M936, M936A1, M936A2 has a 45,000 pound rear winch. All other models has a 20,000 winch on the front according to the 5-ton TM.
All the numbers I us come from the 5-Ton TM. For this "experiment" I am using the M931A2 and the M936 as the 2 example trucks as they are the lightest and heaviest according to the TM.
The outlines for this calculation:
Both trucks are stuck in deep mud, the effort required will be listed with 2 numbers; the pulling effort required at a flat recovery, and the effort required if trying to recover at a 45 degree angle. All numbers in lbs
M931A2
Base weight: 19,895 lbs
Rolling resistance, Deep mud: 19,895 * 0.66 = 13,130.7 lbs
Gradient Resistance (0°) = 2 * 19,895 * 0 / 100 = 0 lbs
Gradient Resistance (45°) = 2 * 19,895 * 45 / 100 = 17,905.5 lbs
Total Pulling Effort required (0°) = 13,130.7 lbs
Total Pulling Effort required (45°) = 31,036.2 lbs
(Rolling + Gradient Resistance)
M936
Base weight: 39,334 lbs
Rolling resistance, Deep mud: 39,334*0.66 = 25,960.44
Gradient Resistance (0°) = 2 * 39,334 * 0 / 100 = 0 lbs
Gradient Resistance (45°) = 2 * 39,334 * 45 / 100 = 35,400.6 lbs
Total Pulling Effort required (0°) = 25,960.44
Total Pulling Effort required (45°) = 61,361.04 lbs
(Rolling + Gradient Resistance)
So as you can see its not a tiny weight we are dealing with one way or the other. And one thing worth noting is that the numbers provided is the
effort required. But that doesn't mean that your 20,000 lbs winch with a snatch-block can simply recover a stuck up to its axles in mud M931A2 up a 45° slope. Theoretically it can, BUT you would need to secure the recovering vehicle in such a way that its anchored weight is sufficient, otherwise all you would simply achieve is to pull the recovering vehicle towards the vehicle being recovered.
So there you have it