What is the correct tension equation for a pendulum at rest?

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    Pendulum Tension
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The discussion centers on determining the correct tension equation for a pendulum at rest when the string breaks. The user questions whether the tension should be calculated using T = mgCos(theta) + mv^2/r or T = mgSin(theta) + mv^2/r, as the latter is indicated in the provided answers. A vector diagram suggests that the correct equation involves the sine component, leading to T = mgSin(theta) + mv^2/r. The importance of measuring the angle from the horizontal is emphasized to avoid confusion. The consensus is that the tension due to the weight of the bob at the original position (theta = 0) should be clarified in relation to the chosen equation.
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Homework Statement


The pendulum cord is released from rest when the angle = 0 (from a horizontal)
If the string breaks when the tension is twice the weight of the bob at what angle does it break?

Im from NZ and our NCEA system is often riddled with mistakes and I want to clear this up.
Is the tension equation this:

T = mgCos(theta) + mv^2/r
or
T = mgSin(theta) + mv^2/r (What the answers say)

By drawing a vector diagram I believe the test is wrong as the component in Fg providing tension is mgCos(theta)

Can anyone confirm for me? and provide reasons why so that I don't confuse myself.
 
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Pochen Liu said:
T = mgCos(theta) + mv^2/r
or
T = mgSin(theta) + mv^2/r (What the answers say)

By drawing a vector diagram I believe the test is wrong as the component in Fg providing tension is mgCos(theta)

Can anyone confirm for me? and provide reasons why so that I don't confuse myself.

Don't get confused...Just make sure that the angle(theta) is measured from the horizontal.
Then by the vector diagram, you will get T=mgSin(theta) + mv^2/r.
 
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What's the tension due to the weight of the bob in the original position (##\theta = 0##)?
 
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