What is the Tension Force of a Swing Set at the Bottom of its Path?

[testme]

Homework Statement

At the playground, a girl gets on a swing. Her mother pulls her back so that the chain on the swing is at an angle θ from the vertical. She holds her daughter and then let's her go, without pushing. Show that the tension in the chain at the bottom of the swing’s path is given by:

T= mg(1-cosθ)

Homework Equations

Fnet = mv^2/r
mgh = 1/2mv^2

The Attempt at a Solution

Let Upwards be positive.

I first found out that the height the swing was moved upwards by, h, is:

h = r - rcosθ

Then I used mgh = 1/2mv^2 to solve for v^2:

mg(r-rcosθ) = 1/2 mv^2
2g(r-rcosθ) = v^2

Knowing v^2 I used Fnet = mv^2/r and tried to solve for the tensional force.

Fnet = mv^2/r
-Fg + T = mv^2/r

-mg + T = mv^2/r
-mg + T = m(2g(r-rcosθ))/r
-mg + T = 2mgr(1-cosθ)/r
-mg + T = 2mg(1-cosθ)
T = 2mg - 2mgcosθ + mg
T = 3mg - 2mgcosθ
T = mg(3-2cosθ)

This isn't the same as the answer it should be though. Someone told me there are no signs involved and everything is positive which does end up giving you the right answer but why would we ignore the direction?

 

[NascentOxygen]

Show that the tension in the chain at the bottom of the swing’s path is given by:

T= mg(1-cosθ)

[PLAIN]https://www.physicsforums.com/images/icons/icon13.gif Were this correct, it's saying tension at lowest point in arc would be less than the person's weight.

T = mg(3-2cosθ)

That is my answer, too.

 

[testme]

So you also think that the teach made a mistake and my answer is correct?

 

[PeterO]

So you also think that the teach made a mistake and my answer is correct?

Cetainly T= mg(1-cosθ) can't be the answer, since that is less than mg alone, and we know the Tension has to be greater than mg.
Indeed, if the swing is is "pulled aside" by 0^o then the Tension supposedly becomes zero?

 

[testme]

I see what you mean, so, the teacher, or whoever made up the question, probably forgot to make either tension or gravity a negative then, thanks.