Calculation of wire rope tension
It is as if each side of the wire rope holds up one half of the weight, sharing the weight between them. However, if the ends were pulled apart, but kept level, the wire's tension would increase.
Each side of the wire rope would no longer be fighting just the gravitational force pulling down on the hanging body, but also the opposing lateral, or horizontal, force from coming from the other side of the wire rope. This is a direct result of the two sides changing from a vertical aspect to a V shape.
Draw a diagram of a weight hanging from the center of a wire rope. Calculate the gravitational force pulling the mass down as F=mg = m x 9.80m/sec^2, where the caret ^ indicates exponentiation. g is the gravitational acceleration constant.Denote the weight's mass with the letter m. Denote the angle from the vertical that each side of the wire makes with the letter θ.
Equate the vertical component of the tension T with which side of the wire rope pulls up with half the hanging weight. Therefore T x cos θ = mg/2. For example, that the angle between one side of the wire rope and its vertical support wall is 30 degrees. Suppose also that the weight has a mass of 5 kg. Then the equation becomes T x √3/2 = [ 5 kg x 9.80 m/s^2 ] / 2.Solve for T, using the equation you just derived, remember to round to the proper number of significant figures. Continuing with the above example, you get tension T = 28.3N.
If a body hangs off the center of a wire rope whose ends attach a negligible distance from each other, then the tension in thewire rope is half the weight of the body.
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