Anchors

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Category: Technical Rescue

Published on Sunday, 30 September 2012 21:16

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Anchors are typically composed of multiple elements, with the load distributed among them.  Here we will look at how that load gets distributed for a two element anchor.  Many rigging guides suggest using a maximum of 90° between elements of an anchor.  We will show that this is a good value to use.  

Forces and the Physics Trick

Vectors as described in the vector decomposition article can be applied directly to many examples.  If we want to look at forces in mechanical systems, we still need one more tool: Newton's second law of motion.  This law states "The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass."  We do not need to worry about masses or accelerations in our treatment of mechanical systems because we will assume that nothing is moving.  This means that the acceleration is zero, so the sum of all forces acting on any part of our system is zero.  Written as an equation,

\( \sum{\textbf{F}}=0. \)

So long as you don't take any falls, this assumption is typically valid for climbing and technical rescue system anchors.

Example - Climbing Anchor

The figure below diagrams the anchor we will be analyzing.  The forces and angles are left as variables, so that we will obtain a generic solution which can be used for a variety of values.  This example is for a two element anchor, with the main load, L, perfectly equalized between two anchor elements.  The load transferred to these elements is represented by A₁ and A₂.

To calculate the forces A₁ and A₂, we first sum the forces in the principal directions.

\( \sum{\textbf{F}_x} = -A_1 \times\sin(\theta_1)+A_2 \times\sin(\theta_2)=0 \)    Eq.1

\( \sum{\textbf{F}_y} = A_1 \times\cos(\theta_1)+A_2 \times\cos(\theta_2)=L \)    Eq.2

Solving Eq.1 for the variable A₁

\( A_1 = A_2 \times\frac{\sin(\theta_2)}{\sin(\theta_1)} \)    Eq.3

Substituting Eq.3 into Eq.2

\( A_2\times\frac{\sin(\theta_2)}{\sin(\theta_1)} + A_2\times\cos(\theta_2) = L \)

Solving for A₂ and simplifying

\( A_2 = \frac{L}{\frac{\sin(\theta_2)}{\tan(\theta_1)}+\cos(\theta_2)} \)    Eq.4

Substituting Eq.4 into Eq.1

\( -A_1\sin(\theta_1)+\frac{L}{\frac{\sin(\theta_2)}{\tan(\theta_1)}+\cos(\theta_2)}\times\sin(\theta_2) = 0 \)

Solving for A₁ and simplifying

\( A_1 = \frac{L}{\frac{\sin(\theta_2)}{\tan(\theta_1)}+\cos(\theta_2)}\times\frac{\sin(\theta_2)}{\sin(\theta_1)} \)

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