Mechanical Engineering: Ch 10: Forces on Cables (10 of 33) Distributed Load: Example (Part 1)

[@paruliansidabutar3488]

In 3:40 you calculate the Weight per feet of the cable without knowing the lenght of the cable( or half of the cable) , in real-life we name it "span" , so you calculate the weight of the half lenght of the cable using half span while length of the cable must be longer, and the weight per feet should be more also.

[@MichelvanBiezen]

It is a close approximation. With these approximation we are only off by about 0.5% (see video 18 in the playlist)

[@haidinh5113]

Dear Mr. Biezen, because you are talking about uniform distributed load on the horizontal line (which leads to your calculation of W = w.x) then it's perhaps clearer to draw the distributed load on the horizontal line (instead of on the cable). Should the load be uniformly distributed on the cable (e.g., the case of cable self-weight), the equivalent horizontal load will be w.x/cos(theta) where theta is the inclined angle, which is changing, which makes the horizontal load not uniform any more. Please correct me if I am wrong. Thanks for your decent series on cable analysis.

[@vigneshbalaji21]

Nice video, Although you approximated the equations ofr parabola instead of catenary. I was able to learn from it. I am trying to learn more on catenary. Can you also please put some videos on catenary ?

[@MichelvanBiezen]

We have a playlist in the machanical egineering section.

[@giulianagrasso5445]

amazing, incredible, so well explained!

[@tmoney780]

your bow tie is amazing.

[@brandonrobertson6586]

Why is tension measured in pounds/'lb'?

[@MichelvanBiezen]

In the US, the unit of force is still the pound.

[@ronnylov]

Thank you very much sir

[@steng156]

thanks

[@MichelvanBiezen]

Welcome