Physics question: Finding 2 angles with given tensions and mass

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Homework Help Overview

The problem involves determining the angles formed by two tension forces acting on a traffic light, given specific values for the tensions and the weight of the light. The context is within the subject area of static equilibrium in physics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the equations for forces in both the x and y directions, questioning the correctness of the calculations and the approach to solving for the angle θ.

Discussion Status

Some participants have provided guidance on the setup of the equations, while others are exploring the implications of the calculations presented. There is an ongoing examination of the steps taken to solve for θ, with no explicit consensus reached yet.

Contextual Notes

Participants are navigating through potential errors in their calculations and the assumptions made in the problem setup, particularly regarding the relationship between the tensions and the angles involved.

MadameCassie
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Homework Statement


Suppose the traffic light (1.00 x 10^2 N) is hung so that the tensions T1 and T2 are both equal to 80.0 N. Find the new angles they make with respect to the x axis. (By symmetry, these angles will be the same.)

Homework Equations


Fx = 0 = -T1 cos theta + T2 cos theta
Fy = 0 = T1 sin theta + T2 sin theta - 1.00 x 10^2

The Attempt at a Solution


Fx = 0 = -80 cos theta + 80 cos theta
Fy = 0 = 80 sin theta + 80 sin theta - 1.00 x 10^2
 
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Welcome to PF!

Hi MadameCassie! Welcome to PF! :smile:

(have a theta: θ :wink:)
MadameCassie said:
Fy = 0 = 80 sin theta + 80 sin theta - 1.00 x 10^2

Yes, you're almost there.

So θ = … ? :smile:
 
Thanks tiny tim! Nice to meet you! :)

It's kinda hard to solve for theta. I've been setting up the problem and it comes out not right.
 
show us your full calculations, and then we'll see what went wrong, and we'll know how to help! :smile:
 
Fy = 0 = 80 sin theta + 80 sin theta - 1.00 x 10^2
0=160 sin θ- 100 N
0=60 sin θ
-60=sin θ
 
MadameCassie said:
0=160 sin θ- 100 N
0=60 sin θ

Noooo :redface:

0=160 sinθ - 100 N

160 sin θ = 100 N :smile:

get some sleep! :zzz:​
 
haha I get it! Thanks so much!
 

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