Totally off topic....... puzzle

[elevmike]

I'm looking for a hint on the 23/30 also..

 

[MarkTTU]

I'm looking for a hint on the 23/30 also..

Same here...

 

[Mickey]

Answer: on a calendar



http://www.janbrett.com/pocket_calendar/pocket_calendar_23_30_24_31.htm

 

[gbradley]

Good Job Mickey!

I noticed 7 apart but didn't think of a calendar.

13/20 is another...

 

[bernie_carlton]

I can't see 13/20 but 24/31 would be another

 

[Steve Bailey]

Well, as long as we're offering puzzles...

You have to tighten a 19 mm hex nut to a specified torque. Unfortunately, it's close to the frame and your 12" torque wrench won't fit. Fortunately, it's near the top of the frame and you can fit a 19 mm open end/box end wrench so that the box end extends above the frame. The distance between the center of the open end and the center of the box end is 9".

You also happen to have a set of hex sockets that fit your torque wrench, but it's an English dimensioned set. Fortunately, you remember that 19 mm is very close to 3/4". You select the 3/4" hex socket, put it on the torque wrench, and put the male hex end through the box end of the wrench. Get the picture?

The qustion now becomes:

1. If the nut is supposed to be tightened to 4 Nt-m, what will the torque wrench read when the nut is properly torqued? BTW, the torque wrench is calibrated in Ft-Lbs.

2. Does it matter whether or not the handle of the torque wrench is parallel to the handle of the open end/box end wrench?

2a. If the answer to qustion 2 is "yes, it matters", what should the torque wrench read if it is at a 30 degree angle to the open end/box end wrench?

3. Does it matter where you grab the torque wrench when you're tightening the nut?

 

[elevmike]

1 ftlb = 1.35582 Nt-m

2) Yes,

2a) too much thinking required for me right now..its Saturday.

3) Yes, thats why theres a piviot on the handle...

You need the answer to 2a before you can calculate the anwer to 1.

 

[gbradley]

I can't see 13/20 but 24/31 would be another

Ok will 21/28 work?

 

[bernie_carlton]

Actually I believe only 23/30 and 24/31 fit this problem.

 

[Terry Woods]

1. If the nut is supposed to be tightened to 4 Nt-m, what will the torque wrench read when the nut is properly torqued? BTW, the torque wrench is calibrated in Ft-Lbs.

A-1 (assuming only the 12-inch torque wrench is applied): 2.948 Ft-Lbs
A-2 (assuming the 12-inch torque wrench is applied aligned with the 9-inch metric wrench): 1.684 Ft-Lbs

2. Does it matter whether or not the handle of the torque wrench is parallel to the handle of the open end/box end wrench?

A: Absolutely!

2a. If the answer to qustion 2 is "yes, it matters", what should the torque wrench read if it is at a 30 degree angle to the open end/box end wrench?

A: 1.939 Ft-Lbs

3. Does it matter where you grab the torque wrench when you're tightening the nut?

A: NO. If you grab-on at the end (assuming a parallel alignment), you only need to apply the actual required pounds of force; 1.684 Pounds of Force. The indicator will show 1.684 Ft-Lbs of Force.

However, if you grab-on closer to the head, then you will need to apply more actual pounds of force. The closer to the head, the more actual force is required. However, the indicator will still indicate the "effective torque". You need to apply whatever force is necessary to produce an indication of 1.684 Ft-Lbs.

I don't think that the mixture of Metric and English makes any difference.

 

[Kent Hostetler]

Got it on my first try (and every one thereafter). I don't agree with the "the smarter you are the longer it will take". I used that line as a hint, and approached it with an empty mind -- and bongo!, er..bingi, uh... bingo! oops- had to refill it up a little!

 

[Steve Bailey]

Terry,

It's true that as you apply the force (grab the torque wrench) closer and closer to the center of the nut, you need to apply more force to create the same amount of torque. What will the torque wrench read when you apply the force twelve inches in from the end of the torque wrench? That would be the center of the connection between the torque wrench and the box end wrench.

 

[Terry Woods]

Let's try this again...

1. If the nut is supposed to be tightened to 4 Nt-m, what will the torque wrench read when the nut is properly torqued? BTW, the torque wrench is calibrated in Ft-Lbs.

(4 N-m = 2.952 Ft-Lbs)

If force is applied at a distance equal to the length of the metric wrench...
2.952 Ft-Lbs / .75-ft = 3.936 Lbs of Force at point "B".
That is, a force of 3.936-Lbs applied at a distance of .75-Ft will produce 2.952 Ft-Lbs at the nut.

Now, if we assume that the nut is already tightened to 2.952 Ft-Lbs (exactly 4 N-m) then, if we were to apply 3.936 Lbs of Force at point "B" there would be no further motion on the nut and no futher increase in the torque applied to the nut. The resisting force at the nut and the force applied at point "B" would be in equilibrium.

Since point "A" and point "B" are in equilibrium... with 3.936-Lbs of force applied at point "B", in a CW direction relative to point "A", thus developing 2.952 Ft-Lbs of CW torque at point "A"... it would be just as true to say that 3.936-Lbs of force are being applied to point "A", in a CCW direction relative to point "B", thus developing 2.952 Ft-Lbs of CCW torque at point "B".

The CW torque applied to point "C" needs to over-come the CCW torque at point "B".

Since the CCW torque at point "B" is 2.952 Ft-Lbs, the torque developed by applying force to point "C" must develop at least 2.952 Ft-Lbs of CW torque to equal the CCW torque. Since the torque wrench is 1-foot long, the force required is equal to the opposing torque... 2.952 Lbs thus developing 2.952 Ft-Lbs.

Any additional force will increase the torque beyond the required 2.952 Ft-Lbs.

2. Does it matter whether or not the handle of the torque wrench is parallel to the handle of the open end/box end wrench?

Absolutely!

2a. If the answer to qustion 2 is "yes, it matters", what should the torque wrench read if it is at a 30 degree angle to the open end/box end wrench?

Effective length of the torque wrench is (12 COS 30)" = 10.392" = .866-Ft

2.952 Ft-Lbs / .866-Ft = 3.408 Ft-Lbs

3. Does it matter where you grab the torque wrench when you're tightening the nut?

A: NO. If you grab-on at the end (assuming a parallel alignment), you only need to apply the actual required pounds of force; 2.952-Lbs of Force. The indicator will show 2.952 Ft-Lbs of Torque.

However, if you grab-on closer to the head, then you will need to apply more actual pounds of force. The closer to the head, the more actual force is required. However, the indicator will still indicate the "effective torque". You need to apply whatever force is necessary to produce an indication of 2.952 Ft-Lbs of torque. The indication is based on the assumption that the user will grab-on at the handle.

As you grab-on closer to the head you have to apply more force. If you grab-on at the common point (point "B") then you have to apply 3.936-Lbs of Force (3.936-Lbs x .75-Ft = 2.952 Ft-Lbs). However, the indicator will show Zero Ft-Lbs because no force is being applied through the torque wrench.


If that ain't right Steve... please don't keep me in suspense.

 

[Steve Bailey]

Using Terry's nomenclature, point A is the nut that we want to tighten to 4 Nt-m torque (2.95 Ft-Lb). Point B is the junction of the metric wrench and the torque wrench. Point C is the place where we apply the force to the torque wrench. We want to apply 2.95 Ft-Lb at point A, but the only way we have available to measure that torque is at point B.

The force required to produce 2.95 Ft-Lb at point A will indeed be 1.68 Lbs applied at a distance of 1.75 Ft (9 inches plus 12 inches). But according to the rules I set forth (it's my puzzle and I get to make the rules), we don't have any way to measure the applied force; we can only measure the torque developed at point B. You still haven't answered what the torque wrench (point B) will read when the nut at point A is properly tightened. Hint: It's not 2.95 Ft-Lb.

A guy from the assembly shop at a company I used to work for came to me with this problem many years ago. The answers to parts 1 and 2 are straightforward Statics, but to me the most interesting aspect of it is part 3. I couldn't believe the conclusion I arrived at, and I had to prove it to myself.

 

[Terry Woods]

3.936 Lbs - 1.684 Lbs = 2.242 Lbs
-OR-
3.936 Lbs + 1.684 Lbs = 5.610 Lbs

hmmm...

I think it's 5.610 Lbs