Question

A 52 kgcircus performer is to slide down a rope that will break if the tension exceeds 425 N. (a) What happens if the performer hangs stationary on the rope? (b) At what magnitude of acceleration does the performer just avoid breaking the rope?

Short answer

(a) When a performer hangs stationary on a rope, the rope will break.

(b) Magnitude of acceleration without breaking the rope is 1.6 $\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Given

1. Mass of the performer is m = 52 kg
2. The maximum tension in the rope is T = 425 N.

Understanding the concept

The weight of the object is equal to the product of mass and gravitational acceleration.

Find the weight of the performer by multiplying by g. Using the maximum tension, we can find the acceleration without breaking the rope.

Formulae:

$$\begin{gathered} \mathbf{1}\mathbf{)}\mathbf{}\mathbf{}\mathbf{W}\mathbf{=}\mathbf{mg} \\ \mathbf{2}\mathbf{)}\mathbf{}\mathbf{}\mathbf{F}\mathbf{=}\mathbf{ma} \end{gathered}$$

Calculate the tension when the performer is stationary on rope

The weight of the performer will cause the tension in the rope.

$$\begin{gathered} W=mg \\ =52\mathrm{kgx}9.8\mathrm{m}/{\mathrm{s}}^2 \\ =510\mathrm{N} \end{gathered}$$

When the performer is stationary on the rope, the tension is 510 N which exceeds the maximum tension so the rope will break.

Calculate the magnitude of acceleration without breaking the rope

Using Newton’s second law we can write,

$ma=mg-T$

Rearranging for a,

$$\begin{gathered} a=g-\left(\frac{T}{m}\right) \\ =9.8\mathrm{m}/{\mathrm{s}}^2-\left(\frac{425\mathrm{N}}{52\mathrm{kg}}\right) \\ =1.6\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration of the performer is $1.6\mathrm{m}/{\mathrm{s}}^2$.