Minimize Tension of Mass m Hanging from Cords

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The discussion focuses on minimizing the tension in cord 2 of a system where a mass m hangs from two cords at specific angles. Participants suggest using a force diagram to analyze the problem, with one user opting for the component method to establish relationships between the tensions in the cords. The user proposes that to minimize tension in cord 2, the angle theta should be 0 degrees, which prompts a correction from another participant who advises against shortcuts and emphasizes proper elimination of variables. The conversation highlights the importance of deriving equations accurately and minimizing tension through systematic analysis. The key takeaway is to approach the problem methodically rather than making assumptions.
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Homework Statement



A package of mass m hangs from a short cord that is tied to the wall via cord 1 and to the ceiling via cord 2. Cord 1 is at angle 40 degrees with the horizontal; cord 2 is at angle, theta.

a) for what value of theta is the tension in cord 2 minimized?
b)in terms of mg, what is the minimum tension in cord 2?

(here's the link of the problem with the picture http://www.unc.edu/~rowan/phys26/P26recit/p26-12-b.htm )

I know I have to figure out the force diagram first. I'm just having trouble with where to start.

I'm not asking for the solution, i just need some direction,
 
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Hi munchy35! :wink:

You can solve this either with a vector triangle, or by taking components in a convenient direction.

Choose one method, and show us how far you get. :smile:
 
i used the component method so

for horizontal component i said t1cos40=t2cosx

(i'm just using x for theta)

t2=t1cos40/cosx

and t1 = t2cosx/cos40

so for part a, i used the t2 equation and just assumed for t2 to be a minimum cos x had to be maximum, the maximum cosx=1, so x= 0 degrees

^ can i just assume that?

for part b, i used vertical component

t1sin40 + t2sinx = mg

after plugging in t1

t2 = mg/ tan40cos0 + sin0

and eventually get 1.192mg

is that right
 
Hi munchy35! :smile:
munchy35 said:
… for horizontal component i said t1cos40=t2cosx

t1sin40 + t2sinx = mg

so for part a, i used the t2 equation and just assumed for t2 to be a minimum cos x had to be maximum, the maximum cosx=1, so x= 0 degrees

^ can i just assume that?

Nooo … stop trying to take short-cuts! :rolleyes:

Your two equations are correct :smile:

now eliminate t1 to find t2 …

then minimise t2. :wink:
 
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