A tightrope walker walks across a 30-m long wire tied between two poles. The center of the wire is displaced vertically downward by 1.0 m when he is halfway across. If the tension in both halves of the wire at this point is $6294 \mathrm {~N}$ what is the mass of the tightrope walker? Neglect the mass of the wire.

Question

Quizplus · Accepted Answer

The tension in the wire creates two equal forces acting upward at an angle. Using trigonometry, the vertical component of the tension is $T \sin(\theta)$, where $\theta$ is the angle of the wire with the horizontal. The vertical components must balance the weight of the walker, $mg$. Given the geometry, $\sin(\theta) = \frac{1}{15}$. Thus, $2T \sin(\theta) = mg$, solving for $m$ gives $m = \frac{2 \times 6294 \times \frac{1}{15}}{9.8} \approx 85 \text{ kg}$.

A Tightrope Walker Walks Across a 30-M Long Wire Tied 6294 N6294 \mathrm {~N}6294 N

A Tightrope Walker Walks Across a 30-M Long Wire Tied $6294 \mathrm {~N}$