Finding Equilibrium: Solving for Unknown Forces and Angles in a Particle System

AI Thread Summary
The discussion focuses on solving for the unknown force magnitude (S) and angle (theta) in a particle system under equilibrium, with two equations derived from horizontal and vertical force balances. The user initially struggles with manipulating the equations to eliminate S and find a solution. Suggestions include using the tangent function by dividing the equations and employing the Pythagorean identity to combine them. The importance of understanding different methods for solving simultaneous equations is emphasized, as well as the value of having various mathematical strategies at one's disposal. The conversation highlights the user's desire to grasp the underlying theory rather than just applying techniques mechanically.
jamie_o
Messages
13
Reaction score
0
Hello, this isn't technically a homework, but its something I'm trying to figure for myself. As shown in the diagram (in the attachment), there are three forces acting on a particle. The particle is in equilibrium. I have to find the magnitude of S and the angle theta (which ill write as x, because I don't know how to type theta on a computer screen) Now what I would do is solve it horizontally and vertically.
so horizontally: 6 - Scosx = 0
vertically : 2.5 - Ssinx = 0

My problem is where do I go from here? I know I could make it 6 = Scosx and likewise for the other, but it won't help me solve it. Since they are both equal to 0 I tried setting them equal to each other in the hope that the S would cancel and I would get sinx/cosx which is equal to tanx = to a number and then work from there. However I can't get the equation into that form. Any ideas on how I would solve this? Any help is very much appreciated :)
 

Attachments

  • eg.jpg
    eg.jpg
    6 KB · Views: 375
Physics news on Phys.org
Try writing your equations like this:
(1) 6 = Scosx
(2) 2.5 = Ssinx

Now how can you get the S to cancel?
 
Thank you for the reply. I can see a link there and use Ssinx/Scosx to cancel S and leave tanx = 2.5/6, then use inverse tan of 2.5/6 = x. Then working out S would be straightforward from there. Why would I do this though? Why would I divide the horizontal equation into the vertical equation? Is there theory behind doing this? I don't like being 'monkey sees, monkey does' :) Thanks for the help.
 
Oh and if possible, could someone please tell me why setting the two equal to each other does not work? I might have a serious hole in my understanding of it all. Thanks.
 
You're just solving two equations with two unknowns. There are many ways to approach it; here are two:
(1) Solve for S in one equation, then plug that into the second. That's a standard approach.
(2) Square both equations and add them. Take advantage of sin^2\theta + cos^2\theta = 1.

Your skill in math often depends on picking up various little "tricks of the trade". Try to have as many "tricks" in your bag as possible.
 
Thank you. I should have spotted that since I am much further on in maths now, from when I got this question. I did try to take advantage of https://www.physicsforums.com/latex_images/35/352887-0.png previous to posting the first time. However I came out with a very large sum, what I had before was incorrect I think.
 
Last edited by a moderator:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top