Newton's 2nd Law on light fixture

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Homework Help Overview

The discussion revolves around a physics problem involving Newton's 2nd Law, specifically related to a lightweight string supporting three equal masses and the angles formed in the setup. The original poster seeks to determine the angle phi and the distance D based on the configuration described.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of free body diagrams and the sum of forces in the vertical direction. There is a focus on the relationship between tensions in the string segments and the angles involved. Some participants question the assumption that tensions are equal across different segments of the string.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the forces at play. Some guidance has been provided regarding the symmetry of the setup and the cancellation of horizontal components of tension at the tie points. However, there is no explicit consensus on the correct approach or solution yet.

Contextual Notes

Participants are working with the constraints of the problem as presented, including the assumption of equal masses and the specific angles involved. There is mention of attachments that may contain additional information relevant to the problem.

jburt
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Homework Statement


A lightweight string of length 60.0 cm is attached to the ceiling at points separated by distance D, as shown on the diagram in my attachment. Three objects of equal mass are hung from the string, separating the string into four equal segments of lenth L. If θ=40°
, find the measure of angle phi and the distance D.

Homework Equations



please see attachment

The Attempt at a Solution



please see attachment
 

Attachments

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I also want to say, I've attempted to use only the equations for the sum of y forces on the knots, since I've ruled the masses out as irrelevant, so those free body diagrams are also irrelevant. I subbed the value of T from the middle knot sumFy equation into the T for the 1st knot's sumFy equation and solved for phi. . . wrong answer resulted
 
L is the length of the pieces of the string, not the tension in them. The tensions are not the same in the different pieces at both sides of a knot.

ehild
 
I'd say that, at each tie point, the two horizontal components of the tensions cancel.
 
Thanks,

So, on my free body diagram on the 1st knot from the left, say I label the mg T, as T1, then T2 (upper left) and T3 (upper right), then for the middle knot, the same T1 from the mg (since equal mass=equal weight), but the tensions on either sides are BOTH T3 (equal, but opposite, right? due to same L)?? That's what I'm going to attempt.
 
It will be all right. The right-hand part is mirror image of the left-hand part because of the equal lengths and masses, therefore the tensions are the same in the bottom pieces of the string.

ehildehild
 

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