Reducing Coplanar Forces To A Single Resultant Example

Reducing coplanar forces to a single resultant example

In engineering and physics, understanding how to simplify a system of forces is essential for analyzing the effects on a structure or object. When dealing with coplanar forces, where all forces lie within the same plane, I can reduce the entire system to a single resultant force acting at a specific point. This reduction not only simplifies the analysis but also provides a clearer understanding of the forces at play.

In this blog post, I will walk through an example of reducing a system of coplanar forces to a single resultant force. The focus will be on calculating the resultant force, determining its line of action, and ensuring that it produces the same moment as the original system.

Understanding the resultant force

Given a system of coplanar forces, the first step is to compute the resultant force $\mathbf{F}_R$. The resultant force is the vector sum of all the forces in the system:

$$\mathbf{F}_R = \sum \mathbf{F}_i$$

The components of this resultant force in the $x$ and $y$ directions are given by:

$$\begin{aligned} & F_{Rx} = \sum F_x \\ & F_{Ry} = \sum F_y \end{aligned}$$

This resultant force represents the combined effect of all the individual forces acting within the plane.

Calculating the moment

Next, I calculate the moment about a specific point, which must account for the sum of the moments of all the forces. The moment about a point $O$ is calculated using:

$$M_O = \sum (\mathbf{r}_i \times \mathbf{F}_i)$$

Here, $\mathbf{r}_i$ is the position vector of the point of application of the $i$-th force relative to $O$. This moment is crucial for determining the line of action of the resultant force.

Applying the example

Let’s apply these concepts to an example where I have already calculated the resultant force $F_R = 403.11$ acting at an angle $\theta = 60.26^\circ$ with the $x$-axis. The moment about point $C$ is $M_C = 1950$ Nm.

To find the line of action for the resultant force, I first need to calculate the perpendicular distance $d$ from point $C$ to the line of action of $F_R$. This is given by:

$$d = \frac{M_C}{F_R} = \frac{1950}{403.11} \approx 4.84$$

Now, I determine the equation of the line parallel to the line passing through point $C(3, 0)$ with a slope of $an(60.26^\circ) \approx 1.75$. The equation of this line is:

$$y = 1.75(x - 3) = 1.75x - 5.25$$

To find a parallel line at a perpendicular distance of $d = 4.84$ from point $C(3, 0)$, I use the distance formula:

$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

After substituting the values and solving, I find the possible lines of action:

1. $y = 1.75x + 4.5$
2. $y = 1.75x - 15$

Given that the desired moment is $+1950$ Nm (counterclockwise), the correct line of action is:

$$y = 1.75x - 15$$

This line intersects the $x$-axis at approximately $x = 8.57$. To confirm, I calculate the moment generated by the resultant force about point $C$ using the position vector $\mathbf{r}_{CA} = (-3, -15)$ and the force $\mathbf{F} = 200\mathbf{i} + 350\mathbf{j}$. The moment is:

$$M_C = \mathbf{r}_{CA} \times \mathbf{F} = (-3 \times 350) - (-15 \times 200) = 1950 \, \text{Nm}$$

This confirms that the system can be reduced to a single resultant force acting along the line $y = 1.75x - 15$, providing the same overall effect on the body.

Conclusion

By reducing a system of coplanar forces to a single resultant force and ensuring it acts at the correct point, I can simplify the analysis and maintain the integrity of the original force system. This process is vital for anyone working with forces in two dimensions, offering both clarity and precision in calculations.

For more insights into this topic, you can find the details of the example here.