Physics question

So for those that say the tension is the same in both cables, at what point is the tension not the same? What if cable A is 5 feet long and cable B is 1 foot long, is the tension still the same?

 

I agree the tension is the same in both cables regardless of the attitude of the beam as long as it is stationary. I misread the equilibrium comment earlier this morning

Edit: And not sure what i was thinking about when i responded.

 

"Righting moment" along with vortex induced vibrations must be considered here.

 

When the difference in cable lengths is equal to the length of the beam, then they'll have different tensions.

 

does not matter. still the same. your original post is somewhere between the following extreme.

5 ft cable and 5 ft cable (horizontal bar) > same tension

5 ft cable and 0.00000...1 ft cable (vertical bar)> same tension

Do not worry. I have already considered 14th perturbation theory on this subject.

 

Yes, as long as the cables or at 90 degrees from the ceiling or parallel to each other

 

Angle to the ceiling doesn't matter it's a pitched roof. Only thing that matters is that the lines are parallel. You could hook one line to a pitched part of a roof and the other to a flat roof overhang and they are still equal. All that assumes the center of gravity is equally spaced between the lines. The flaw in this study is everyone is assuming a beam. You are going to pick up a hard top that isn't balanced. In this case the weight on each cable will shift more toward the cable below the center of gravity in some increasing percentage regardless of the length of the cable. Hope that made sense the way i said it. I am not a physics guy. But i have picked up 20 to 30 18000 to 30000 pound boats with scales hooked to each cable every day for the last 25 years.

 

Hilarious mental wrangling going on here. Now you know how beginning engineering students about to flunk statics feel.

Free Body Diagram, make one so you can formally understand all the forces involved, and the directions they act in.

Another Hint: each tension is "unknown". so, take the Sum of the moments centered at the lower string. Then you know the 50lbs or whatever it was of the beam is centered at the center of the bar, a known distance (moment arm) from the lower string. The bar doesn't rotate, it is static, so you know the direction and length of the moment at the upper string.

This moment - that moment = 0
Force times distance - another force times another distance = 0
You know one force, both distances, have one unknown force (the string tension), you know the directions of the force, so solve for the one unknown force with your one equation.

"what could go wrong?"

 

You can never get a straight answer from the damn internet. Is what this thread is.

 

This post sums it up nicely
https://www.tacomaworld.com/threads...s-it-take-to-change-oil.603575/#post-20322906

 

People wanna be handed the answer, w/o doing the math.

do the math. I tol you everthing you needed to know to do it.

"Math class is hard," Barbie.

I know, girl. That's why I'm going wheeling and shooting. see ya

 

I welcome the math, it's all I've been doing for the last 3 years in school. At least this is easy math once a FBD is put together correctly. I just don't completely understand how to interpolate the forces in the -y direction. Since there's nothing acting in the x or z direction, it's all in the -y aside from the normal force in the +y. Since it's all in the -y direction, I just don't know how to calculate how much force is on each cable with the beam at an angle. If the beam was attached to a wall in one point I can tell you many things about it but I am drawing blanks here. I haven't had time to dig into my physics or statics books. I may just wait until I can see my physics professor and have him explain the forces to me.

There's a lot of smart people here chiming in but there's 2 directions of thought and no math here to prove anything. There was one guy on the first page of this post that posted some math but I don't agree with his approach and I can't visualize his numbers on a FBD.

 

The beam in question has a center of gravity in the center of the beam . Therefore no matter the angle of the beam the distance to the lift points remain the same , keeping the load equal at the lift points .

 

In the original post, the CG will be in the center of the beam.

 

Isn't this what i said but with numbers and stuff?!?! Lol

 

Sometimes it's the question that's wrong. Not the answers.

 

I didn't have time to read all the replies . I just posted those thinking they might help .

 

The CG is offset in that pic because of the angle it's at.

 

You missed the bug, haterz!




If you want to solve this problem, you hafta understand:
algebra
sine and cosine
force as a vector


Peeps who light up math books are why we can't have nice things.