One end of a nylon rope is tied to a stationary support at the top of a vertical

Question

One end of a nylon rope is tied to a stationary support at the top of a vertical mine shaft of depth 76.0 m . The rope is stretched taut by a box of mineral samples with mass 22.0 kg attached at the lower end. The mass of the rope is 1.60 kg . The geologist at the bottom of the mine signals to his colleague at the top by jerking the rope sideways. (Do not neglect the weight of the rope.) What is the wave speed at the bottom of the rope? What is the wave speed at the middle of the rope? What is the wave speed at the top of the rope?

Explanation / Answer

V = sqrt(T/mu)

mu = mass per unit length = 1.60/76 = 0.0210 kg/m

A.

At the bottom of the rope total tension will be

T = Mb*g = 22*9.81

v = sqrt(22*9.81/0.0210) = 101.37 m/sec

B. At the middle of the rope

T = Mb*g + Mr*g/2 = 22*9.81 + 1.6*9.81/2 = 223.66

v = sqrt(223.66/0.0210) = 103.20 m/sec

C.

At the top of the rope

T= Mb*g + Mr*g = (22 + 1.6)*9.81

v = sqrt(T/mu) = sqrt((22 + 1.6)*9.81/0.0210) = 104.99 N

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.