Exercise S-1.1.1 TechMech 1 Rigid Body Statics

Practice Exercise S-1.1.1

Central Force Systems - First Basic Task: Reduction to a Single Force

Problem Statement

Graphically and analytically determine the magnitude and direction of the resultant force from forces F₁ to F₆.

Central Force System — Fig. 1: Layout Diagramm

$$ \begin{alignat}{5} F_1 &= 1000~\mathrm{N} \quad && \alpha_1 &&=0° \\[7pt] F_2 &= 1280~\mathrm{N} \quad && \alpha_2 &&=30° \\[7pt] F_3 &= 1750~\mathrm{N} \quad && \alpha_3 &&=50° \\[7pt] F_4 &= 840~\mathrm{N} \quad && \alpha_4 &&=90° \\[7pt] F_5 &= 220~\mathrm{N} \quad && \alpha_5 &&=100° \\[7pt] F_6 &= 760~\mathrm{N} \quad && \alpha_6 &&=150° \end{alignat} $$

Before You Embark on This Quest: A Pre-Mission Briefing

Before you dive into this adventure, let's do a quick check to see if you're ready for the challenge.

What's the Big Deal?

What topic are we dealing with here? Can you explain why?

All the lines of action of the attacking forces intersect at one point. So, we're dealing with a central force system.

Need a Little Refresher?

If you're not quite familiar with central force systems, no worries! Here are some refresher courses to get you up to speed:

Graphical Solution
Analytical Solution

Alright, You're Ready for the Task!

Good luck!

Cracking the Case of the Central Force System: A Graphical Solution

Alright, let's unravel the mystery of the resultant force in your central force system in a lighthearted way. Imagine you're a detective on a thrilling case: Where is the resultant force hiding, and how strong and in what direction does it act?

Step 1: Scene of the Crime - The Layout Diagram

Lucky you! The position diagram of the system is already given. So, you can see where all the forces are acting, but there's no actual object. Instead, they all act at the origin of an x,y coordinate system. It's like a meeting place for the forces!

Step 2: Force Diagram - Profiling the Perpetrators

Now it gets exciting: Create a force diagram that tells us more about the perpetrators (the forces).

Choose a scale that fits the given forces. $1~\mathrm{LU}~\widehat{=}~ 200~\mathrm{N}$ might be a good fit.
Convert the magnitudes of the forces into lengths. This will give us a feel for their strength.
$$ \begin{alignat}{3} \tag{1} F_1 &= 1000~\mathrm{N}~ &&\widehat{=}~ 5~\mathrm{LU} \\[7pt] \tag{2} F_2 &= 1280~\mathrm{N}~ &&\widehat{=}~ 6.4~\mathrm{LU} \\[7pt] \tag{3} F_3 &= 1750~\mathrm{N}~ &&\widehat{=}~ 8.75~\mathrm{LU} \\[7pt] \tag{4} F_4 &= 840~\mathrm{N}~ &&\widehat{=}~ 4.2~\mathrm{LU} \\[7pt] \tag{5} F_5 &= 220~\mathrm{N}~ &&\widehat{=}~ 1.1~\mathrm{LU} \\[7pt] \tag{6} F_6 &= 760~\mathrm{N}~ &&\widehat{=}~ 3.8~\mathrm{LU} \end{alignat} $$

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Calculating the Resultant Force

To solve this problem mathematically, you'll need the angle $\alpha$ from the positive x-axis to the force vector (counterclockwise is positive) for each force in the central force system, as well as the magnitude of each force.

This figure 5 shows a force vector in 3 positions in an x,y-coordinate system. — Fig. 5: General force vector in the plane

Luckily, the system's layout is already given to you along with the task. This includes the required magnitudes and angles of the individual forces:

Given:

$$ \begin{alignat}{5} F_1 &= 1000~\mathrm{N} \quad && \alpha_1 &&=0° \\[7pt] F_2 &= 1280~\mathrm{N} \quad && \alpha_2 &&=30° \\[7pt] F_3 &= 1750~\mathrm{N} \quad && \alpha_3 &&=50° \\[7pt] F_4 &= 840~\mathrm{N} \quad && \alpha_4 &&=90° \\[7pt] F_5 &= 220~\mathrm{N} \quad && \alpha_5 &&=100° \\[7pt] F_6 &= 760~\mathrm{N} \quad && \alpha_6 &&=150° \end{alignat} $$

Step 1: Add the components of the individual forces

You can determine the components of the resultant force by adding the components of the individual forces using Equation (2.4):

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