This exercise shows how to calculate the resulting moment from several forces acting under different angles in a plane state.

## Task

There are three forces attacking on a cranked body. Which moment results from the forces F_{1}, F_{2} and F_{3} regarding to point P?

## Solution

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

### Solution sketch

### Approach

In the following solution, counterclockwise torques are defined as positive. In the solution sketch, the lever arms for the forces F_{2} are marked in green (distance e) and F_{3} light orange (distance f).

The resulting moment in point P is

\[\tag{1} M_P = F_1 \cdot a -- F_2 \cdot e + F_3 \cdot f \]

The lever arm e results from:

\[\tag{2} e = d \cdot \sin \beta \]

The distance d results from

\[\tag{3} d = \frac{b}{\cos \alpha} \]

Inserted in equation 2 it gives

\[\tag{4} e = \frac{b}{\cos \alpha} \cdot \sin \beta \]

The segment f results from

\[\tag{5} f = \frac{b+c}{\cos \alpha} \]

The resulting moment at point P is finally

\[\tag{6} M_P = F_1 \cdot a -- F_2 \cdot \frac{b}{\cos \alpha} \cdot \sin \beta + F_3 \cdot \frac{b+c}{\cos \alpha} \]

So this is the resulting moment from several forces regarding point P. Maybe you want to try for yourself to define another point, maybe the base of F_{1} and calculate the resulting moment for this case. Is it the same value?

There is another exercise for the three-dimensional calculation of resulting moments and forces where we show the calculation of the moments for two different points at a frame. You can find this exercise here.

Don’t miss the other exercises regarding Engineering Mechanics 1!