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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 floor
with a constant horizontal force F = [dALg]/2
applied at one end. Find the strain at the
midpoint.
(b) If the steel rope is vertical and moving with
the force acting vertically upward at the upper,
end. Find the strain at a point L/3 from lower
end
  

Asked by Shanthiswaroop Srinivas 4th November 2017, 6:05 PM
Answered by Expert
Answer:
(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
 
begin mathsize 12px style S t r a i n space equals space fraction numerator S t r e s s over denominator Y o u n g s space M o d u l u s end fraction space equals space fraction numerator begin display style bevelled fraction numerator F o r c e over denominator A r e a end fraction end style over denominator Y o u n g s space M o d u l u s end fraction

S t r a i n space equals space fraction numerator begin display style bevelled fraction numerator d A L g over denominator 4 A end fraction end style over denominator Y end fraction space equals space fraction numerator d L g over denominator 4 Y end fraction end style
 
 
(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

begin mathsize 12px style Strain space equals space minus fraction numerator dlg over denominator 6 straight Y end fraction
end style

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