Using the graph of the dependence of the average elastic force of a seat belt on stretching, determine its stretching under a load of 24.5 kN.

Solution:

The problem asks us to find the stretching (x) of a seat belt when the elastic force (F_upr) is 24.5 kN. We are given a graph that shows the relationship between F_upr and x.

First, let's examine the graph. The y-axis represents the elastic force in kN, and the x-axis represents the stretching in mm.

The graph shows a linear relationship, which means we can find the stretching by locating 24.5 kN on the y-axis and then finding the corresponding value on the x-axis by tracing a horizontal line to the graph and then a vertical line down to the x-axis.

However, the provided graph only shows values up to 17.5 kN on the y-axis.

Let's assume the graph is a straight line passing through the origin (0,0) and the point (0.5 mm, 17.5 kN). We can find the slope of this line, which represents the stiffness of the seat belt.

The slope (k) is calculated as: $$k = \frac{\Delta F}{\Delta x} = \frac{17.5\ \text{kN} - 0\ \text{kN}}{0.5\ \text{mm} - 0\ \text{mm}} = \frac{17.5}{0.5}\ \text{kN/mm} = 35\ \text{kN/mm}$$

Now we can use this slope to find the stretching (x) for a force of 24.5 kN using the formula $$F_{\text{upr}} = k \times x$$.

$$24.5\ \text{kN} = 35\ \text{kN/mm} \times x$$

To find x, we rearrange the formula:

$$x = \frac{24.5\ \text{kN}}{35\ \text{kN/mm}}$$

$$x = 0.7\ \text{mm}$$

Let's re-examine the provided graph. It seems the question expects us to extrapolate beyond the given y-axis limits. If we extend the line, we need to find where 24.5 kN intersects it.

Looking closely at the graph, the points seem to be:

• (0, 0)
• (0.1, 3.5)
• (0.2, 7.0)
• (0.3, 10.5)
• (0.4, 14.0)
• (0.5, 17.5)

This confirms a linear relationship where $$F_{\text{upr}} = 35x$$.

Now we need to find $$x$$ when $$F_{\text{upr}} = 24.5$$ kN:

$$24.5 = 35x$$

$$x = \frac{24.5}{35}$$

$$x = 0.7$$

The unit for x is mm as indicated on the graph.

The Answer

Ответ: 0.7 мм.