A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are attached to fixed supports. What is the tension in each half of the rope?

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, = 0

The vertical component of the tension, , in each half of the rope is given as follows;

= T × sin(θ)

∴ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

"i have always loved contemporary art and made it a general goal to work within the field. after a few nonprofit internships at museums and galleries, i decided to try the for-profit art world and was offered a job as the ad assistant at the magazine. i much prefer the pace of the for-profit art community and have stayed with the company for three years, with a promotion to an account manager last spring."

i just did this and got a 100.

okay i have so many questions. but the big one here is: um wt..f is this mess? ?