Weight & Tension Helo 2 Tension

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    Tension Weight
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Homework Help Overview

The discussion revolves around a statics problem involving weight and tension in a system. The original poster seeks assistance in determining the weight necessary to solve the equations of static equilibrium, specifically focusing on the sum of forces in both the X and Y directions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss resolving tensions into their X and Y components and emphasize the importance of the static condition where the sum of forces must equal zero. There are questions regarding the correctness of the equations set up by the original poster and the implications of the tensions involved.

Discussion Status

Participants are actively engaging with the problem, questioning the original equations and exploring the relationships between the tensions and the weight. Some guidance has been offered regarding the critical nature of the tensions and their impact on the system, but no consensus has been reached on the correct approach to finding the weight.

Contextual Notes

There is mention of specific tension values and the need to consider which tension is critical for the system's stability. The original poster expresses uncertainty about their calculations and the overall setup of the problem.

d1343
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Homework Statement



How to find Weight in this problem?
I am trying to do this statics problem
but I don't know how to find the weight
so I can solve for
Sum Fy = 0
here is the picture link
vwu4y0.jpg


Homework Equations


TI
sum of Fx = 0
480 cos(30) - 450 = 0
-34

T1
sum of Fy = 0
480 sin(30) - W = 0
how do I find the weight to substract it

The Attempt at a Solution



I don't know I am like really stuck
 
Last edited:
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Resolve the tensions into their X and Y components.

The system is static, so the sum of the X forces must equal 0 and sum of the Y as well.

Be careful. One of the tensions may not be critical to holding up the weight.
 
wht the equation I wrote is not right?
Sum of the Force in X = 480 cos(30) - 450 = 0
I get = 415 - 450
= -34
Sum of the Force in y = 480 sin (30) - W = 0
240 -w = 0

I how do I solve for weight
the answer is about 259.81 lb
& there is also for t2


ΣFx = 0; 480cos(30) − 450 = 0

ΣFy = 0; 480sin(30) − W = 0

(F ab2 W2) = Find (FAB W)

W2 = 240.00 lb
W= min (W1 W2) = W 240.00 lb
 
You have 2 tensions to worry about.

Like the weakest link - that's where it will break right?

First figure out which is the critical constraint.

If T1 = 450 max, what does that translate to for the chain that can withstand 480? Will the 450 break if it's 480, or will the 480 break if it's 450?

Then you are armed with what you need to know, because from your equations, you can see that whatever the T1, the weight it can support is 1/2 T1.
 

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