Finding the tension in the cable

[tjohn101]

Homework Statement

A ski gondola is connected to the top of a hill by a steel cable of length 580 m and diameter 1.5 cm. As the gondola comes to the end of its run, it bumps into the terminal and sends a wave pulse along the cable. It is observed that it took 19 s for the pulse to travel the length of the cable and then return.

(a) What is the speed of the pulse?
61.05263158 m/s
(b) What is the tension in the cable?
? N

Homework Equations

I would assume you would sum the forces, but I don't know what to put together to reach a tension.

The Attempt at a Solution

I do not know where to begin.

Help is greatly appreciated! Thank you!

 

[kuruman]

How is the speed related to the tension and the linear mass density? You will have to look up the density of steel to figure out the linear mass density of this cable.

 

[kerimek]

I would approach this problem by using the basic principles of physics and applying them to the given information. First, we can use the formula for wave speed, v=λf, where v is the speed of the wave, λ is the wavelength, and f is the frequency. In this case, we are given the length of the cable (l=580 m) and the time it took for the pulse to travel the length of the cable and back (t=19 s). Therefore, we can calculate the wavelength of the wave pulse using the formula λ=vt=580 m/19 s=30.52631579 m.

Next, we can use the formula for tension in a string or cable, T=μv^2, where T is the tension, μ is the linear mass density of the cable, and v is the speed of the wave. The linear mass density can be calculated by dividing the mass of the cable by its length, μ=m/l.

To find the mass of the cable, we can use the formula for the volume of a cylinder, V=πr^2h, where V is the volume, π is the constant pi, r is the radius, and h is the height. We are given the diameter of the cable (d=1.5 cm), so we can calculate the radius as r=d/2=0.0075 m. The height of the cable is the length of the cable, 580 m. Therefore, the volume of the cable is V=π(0.0075 m)^2(580 m)=0.0002025 m^3.

Now, we can use the density of steel, ρ=7850 kg/m^3, to find the mass of the cable, m=ρV=7850 kg/m^3 x 0.0002025 m^3=1.592 kg.

Finally, we can plug in all the values into the formula for tension, T=μv^2, to find the tension in the cable, T= (1.592 kg/580 m) x (61.05263158 m/s)^2= 106.941 N. Therefore, the tension in the cable is approximately 106.941 N.