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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 for 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

Respuesta :

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

What is the Net horizontal force?

The net horizontal force of the cylinder when it is at equilibrium position can be found by applying Newton's second law of motion. Thus;

∑F = 0

F - μF_n = 0

We are given;

F_n = 5 kN = 5000 N

μ = 0.2

Thus;

F - 0.2(5,000) = 0

F - 1,000 = 0

F = 1,000 N

The strength of the applied force will be increasing as the number of turns of the rope increases. Thus;

Minimum force = Total force/number of turns of rope

Since rope is wrapped three and half times, then;

number of turns = 3.5

Thus;

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.

Read more about net horizontal force at; https://brainly.com/question/26957287

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