1. The problem statement, all variables and given/known data 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 lets 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) 2. Relevant equations Fnet = mv^2/r mgh = 1/2mv^2 3. 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?