Start » Exercises » Engineering Mechanics I » Reaction forces of a hook wrench

# Reaction forces of a hook wrench

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:

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

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

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