How can I calculate the load on each line in a pulley system?

[Redliner56]

Hello, I'm working with a pully system and can't find the answer for this question on Google. I'm wondering how to calculate the load on the rope in order to determine what grade rope I should use. For example let's say I'm picking up a 100 lb box using a 4 to 1 system. Not accounting for friction gains I know I'd need a +/- 25lbs to pick up the box. I'm wondering how much stress would be placed onto the each of the 4 lines. Is there a simple formula I could follow?

 

[A.T.]

Hello, I'm working with a pully system and can't find the answer for this question on Google. I'm wondering how to calculate the load on the rope in order to determine what grade rope I should use. For example let's say I'm picking up a 100 lb box using a 4 to 1 system. Not accounting for friction gains I know I'd need a +/- 25lbs to pick up the box. I'm wondering how much stress would be placed onto the each of the 4 lines. Is there a simple formula I could follow?

If you ignore friction and the mass of the pulleys, the tension in the rope is constant. So if you pull with 25lbs the entire rope has that tension.

 

[Redliner56]

Thanks for the help, so this should mean that if we ignore adding in a safety factor a rope with a +/- 30lb breaking strength would be capable of picking up a 100lb box using 4 pullys. And following that line of thinking, if the pully breaking strength is +55lbs they would be enough to pick up the load since 2 lines connect to each pully?

 

[berkeman]

if we ignore adding in a safety factor

Welcome to PF. What safety factor are you planning on using? What things should go into your choice of safety factor beyond the bearing friction?

 

[A.T.]

Thanks for the help, so this should mean that if we ignore adding in a safety factor a rope with a +/- 30lb breaking strength would be capable of picking up a 100lb box using 4 pullys. And following that line of thinking, if the pully breaking strength is +55lbs they would be enough to pick up the load since 2 lines connect to each pully?

Keep in mind that those simple calculations only apply for static situations and steady motion. How much reserve you need depends on how quickly you want to accelerate the load. If you are using a motor, you can look at the max torque it outputs. If you are pulling by hand, the rope should at least support one bodyweight, in case someone hangs on to it.

 

[Redliner56]

All hardware I use will be rated for at least a safe working load equal to the weight of what I'm picking up. I just wanted to ignore the safety factor for the sake of keeping it simple. Know I know what I need to know. But if anyone wants to explain it to me. I don't understand how one rope, when dividing it's load over the wheel in the pully can withstand close to double what it could as an individual. Is this due to the friction between the rope and the wheel? That would make since but I expected it have more loss and needing something between a 30-45lb breaking strength

 

[Elquery]

I don't understand how one rope, when dividing it's load over the wheel in the pully can withstand close to double what it could as an individual.

But the rope tension is not 'divided,' it adds. Meaning with a given tension of 10lbs on the roope, when through a pulley, will add to 20lbs of supporting force.

As said before, the rope tension is (theoretically) constant throughout the system when going through pulley's. So it is not 'withstanding double' of anything.

Just think of a 100 lb weight dangling from 4 equalized ropes. Would you expect all 4 ropes to experience 100 lbs of tension? Or 25.

Perhaps what you are thinking is that since the axle of a pulley would see twice the rope tension, somewhere the rope must also 'see' twice it's own tension since it's wrapped around that axle.

What might help you here is to consider vectors. In the line created by the pulley axle and the load, the rope is normal (perpendicular) to that force. I.e. there is no tension component on the rope at that point. But there is compression.

You could also—instead of a wheel—picture a spreader bar with the load in the center and the two supporting ropes on either side. If you cut one supporting rope, the system will become unbalanced and the load will seek to align with the one remaining rope.
In the case of a pulley, this means you could wrap the rope around the pulley and tie it back onto itself and the axle would then align between the load and rope. The rope would then be experiencing the full load under tension (though split along the looped portion according to the vector forces based on the angles created by where the knot is tied). That's when you need trig to access those forces.

 

[Arjan82]

I think it is useful if you think in terms of free body diagrams like the one below.

Think of this upside down (I couldn't find a better picture) and $T_1$ is the weight you are trying to lift and $T_2$ is the tension in the cable. In here $T_2 = T_1/2$ (otherwise you have an imalance) so the force on the rope is half that of the weight.

It is explained here

 

[Lnewqban]

Please, see:
https://roperescuetraining.com/physics_friction_raising.php

https://roperescuetraining.com/physics_cod.php

https://roperescuetraining.com/physics_safety_factors.php

 

[A.T.]

Is this due to the friction between the rope and the wheel?

No. The force on a friction-less pulley is also the sum of the rope tensions in the straight parts (for a 180° wrap around).