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Exercise S-1.1.6

Practice Exercise in Technical Mechanics 1, Rigid Body Statics.

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

Practice Exercise S-1.1.6

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

Problem Statement

Four people are pulling a cart with ropes that are hooked into the towing eye of the drawbar according to the sketch.

Central Force System
Fig. 1: Task Description

Given:

F1=400 N, F2=350 N, F3=300 N, F4=500 N

  1. Determine the magnitude of the resultant force R.
  2. Determine the direction angle \(\alpha_R\) of the resultant force relative to the horizontal line through the drawbar.
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:

Alright, You're Ready for the Task!

Good luck!

Calculating the Resultant Force

To solve this problem computationally, you need the angle \(\alpha\) for all forces in the central force system from the positive x-axis to the force vector (counted positively counterclockwise) and the magnitude of each force.

This illustration shows a force vector in 3 positions in an x,y coordinate system.
Fig. 2: General force vector in the plane

First, visualize the central force system of this problem. Draw a Cartesian coordinate system with the origin (0,0) at the towing eye of the drawbar. This is the intersection point of the lines of action of the four cable forces.

Then draw the four forces F1, F2, F3, and F4 acting at this intersection point.

Consider the direction angles \(\alpha_1\), \(\alpha_2\), \(\alpha_3\), and \(\alpha_4\) of the forces, and mark them in your sketch. The direction angle of a force is counted positively counterclockwise from the positive x-axis. This also means that you can mark the direction angle negatively clockwise from the positive x-axis. This makes no difference in the calculation of the components.

Your sketch should look something like this:

Central force system
Fig. 3: Sketch

The given values for F1, F2, F3, F4, \(\alpha_1\), \(\alpha_2\), \(\alpha_3\), and \(\alpha_4\) are:

$$ \begin{alignat}{5} F_1 &= 400~\mathrm{N}\qquad \alpha_1 &&= 40° &&\\[7pt] F_2 &= 350~\mathrm{N}\qquad \alpha_2 &&= 0° &&\\[7pt] F_3 &= 300~\mathrm{N}\qquad \alpha_3 &&= 330° &&= -30°\\[7pt] F_4 &= 500~\mathrm{N}\qquad \alpha_4 &&= 320° &&= -40°\\[7pt] \end{alignat} $$
  1. Determine the magnitude of the resultant force R.

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