## Task

A hook wrench is attached to a stationary shaft. Which reaction forces arise in A and B under the assumption that the contact in B is frictionless?

## Solution

The following video is in german language. Please scroll down for the written solution.

\[ \require{cancel} \] \[ \newcommand{\myvec}[1]{{\begin{pmatrix}#1\end{pmatrix}}} \]The reaction forces on the shaft are shown in the sketch of the solution. The hook wrench touches the shaft at points A and B. On a spatial shaft a line contact would occur.

The equilibrium of forces in the x-direction is

\[\tag{1} \sum F_x = 0 = F_B - F_{Ax} \]

\[\tag{2} F_{Ax} = F_B \]

The equilibrium of forces in the y-direction is

\[\tag{3} \sum F_y = 0 = F_{Ay} - F \]

\[\tag{4} F_{Ay} = F \]

The equilibrium of moments is established around the axis of rotation of the shaft. The moment applied by the force F is referred to as M _{ F }, so that the moment equilibrium reads

\[\tag{5} \sum M = 0 = - M_F + F_A \cdot \frac{d}{2} \]

\[\tag{6} F_A \cdot \frac{d}{2} = F \cdot l \]

\[\tag{7} F_A = \frac{2 \cdot F \cdot l}{d} \]

The still unknown horizontal forces are calculated using the Pythagorean theorem:

\[\tag{8} F_A^2 = F_{Ax}^2 + F_{Ay}^2 \]

\[\tag{9} \left( \frac{2 \cdot F \cdot l}{d} \right)^2 = F_{Ax}^2 + F^2 \]

\[\tag{10} F_{Ax} = \sqrt{\left( \frac{2 \cdot F \cdot l}{d} \right)^2 - F^2} \]