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Equilibrant force 
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#1
Oct2306, 07:10 PM

P: 2

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? 


#2
Oct2306, 07:19 PM

HW Helper
P: 3,220

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.



#3
Oct2306, 08:18 PM

P: 2

Thank you for your input...I think I have it.



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