another physics question! plz help!!?

Question

a trapeze artist weighs 8.00 * 10^2 . The artist is momentarily held to one side of a swing by a partner so that both of the swing ropes are at an angle of 30.0 degrees with the vertical.In such a condition of static equilibrium, what is the horizontal force being applied by the partner?

Answer 1

It is hard to imagine a trapeze artist weighing 800 anything - pounds, kg, gm, stone, etc. so I'll ignore the numbers.

On the trapeze artist we have three forces:

W - the weight of gravity, down
P - the horizontal pull of the partner
T - the combined tension of the ropes at an angle D from the vertical

Since the artist is not moving and not acclerating, all the forces must balance.

We can split the forces up into vertical and horizontal forces and both have to balance.

To split up the T into Tv and Th we use Th = T sin D and Tv = T cos D

Then we have:

W = Tv
P = Th

We know W so we know Tv; we know cos D so we can compute T; we know sin D so we can compute Th; which gives us P, the value we want.