A tightrope walker of mass 70.0 kg stands at the middle of a wire that sags by 5.0 degrees on each side. What is the tension in the wire?

1. The walker is stationary, so the net external force on him is zero. 2. The horizontal components of the two tensions cancel, so the two tensions are equal in magnitude. 3. Vertically, the two upward components must together balance his weight. 4. Each vertical component is the tension times the sine of 5.0 degrees, so twice that equals the weight. 5. Rearranged, the tension equals the weight divided by twice the sine of 5.0 degrees. 6. Substitute the values: (70.0 kg)(9.80 m/s^2) / (2 x 0.0872). 7. That is 686 N divided by 0.1744, which gives about 3900 N. 8. The tension is almost six times his 686 N weight, because the wire is nearly horizontal. _Source: OpenStax College Physics (CC BY 4.0), Ch 4 "Dynamics: Force and Newton's Laws of Motion", section 4.5 Normal, Tension, and Other Examples of Forces_