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Llana [10]
2 years ago
11

A rope is wrapped three and a half times around a cylinder. Determine the range of force T exerted on the free end of the rope f

or maintaining equilibrium that is required to just support a 5 kN weight. The coefficient of friction between the rope and the cylinder is 0.2
Physics
1 answer:
Taya2010 [7]2 years ago
8 0

The range of force exerted at the end of the rope is 285.7 N to 1,000 N.

<h3>Net horizontal force of the cylinder</h3>

The net horizontal force of the cylinder when it is at equilibrium position is determined by applying Newton's second law of motion.

∑F = 0

F - μFn = 0

F - 0.2(5,000) = 0

F - 1,000 = 0

F = 1,000 N

The strength of the applied force increases as the number of turns of the rope increases.

minimum force = total force/number of turns of rope

minimum force = 1,000/3.5

minimum force = 285.7 N

Thus, the range of force exerted at the end of the rope is 285.7 N to 1,000 N.

Learn more about Newton's second law of motion here: brainly.com/question/3999427

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