Sizing a winch for loading onto a ramp

[prefetch]

Homework Statement

this isn't a homework problem, but i guess it's a basic enough question that i figured i'd post it here.

okay, so I'm trying to figure out how big of a winch to buy for a very large tow truck i am building. winches are rated based on how much they can pull. eg. a 20,000 lb winch can pull 20,000 lbs at a 0 degree incline (flat surface.)

so if i have a vehicle that weighs 40,000 lbs, and a ramp angle of 30 degrees, and a drag coefficient of .5 (if I'm towing a disabled tracked vehicle, I'm guessing a drag coefficient as high as .5 isn't unreasonable) then how big of a winch would i need to pull this up the ramp?

i'm looking for a formula, so i can change the values of my variables: weight of vehicle, drag coefficient of vehicle, winch size, angle of ramp.

 

[prefetch]

k, i found the answer out.

here it is:

The first thing you are interested in is the winch pull required to
overcome the 30 degree slope. In the explanation that is called Fp. In your
case:

Fp = Fg(weight of vehicle) x sin(30 degrees) = 40,000# x 0.5 = 20,000#

Normally a coefficient of friction related to the two surfaces in contact
is multiplied by Fn to determine the resistance in dragging the object up
the incline. However, in your case you primarily have the drag caused by
having to overcome the mechanical friction associated with turning the
track drive. I have towed track equipment before and this is a substantial
drag. You mention a drag coefficient of .5 and I wouldn't dispute this
value at all. So, the drag would be:

Fd = Fn x 0.5 = Fg x cos(30 degrees) x 0.5 = 40,000 x 0.866 x 0.5 =
17,320#

Then your total winch pull would be:

Ft = Fp + Fd = 20,000 + 17,320 = 37,320#