Question

A 70 kg tightrope walker is the exact middle of a span of her rope. At that point each rope makes an angle of 3 degrees below horizontal with its supporting column. What is the TENSION in the ropes.

Answer 1

Answer 2

To find the tension in the ropes, we can use the concept of equilibrium for the tightrope walker. At the middle of the span, the forces acting on the tightrope walker are balanced, so the vertical forces and horizontal forces must cancel out.

Let's analyze the forces acting on the tightrope walker at the middle of the span:

1. Vertical forces: The vertical forces acting on the tightrope walker are the weight (mg) acting downwards and the tension (T) in each rope acting upwards.

2. Horizontal forces: There are no horizontal forces acting on the tightrope walker at the middle of the span because the ropes make an angle of 3 degrees below horizontal, so there is no horizontal component of tension.

Now, let's write the equilibrium equations:

Vertical equilibrium: T + T - mg = 0

Horizontal equilibrium: Tension_horizontal = 0

Since the ropes are making an angle of 3 degrees below horizontal, the vertical component of tension (T) is T * sin(3°), and the horizontal component of tension is T * cos(3°). However, the horizontal component of tension is not relevant in this case because there are no horizontal forces to balance.

Now, we can solve for T:

T * sin(3°) + T * sin(3°) - mg = 0 2T * sin(3°) = mg T = mg / (2 * sin(3°))

Given: m (mass of tightrope walker) = 70 kg g (acceleration due to gravity) = 9.8 m/s^2

T = 70 kg * 9.8 m/s^2 / (2 * sin(3°))

Now, calculate T:

T ≈ 670.78 N

So, the tension in each rope is approximately 670.78 Newtons.