Solving a Homework Statement: Finding Forces and Distances

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SUMMARY

The discussion centers on solving a physics homework problem involving forces and distances, specifically using the equations of motion. The user successfully calculated the radius (r = 1.323m) and angle (θ = 41.41°) using the equations ΣFx = Tuppersinθ + Tlowersinθ + mgsinθ = mv²/r and ΣFy = Tuppercosθ - Tlowercosθ - mg = 0. The final tension values were determined as Tlower = 77.3N and Tupper = 135.45N. The user clarified the direction of gravitational force in relation to the problem, concluding that it does not affect the horizontal force calculation.

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Homework Statement


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**Hope I don't get in trouble for posting a screenshot of this online homework question...**

Homework Equations


[itex]\Sigma[/itex]Fx = marad = [itex]\frac{mv^2}{r}[/itex]
[itex]\Sigma[/itex]Fy = mg

The Attempt at a Solution


Simple equations allowed me to find r = 1.323m and θ = 41.41°
[itex]\Sigma[/itex]Fx = Tuppersinθ + Tlowersinθ + mgsinθ = [itex]\frac{mv^2}{r}[/itex]
Tuppersinθ + Tlowersinθ + 28.85N = 169.57N
Tupper + Tlower = 212.75N

[itex]\Sigma[/itex]Fy = Tuppercosθ - Tlowercosθ - mg = 0
Tuppercosθ - Tlowercosθ = mg
Tupper - Tlower = 58.15N
Tupper = 58.15N + Tlower

Substituting Tupper,
58.15N + Tlower + Tlower = 212.75N
58.15N + 2(Tlower) = 212.75N
Tlower = 77.3N
Tupper = 58.15N + 77.3N = 135.45N
 
Last edited:
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StrawHat said:
Simple equations allowed me to find r = 1.323m and θ = 41.41°
Good.
[itex]\Sigma[/itex]Fx = Tuppersinθ + Tlowersinθ + mgsinθ = [itex]\frac{mv^2}{r}[/itex]
In what direction does the weight act?
 


Doc Al said:
In what direction does the weight act?

Away from the pole, so I guess mgsinθ is negative as well?

EDIT: Ah, nope. Fx is only mv^2/r, so I don't need to take into account the gravitational force. Got the answers, thanks for your help!
 
Last edited:

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