To better understand camming devices, the forces between the cam and the rock wall need to be well understood. Assume a cam with camming angle theta is placed in a perfectly parallel crack. If a downward force, T, is placed on the axles of the cam, what are the resulting forces?

Let us assume we are analyzing a cam with four lobes (in the image above we can only see two lobes, but there is another identical pair of lobes behind the pair we see).

To simplify the problem, let us look at the forces from the cam lobe onto the rock at a single lobe-rock interface as shown below.

Let us call FT the force from the cam onto the rock applied tangentially to the rock face and let us call FN the force from the cam onto the rock applied perpendicular to the rock face. By symmetry and trigonometric identities, we know:

FT = T/4

FN = FT / Tan(θ) = (T/4) / Tan(θ)

Now let us analyze the reaction forces from the rock face and use force balance to find implications of this system. In the diagram below, the central origin represents the point at which a single lobe and rock make contact.

In the x-axis, we know that the rock face will apply an equal and opposite reactionary force to the cam lobe, so that the normal force, N, equals FN.

In the y-axis, we know that the frictional force, FF, equals μN, where μ equals the coefficient of friction between the lobe-rock interface; therefore,

FF = μN = μFN = μ(T/4) / Tan(θ)

In order for the cam to hold, FF ≥ FT; therefore, μ(T/4) / Tan(θ) ≥ T/4

This simplifies to the design requirement that μ ≥ Tan(θ).