Equilibrio en 2 dimensiones

How to solve an equilibrium problem in two dimensions?

When we face everyday problems about forces and equilibrium, we often need to simplify to obtain more manageable solutions. Today we will see a clear example: reducing the forces exerted on a three-dimensional table to a two-dimensional problem. In this way, we can apply basic principles of the equations of equilibrium and solve them with ease.

What are the steps to reduce a three-dimensional problem to two dimensions?

First, we consider the dimensions of the table: width, length and depth. For simplicity, we will work with only two dimensions, which means that we will solve only three essential equations:

1. Sum of vertical forces - Must equal zero to avoid upward or downward motion.
2. Sum of horizontal forces - Must also sum to zero to avoid lateral movement.
3. Sum of moments with respect to a point - Ensures that there is no rotation.

What are the support reactions in the system?

The supports are vital for equilibrium. In this exercise, we have:

• Double support: Provides both vertical and horizontal reaction.
• Single support: Provides only a vertical reaction.

Our task is to find these reactions, considering that we have three unknowns: the vertical reactions at points A and B, and the horizontal reaction at A.

How do we apply the equilibrium equations?

Let's see how to solve these equations step by step:

Horizontal force summation.

The simplest equation, since there is only one horizontal force, the reaction at A horizontal:

[ \sum F_x = A_h = 0 ]

This is because there are no other horizontal forces in the system.

Sum of vertical forces

We represent all vertical forces by summing the reactions at A and B and subtracting the applied external forces:

[ A_v + B_v - 15 - 6 - 6 - 6 = 0 ]

Adding the forces:

[A_v + B_v = 27

Moment summation

The use of the summation of moments helps us to solve the undecided equation by choosing a point, in this case point A, and solving for the forces:

Considering only the forces perpendicular to the pivot point, and assuming counterclockwise moments as positive:

[ -15N \times 3m - 6N \times 11m - 6N \times 13m + B_v \times 9m = 0 ]

Simplifying:

[ -45 - 66 - 78 + 9B_v = 0 ]

Results in:

[ 9B_v = 189 \Rightarrow B_v = 21N ]

Finally, having the value of (B_v), we can find (A_v):

[ A_v + 21 = 27 \Rightarrow A_v = 6N ]

Are units relevant in this analysis?

Checking units is crucial. In our case, by dividing a (\text{Newton}) meter by meters, we are left with units of Newton, which is correct since we are evaluating forces. Making sure to keep the units consistent is indispensable for the correct interpretation of the result.

Are we ready to face other equilibrium problems?

Yes, knowing that we have successfully solved the reactions for this problem, we can extend this methodology to other beam systems or structures in equilibrium. The process is fundamentally the same and will allow us to gain confidence and efficiency in tackling increasingly complex problems.

Keep exploring and practicing! Experience is key to mastering systems analysis and ensuring that our solutions are both accurate and effective.