What Happens to Torque as the Wrench Lowers?

  • Thread starter Thread starter epheterson
  • Start date Start date
  • Tags Tags
    Bolt Torque
AI Thread Summary
The discussion centers on calculating the angle through which a bolt will turn when torque is applied using a wrench. A 6-foot arm is producing 1800 ft-lbs of torque, but clarification is needed on how to measure the degrees the bolt turns with the remaining 200 ft-lbs after overcoming the initial 1600 ft-lbs required to loosen it. Participants suggest that the angle should be determined when the torque drops below 1600 ft-lbs. The problem involves two individuals applying force at different positions, which complicates the calculation. Overall, the thread seeks assistance in solving this torque-related question, emphasizing the need for more specific information.
epheterson
Messages
22
Reaction score
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!
 

Attachments

  • tractor.gif
    tractor.gif
    11.1 KB · Views: 488
Physics news on Phys.org
Looks like you aren't given enough information. Did you post the question word for word as it was given?

Perhaps they are looking for the angle at which the torque drops below 1600 ft-lbs.
 
I was getting the feeling that I'm lacking information too.

Verbatim: Assume that “x” = 6 ft. If Ed stands at “x” and Schueller at “x/2” and they start at a horizontal position, how far (phi in degrees) will they be able to turn the bolts?

Ed weighs 250 lb, and Schueller weighs 100

Is this what you were talking about, Doc Al?


Attached PDF so it can be very clear.
 

Attachments

OK, the best you can do is what I had suggested. As the bolt begins to turn and the wrench lowers, what happens to the torque that is exerted by the men? Find the angle at which the torque is no longer enough to turn the bolt.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top