A three-dimensional solid acted on by four constant forces is in equilibrium. No two lines of force are in the same plane. Show that the four lines of force are rulings on a hyperboloid.

**Solution**

Let the lines be L_{1}, L_{2}, L_{3}, L_{4}. If the line L intersects each of L_{1}, L_{2}, L_{3}, then the moment of the first three forces about L is zero. Hence the moment of the fourth force about L is also zero and so L must also intersect L_{4}. This is sufficient to prove that L_{1}, L_{2}, L_{3}, L_{4} lie in a ruled quadric.

In fact the union of all the lines which intersect L_{1}, L_{2}, L_{3} is a ruled quadric. Let P be a point on L_{4}. For a point X and a line L not containing it we take XL to be the unique plane containing X and L. If L and L' are skew lines and X does not lie on either, then XL ∩ XL' is the unique line through X which meets L and L'. So PL_{1} ∩ PL_{2} is the unique line through P which meets L_{1} and L_{2}. Suppose it meets L_{1} at Q. Now QL_{2} ∩ QL_{3} meets L_{1}, L_{2} and L_{3} so it must also meet L_{4}. But there is a unique line through Q meeting L_{2} and L_{4}, so QL_{2} ∩ QL_{3} = PL_{1} ∩ PL_{2}. Hence P lies on the line QL_{2} ∩ QL_{3} which is a line of the quadric. P was arbitrary, so we have shown that L_{4} lies in the quadric.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002