This exercise is our introductory exercise in strength theory and deals with the following questions:

- How to calculate the normal force in a member?
- How to establish an equilibrium of forces with internal forces?

## Task

A weight G_{2} is attached to a rod. The weight G_{1} is applied at connection point C. The members 1 and 2 are fastened with fixed bearings. The members are without mass.

The normal forces that occur in members 1 to 3 have to be calculated.

## Solution

The normal forces are referred to as N and the index of their member number. The normal force in member 1 is the easiest to calculate.

The equilibrium of forces in the y-direction yields

\[ \tag{1} \sum F_y = 0 = N_1 - G_2\]

\[ \tag{2} N_1 = G_2\]

Now, starting from point C, mentally cut through the members and apply their respective normal forces. A small trap lurks here in this task, because the force G_{1} must of course not be missing in this free cut.

The two normal forces in members 2 and 3 can now be calculated using the force equilibria in the x and y directions.

\[ \tag{3} \sum F_x = 0 = - N_3 \cdot sin \alpha + N_2\]

\[ \tag{4} \sum F_y = 0 = N_3 \cdot cos \alpha - G_1 - N_1\]

or converted to N_{3} and the normal force N_{1} replaced by its value:

\[ \tag{5} N_3 = \frac{G_1 + G_2}{cos \alpha} \]

Finally, N_{2} follows from equation 3

\[ \tag{6} N_2 = \frac{G_1 + G_2}{cos \alpha} \cdot sin \alpha\]

or in other words:

\[ \tag{7} N_2 = (G_1+G_2) \cdot tan \alpha\]

The member lengths given in the sketch were not relevant for the calculation of the normal forces.