Question

An adventurous archaeologist crosses between two rock cliffs by slowly going hand over hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope (Fig. P5.60). The rope will break if the tension in it exceeds 2.50 × 10^4 N, and our hero's mass is 90.0 kg. (a) If the angle is 10.0°, what is the tension in the rope? (b) What is the smallest value can have if the rope is not to break?

Answer 1

The tension in the rope is 1323 N. The smallest tension that the rope can have without breaking is 2.50 × 10^4 N.

Step-by-step explanation:

(a) To find the tension in the rope, we can use the following equation:

T = mg + Fapplied,

where T is the tension, m is the mass of the adventurer, g is the acceleration due to gravity, and Fapplied is the force applied by the adventurer. Since the adventurer is resting in the middle of the rope, the force applied is equal to half of his weight:

Fapplied = (1/2)mg.

Plugging in the values:

T = mg + Fapplied = mg + (1/2)mg = (1 + 1/2)mg = (3/2)mg.

Substituting in the given values, m = 90.0 kg and g is approximately equal to 9.8 m/s2:

T = (3/2)mg = (3/2)(90.0 kg)(9.8 m/s2) = 1323 N.

(b) The smallest value the tension can have without breaking the rope is equal to its breaking tension, which is 2.50 × 10^4 N.