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A close-coiled helical spring has a stiffness of 10 N/mm. Its length when fully compressed,with adjacent coils touching each other is 40 cm. The modulus of rigidity of the materialof the spring is 0.8x10 N/mm²(i) Determine the wire diameter and mean coil diameter if their ratio is 1/10. |
Answer» The VALUE of diameter is 20 cmExplanation:
Solution: Now d / D = 1 /10 Length of solid = n x d 400 = n x d n = 400 / d Now d^2 = 0.4 d^2 = 0.4 x 10^3 = 400 d = √400 = 20 mm = 2 cm D = 10 x 2 = 20 cm Thus the value of diameter is 20 cm
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