The Tripod Problem: An Example of Vector Forces

This is a simple problem which we will use to illustrate the application of three-dimensional vector statics. The problem comes from Movnin and Izrayelit (1970).

Consider the tripod above. The legs are evenly spaced (120 degree spacing) and are angled at 30 degrees from the vertical (z-axis.) A weight of 10 kN is hung from the top of the tripod, as shown. Determine the force transmitted through each of the legs.

We need to start by adding a proper coordinate system to the system, which is done at the right. We “cleverly” orient the coordinate system so that one of the x-axes runs through one of the legs, in this case the “C” leg. We’re not told how tall the tripod is but we really don’t need it; since we’re using formal Cartesian vector mechanics, we’ll normalise the force triangles for a unit vector. So let’s start with the “C” leg and note the following:

The distance from the origin to point C is unity (1).
Since we have a right triangle with the origin, Point C and Point B as corners, the hypotenuse (leg BC) has a length of 2 and the z-axis distance from B to the origin is $\sqrt{3}$ . The x,y and z coordinates of Point C are thus $(1,0,0)$ .
The other two triangles have the same geometric structure, the same hypotenuse length and the same distance from the origin to Point B. For leg BD, the coordinates for Point D are $-0.5, \frac{\sqrt{3}}{2},0$ . For leg BA, the coordinates for Point A are $-0.5,-\frac{\sqrt{3}}{2},0$ . The coordinates of Point B are $0,0,\sqrt{3}$ .
Because of the symmetry of the tripod, the three leg forces have the same modulus/magnitude.

To obtain the unit vectors for the three tripod leg forces, we must employ the following formula:

u_n = ((x₂-x₁)i + (y₂-y₁)j + (z₂-z₁)k)/l_hypotenuse (1)

The summation of forces at Point B is

S_BA + S_BC + S_BD – P = 0 (2)

If we compute the unit vectors from Equation (1) from the geometric data given and substitute them into Equation (2), the result is

S_BC(-0.5i + 0.866k) + S_BA(0.25i – 0.433j + 0.866k) + S_BD(0.25i + 0.433j + 0.866k) = (3)(S)(0.866k) = 10k (3)

Note the difference between the scalar moduli of the three leg vectors in Equation (3) and the vector values for those in Equation (2). Note also that all three of these moduli are equal by symmetry, so they can be expressed by the scalar S. The result comes as the i and j components cancel each other out and k divide out, yielding S = 3.85 kN, a compressive force on each leg.

A similar problem can be found in the post An Example of 3D Vector Statics With a Simple Truss.