If a body hangs off the centre of a wire whose ends attach a negligible distance from each other, then the tension in the wire is half the weight of the body. It is as if each side of the wire holds up one half of the weight--as if they attached to the body at two places, 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 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. This is a direct result of the two sides changing from a vertical aspect to a V shape, as discussed in Halliday and Resnick’s introductory “Fundamentals of Physics.”

Draw a diagram of a weight hanging from the centre of a wire. Denote the weight’s mass with the letter m. Denote the angle from the vertical that each side of the wire makes with the Greek letter ?.

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.

Equate the vertical component of the tension T with which side of the wire pulls up with half the hanging weight. Therefore T x cos ? = mg/2.

Suppose, for example, that the angle between one side of the wire and its vertical support wall is 30 degrees. Suppose also that the weight has a mass of 5kg. Then the equation becomes T x ?3/2 = [ 5 kg x 9.80m/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.

#### Tip

If the weight was not in the centre, you’d have to set the sum of the vertical force components equal to the gravitational force on the hanging weight to solve for tension T. The formula would look like this: T x cos ?1 + T x cos ?2 = mg, where ?1 and ?2 are the angles the two sides of the wire made from vertical.