Chapter 5: Q85P (page 123)

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?

Expert verified

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

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

Step by step solution

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:

$1) W = mg 2) F = ma$

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

$W = m g = 52 kgx 9.8 m / s^{2} = 510 N$

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

Using Newton’s second law we can write,

$m a = m g - T$

Rearranging for a,

$a = g - (\frac{T}{m}) = 9.8 m / s^{2} - (\frac{425 N}{52 kg}) = 1.6 m / s^{2}$

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

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