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
(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/s2
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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