Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Bulldozer Statics Question?
Basically with this problem I have to figure a way to start with the bulldozer track, work back through the gear box and torque converter to the engine to find some sort of ration. Then work back to find the torque of the bulldozer track. I am having trouble with calculations and is hoping someone can shed some light on this problem . . .
A bulldozer is equipped with a Diesel engine and drive train (Torque Converter, Gear Box, and Bulldozer Track w/ Drawbard). With the transmission in low gear, find the force at the drawbar, Fd , when the bulldozer is traveling at 1.5 mph and the engine speed is 1500 rpm.
Gear box:
Mech. efficiency, G = 0.90
Gear ratios of gearbox:
High, 1 to 1
Low, 2 to 1
Reverse, 2 to 1
Differential:
Mechanical efficiency, d = 0.95
Gear ratio: 5 to 1
Diameter of drive shafts:
Engine to torque converter, 2 in.
Torque converter to gearbox, 3 in.
Gearbox to differential, 3 in.
Rear axle, 3 in.
Diameter of rear sprocket wheel for bulldozer track, 3 ft.
Height of drawbar from ground, 1 ft.
Dia. of knob on gearbox handle: 1.5 in.
But the two speeds are relating to two different parts right? The 1.5 mph to the rim, and the 1500 rev/min to the engine speed respectively. What I had done originally before starting was converting the 1.5 mph to 2.2 ft/sec.
2 Answers
- 1 decade agoFavorite Answer
Why don't you just ask Dr. Shen for help? However, the report is due tomorrow. So, you might be out of luck.
Or ask Dr. Obie, but he probably won't be able to help much!
- wingstwoLv 71 decade ago
The problem statement is incorrect and incomplete.
But there are two equations: First determine the speed. Then relate Power = Force * speed to engine power or torque, considering efficiency.
Speed at rim of rear sprocket wheel:
1500 rev/min * 1/2 tranny * 1/5 differential * 3Ïft/rev * 60min/hr * mile/5280ft = 16.065 mph, not the 1.5mph stated.
So then, the power:
Power out = Force * speed = P_engine * ε_tranny * ε_diff
Also, power at engine = torque * angular velocity
You need more info.
edit:
yes, the speeds are two different ends of the drive chain. However, since the speed computed does not match the actual speed, there is another >10x reduction that is incorrectly left out of the problem statement. Further, without force, power, or torque specified, no way to detemine any of the others.
