- #1
epheterson
- 22
- 0
1. The Funny Intro:
The summer before he entered high school, Prof. Schueller’s had to run the farm because
his father had an emergency appendectomy. To adjust the “thread” (the width between
the wheels) of a John Deere 4020 tractor, a torque of about 1600 ft-lbs. was needed to
loosen the rusted clamping bolts.
Real Question: With given force and torque arm, how far many degrees past horizontal axis will the bolt turn?
2. What I got
I'm on the last part of the problem here, I have a 6-ft. arm producing 1800 ft.-lb. of Torque.
3. Where I am
I know the 1600 ft-lbs. are used simply to loosen the bolt, but I'm unclear exactly how to measure the degrees the resulting 200 ft-lbs. turns the bolt.
I'm hoping this is an easy one for you guys but he didn't mention a situation like this in class and I don't see the solution in the book. Any help is appreciated :).
For the fun of it, I included the picture my professor drew.
As always, much thanks!
The summer before he entered high school, Prof. Schueller’s had to run the farm because
his father had an emergency appendectomy. To adjust the “thread” (the width between
the wheels) of a John Deere 4020 tractor, a torque of about 1600 ft-lbs. was needed to
loosen the rusted clamping bolts.
Real Question: With given force and torque arm, how far many degrees past horizontal axis will the bolt turn?
2. What I got
I'm on the last part of the problem here, I have a 6-ft. arm producing 1800 ft.-lb. of Torque.
3. Where I am
I know the 1600 ft-lbs. are used simply to loosen the bolt, but I'm unclear exactly how to measure the degrees the resulting 200 ft-lbs. turns the bolt.
I'm hoping this is an easy one for you guys but he didn't mention a situation like this in class and I don't see the solution in the book. Any help is appreciated :).
For the fun of it, I included the picture my professor drew.
As always, much thanks!