Find Equilibrant Force Using Trig Method

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SUMMARY

The discussion focuses on calculating the equilibrant force using trigonometric methods for three different force systems. In force system A, the resultant force is calculated as \(\vec{R} = 5\vec{i} + 5\vec{j}\), leading to an equilibrant force of equal magnitude but opposite direction. The method involves summing the components of each force using cosine and sine functions based on their angles. This approach is applicable to all provided force systems, allowing for the determination of equilibrant forces effectively.

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Students in physics, engineers working with force systems, and anyone interested in understanding equilibrium in mechanics will benefit from this discussion.

nikmar
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force system a:
force no. 1: 5.0N at 0 deg
force no. 2: 5.0N at 90 deg
force system b:
force no. 1: 5.0N at 0 deg
force no. 2: 5.0N at 60 deg
force system c:
force no. 1: 5.0N at 30 deg
force no. 2: 5.0N at 150 deg

how do you find the equilibrant force using the trigonometric method? Do you multiply the force 1 (5.0) times cos 0 deg; force 2 (5.0) times cos 90 deg; etc?:confused:
 
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Well, you have a few systems consisting of two forces here. Now, for every system, i.e. for every couple of forces, you must consider the resultant force, which is the sum of the two forces of a system. For example, for the system a, you have [tex]\vec{R}=\vec{F}_{1}+\vec{F}_{2}=5\vec{i}+5\vec{j}[/tex]. So, the force which will set the system of equilibrium is a force with the same magnitude as the resultant force R, but pointing in the opposite direction. Now you have a clue.
 
Thank you for your input...I think I have it.
 

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