A ball of mass 1.6 kg is attached to the end of a massless string. A circus clown twirls the string so that the ball travels in a horizontal circle of radius 180 cm. The ball completes one
revolution every 2.3 seconds. What is the tension in the string?

Compute the ball's angular speed v :

v = (1 rev) / (2.3 s) • (2π • 180 cm/rev) • (1/100 m/cm) ≈ 4.917 m/s

Use this to find the magnitude of the radial acceleration a :

a = v ²/R

where R is the radius of the circular path. We get

a = v ² / (180 cm) = v ² / (1.8 m) ≈ 13.43 m/s²

The only force acting on the ball in the plane parallel to the circular path is the tension force. By Newton's second law, the net force acting on the ball has magnitude

∑ F = m a

where m is the mass of the ball. So, if t denotes the magnitude of the tension force, then

t = (1.6 kg) (13.43 m/s²) ≈ 21 N

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this