Finding angles of Phi. Very challenging.

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The discussion revolves around solving for two angles, PHI1 and PHI2, in a system where a 50N weight is suspended by two ropes with tensions of 80N and 70N, respectively, and the angles sum to 60°. The initial approach using vertical force balance led to PHI values of 10° and 50°, which were deemed nonsensical. It was clarified that both vertical and horizontal force balances are necessary to find a valid solution, and the equations must account for the components of the tension vectors. Ultimately, it was concluded that the given conditions cannot satisfy both force balances, indicating the problem may be flawed.
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1) Given: two ropes of negligible weight suspend an object weighing 50N from the ceiling There are two angles of PHI that form from as a measure from the vertical. PHI1 + PHI2= 60°. The tension of the left rope is given to be 80N and the tension of the left rope is given to be 70N. The entire system is stationary.

2) Find the measures of each angle PHI as well as each complimentary θ.

3) I tried to use ƩF=0=T1 + T2 + (-Fg)

∴ 50N = 80N*cos(60-PHI1) + 70N*cos(60-PHI2)

I couldn't think of a way to solve this by hand so I tired to use my Ti-84 calculator's equation solver and got PHI values of 10° and 50°. Logically, this doesn't make sense. My teacher decided to just omit the question, but I am still curious as to how it would actually be done.
 

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Why do you use ## 60 - \phi_1 ## and ## 60 - \phi_2 ##? Why not simply ##\phi_1## and ##\phi_2##?

Secondly, the equation you got is for the vertical components of the forces. You should obtain one for the horizontal components. That will give you a system of two linear equation for two unknowns, which you can then solve.
 
It's possible that the given conditions cannot be satisfied and that there is no solution to the problem as given.

Any solution would have to satisfy the conditions for stasis for horizontal and vertical components separately. So set ##\phi_2 = 60° - \phi_1## and use the horizontal force balance to find all solutions for ##\phi_1##. Check to see if any value you found satisfies the vertical force balance. If yes, huzzah! If no, the problem is broken :smile:
 
voko said:
Why do you use ## 60 - \phi_1 ## and ## 60 - \phi_2 ##? Why not simply ##\phi_1## and ##\phi_2##?

Secondly, the equation you got is for the vertical components of the forces. You should obtain one for the horizontal components. That will give you a system of two linear equation for two unknowns, which you can then solve.

Which would be the horizontal components equation? I'm confused. I don't want to trouble you any further, but could you possibly provide me with the two equations you are referring to? I'm a little frustrated so my mind is unclear.
 
The tensions in the ropes are vectors. Your general equation is ##\newcommand{b}[1]{\boldsymbol{#1}} \b {T_1} + \b {T_2} + \b {W} = 0 ##, where ## \b {T_1}, \ \b {T_2}, \ \b {W} ## are the left rope's tension, the right rope's tension and the weight. Each of those vectors has two components: vertical (y) and horizontal (x), so that one general equation gives you two equations for the components: $$

T_{1x} + T_{2x} + W_{x} = 0

\\

T_{1y} + T_{2y} + W_{y} = 0

$$ All that you need to do is find out these components, some of which will depend on ##\phi_1## and ##\phi_2##, and solve for ##\phi_1## and ##\phi_2##.
 
The problem doesn't have a solution that can satisfy the specified input data. No pair of angles summing up to 60 degrees can even balance the vertical force, let alone match the horizontal force balance.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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