 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. 

Answer 1

Given data; Weight of elevator W_E = 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 

= W_E – 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