# Another Tension problem

## Homework Statement

A child’s indoor swing consists of a rope of length L anchored to the ceiling, with a seat at the lower end. The total mass of child and seat is m. They swing in a horizontal circle with constant speed v, as shown in Fig. 6-2; as they swing around, the rope makes a constant angle 0 with the vertical. Assuming the time τ for one revolution (i.e., the period) is known, find the tension T in the rope and the angle θ.

ƩF=mac=m(v2/r)
ƩF=mg

## The Attempt at a Solution

I decided the force from the child on the swing would be Ftotal=( ƩF=mg and ƩF=mac=m(v2/r)).
And the tension of the line is Ty= Ftotal cosθ and Tx= Ftotal sinθ

Therefore T=Ftotal(cosθ+sinθ)
where Ftotal=m(v2/r+g)
Rewritten: T=(m(v2/r+g))(cosθ+sinθ)

And, rearranging this equation would give me θ if I so desired.

Am I doing this right??

Thanks for help

Draw a simple FBD.
In any istant, the (rope tension) x (cos θ) must equal the gravity force.
Do you agree ?
Then you easily have the rope tension.

Since, Tycosθ=mg then Ty=mg/cosθ. Then Tx =m(v2/r)sinθ

Is this correct?