# Geometrical solution to static problem without friction

• maros522
In summary, we have a triangle with sides d1, d2, l and angle α between d1 and d2. We are assuming a small change of Δα in α. Using the cosine rule and simplifying for small values of Δα, we can prove that (l+Δl)^2 = d1^2 + d2^2 - 2d1d2cos(α+Δα) by substituting for cos(α+Δα) and simplifying.
maros522

## Homework Statement

We have triangle with sides d1,d2,l and angle $\alpha$ between d1 and d2.
Assume small change $\Delta\alpha$ of $\alpha$.

## Homework Equations

Then we can write for $\Delta$l equation
$\Delta$l=(d1*d2)/(l) * sin$\alpha$$\Delta\alpha$.

How can I prove that?

## The Attempt at a Solution

I start with function cos($\alpha$+$\Delta\alpha$)=cos$\alpha$ * cos$\Delta\alpha$ - sin$\alpha$ * sin$\Delta\alpha$
and laws of cossines $(l+\Delta l)^2$=$d1^2$+$d2^2$-2d1d2*cos($\alpha$+$\Delta\alpha$)
But I don't know how to continue. :(

maros522 said:

## The Attempt at a Solution

I start with function cos($\alpha$+$\Delta\alpha$)=cos$\alpha$ * cos$\Delta\alpha$ - sin$\alpha$ * sin$\Delta\alpha$
and laws of cossines $(l+\Delta l)^2$=$d1^2$+$d2^2$-2d1d2*cos($\alpha$+$\Delta\alpha$)
But I don't know how to continue. :(

So if Δα is small, then sin(Δα) = Δα and cos(Δα) =1.

So you can simplify cos(Δα+α) using that fact.

Now if you use the original triangle with sides d1,d2,l and angle α, what is l2 using the cosine rule?

$(l+\Delta l)^2$=$d1^2$+$d2^2$-2d1d2*cos($\alpha$+$\Delta\alpha$

If you expand the left side and substitute for cos(α+Δα) and you know that Δl is small so (Δl)2 is negligible you should get it easily.

Thank you for answer I get it now.

## 1. What is the purpose of using geometrical solution in static problems without friction?

The purpose of using geometrical solution in static problems without friction is to determine the equilibrium of forces acting on a body or structure. It helps to find the various forces and their magnitudes required for the body to be in a state of rest.

## 2. How does geometrical solution differ from analytical solution in static problems without friction?

Geometrical solution relies on graphical methods such as vector diagrams, while analytical solution uses equations and mathematical calculations. Geometrical solution provides a visual representation of forces acting on a body, making it easier to understand and visualize the problem.

## 3. What are the limitations of using geometrical solution in static problems without friction?

The main limitation of geometrical solution is that it is not as accurate as analytical solution. It also cannot be used for complex structures or systems with multiple forces acting in different directions. Geometrical solution also does not take into account other factors such as material properties or friction, which may affect the stability of the structure.

## 4. Can geometrical solution be applied to real-life engineering problems?

Yes, geometrical solution can be applied to real-life engineering problems, but it is mostly used for simple structures or systems. It is commonly used in civil engineering for analyzing trusses, frames, and other simple structures. However, for more complex problems, analytical methods are preferred for their accuracy.

## 5. What are the advantages of using geometrical solution in static problems without friction?

Using geometrical solution can help in visualizing the problem and understanding the forces acting on a body. It also provides a quick and easy way to solve simple static problems without the need for complex mathematical calculations. Geometrical solution is also useful for checking the accuracy of analytical solutions.

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