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# A steel rope has length L, area of crosssection A, Young’s modulus Y. [Density=d](a) It is pulled on a horizontal frictionless floorwith a constant horizontal force F = [dALg]/2applied at one end. Find the strain at themidpoint.(b) If the steel rope is vertical and moving withthe force acting vertically upward at the upper,end. Find the strain at a point L/3 from lowerend

Asked by Shanthiswaroop Srinivas 4th November 2017, 6:05 PM
(a)

Acceleration = Force / mass = [dALg]/(2[dALg]) = 1/2 m/s

Force at mid point = (m/2) × (1/2) N = ([dALg]/2) × (1/2) N = [dALg]/4 N

(b)
For part(b) it is assumed same force given for part(a) acting vertically.
By taking account of gravitatational force [dAlg], the net downward force for the
whole steel rope is [dAlg]/2 and this gives net downward acceleration 1/2  m/s2

Force for the lower L/3 part is  ( [dAlg]/3 ) × (1/2) N

By working out for strain as given in part(a) we will get

Answered by Expert 13th December 2017, 7:42 PM
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