Tension Change When Spreading Scales: What Happens and Why?

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As the scales are spread farther apart, the tension in each scale increases due to the change in angle, which affects the vertical and horizontal components of the forces involved. The tension remains equal in both scales as they are symmetrically positioned, but the overall tension increases to balance the gravitational force acting on the object. Understanding the role of the x and y components is crucial for analyzing the forces accurately. The relationship between the angles and the tension can be derived using trigonometric functions. Clarifying the complete problem will help in providing a more detailed explanation.
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Homework Statement



physicstension.jpg

Continue to spread the scales farther apart. What happens to the tension in each scale? Why? [Quick question: There are two scales, so would I have to draw two boxes?]

Homework Equations


Fg=mg
sin70degrees=FTy/FT ---> FTy/sin70degrees

The Attempt at a Solution

So, I understand that when the two angles are the same, then the sum of both FTen would equal to Fg. As the scales are moved further apart, then the tension of one scale would still be equivalent to the other. That must mean the tension increases as they move away, but as I thought before, they still equal to each other. I don't really understand what happens to the tension and why... I also don't quite understand the role of the x and y components besides solving for missing angles/sides and such.
 
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Hi mouasee, welcome to PF.
Please post the complete problem.
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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