Confused about resolving Tension and Weight

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SUMMARY

The discussion centers on the resolution of tension and weight in a static equilibrium system. The user presents two sets of equations involving tension (T1, T2) and weight (W) at an angle (theta), specifically questioning the validity of their calculations. Key errors identified include the incorrect assumption that T2 is perpendicular to T1 and the misapplication of trigonometric relationships, particularly in the equations Wsin(θ) = T1 and Wcos(θ) = T2. The correct approach requires acknowledging the components of tension and their relationship to the angle of inclination.

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  • Understanding of static equilibrium principles
  • Knowledge of trigonometric functions and their applications in physics
  • Familiarity with vector components in force analysis
  • Ability to manipulate and solve algebraic equations
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Students studying physics, particularly those focusing on mechanics, as well as educators and tutors who assist with understanding tension and equilibrium in physical systems.

laser
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Homework Statement
Confused about resolving Tension and Weight
Relevant Equations
uhh
1700939433965.png


Calculations with 1:
T1sintheta + T2sintheta = W
T1costheta = T2costheta

Calculations with 2:
Wsintheta = T1
Wcostheta = T2

These are not equivalent. Can someone point out the flaw in my logic?

Edit: System is in equilibrium!
 
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laser said:
Homework Statement: Confused about resolving Tension and Weight
Welcome to PF!
When giving the Homework Statement, please give the full statement exactly as given to you.

laser said:
Relevant Equations: uhh
Can you list any relevant equations for the forces when you have static equilibrium?

laser said:
Calculations with 1:
T1sintheta + T2sintheta = W
T1costheta = T2costheta
These look right.

laser said:
Calculations with 2:
Wsintheta = T1
Wcostheta = T2

These are not equivalent. Can someone point out the flaw in my logic?

It's hard to follow your logic based on the little that you have written down. The equation ##W \sin \theta = T_1## is incorrect. I'm guessing that you neglected the fact that ##T_2## has a component parallel to ##T_1##. That is, ##T_2## is not perpendicular to ##T_1## for general values of ##\theta##. Likewise, your equation ##W \cos \theta = T_2## is incorrect.
 
TSny said:
When giving the Homework Statement, please give the full statement exactly as given to you.
Oops, probably posted in the wrong forum. This isn't a homework, question, just something I was wondering about.
TSny said:
It's hard to follow your logic based on the little that you have written down. The equation Wsin⁡θ=T1 is incorrect. I'm guessing that you neglected the fact that T2 has a component parallel to T1. That is, T2 is not perpendicular to T1 for general values of θ. Likewise, your equation Wcos⁡θ=T2 is incorrect.
Fair point, I agree with you.

Let's say theta = 45 degrees. That makes them perpendicular, but the equations still don't work out.

From calculation 1:
We get W = Tsqrt(2)

From calculation 2:
We get W=T/sqrt(2)
 
laser said:
Let's say theta = 45 degrees. That makes them perpendicular, but the equations still don't work out.

From calculation 1:
We get W = Tsqrt(2)

From calculation 2:
We get W=T/sqrt(2)
Check your equation for calculation 2.
 
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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