I have hemp rope which is hanging horizontally between two winches 140 m far away from each other.
Is it physically possible that the rope won't rip when I turn the winches so the rope will be even (or hanging only 20 cm in the middle)?
I am looking for a formula which considers the weight of the rope.
My attempt so far is this with a formula which doesn’t consider the weight of the rope, only additional weight in the middle of the rope.
F(o) = Tension in Newton
F(g) = weight in the middle of the rope
F(z) = tensile force for two ends
F(s) = initial tension
l = 140 m (Length of the distance between winches)
h = 0,2 m (maximal sag)
m = 0 kg (additional weight onto the rope)
F(s) = 6000 N
My problem is that I always get 18.000 N (1835.489178 kg)
since F(z)
is always 0
.
Because with this formula my hemp rope won’t rip since a hemp rope with a diameter of 4,6 cm
(which weighs 1,5 kg/meter
) can hold 1870 kg
. So theoretically the rope can be tied horizontally.
But the rope itself weighs 210 kg
and if I assume only 1,5 kg
in the middle of the rope I get F(o) = 25.725 N (= 2620 kg)
- my rope rips.
How can I consider the weight of the rope in the formula? Either the total weight (as
210 kg
) or the weight of1 m
rope (as1,5 kg
).