It is possible to solve this question without F given to me?

  • #1
Femme_physics
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Problem Statement

http://img715.imageshack.us/img715/6372/galde.jpg [Broken]


A wheel is set upon a shaft in a small gap. At point C of the wheel is acting horizontal force F, as described in drawing. The needed measurements are given in a general manner. The friction coeffecient between the wheel and shaft is mieu. What's the max distance (e) where there would still be forward horizontal movement of the wheel on the shaft (presuming F remains constant)

Homework Equations





The Attempt at a Solution



I've no clue, I can write the equations, but what's the point if F can be a million Newtons of 1 Newton?

http://img10.imageshack.us/img10/767/muchraz1.jpg [Broken]

http://img707.imageshack.us/img707/6277/muc2.jpg [Broken]
 
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Answers and Replies

  • #2
I like Serena
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Good evening Fp! :smile:

Good you already got these equations! :cool:

Don't worry about the size of F, we'll get to that later.

First, let's try to complete the set of equations.
Then, let's see if we can quickly eliminate a couple of variables.

Your sum of moments has the wrong distance for F.
Can you correct that?

And I'm missing the equations that relates friction to normal force.
Can you add those?


Running ahead and assuming you got the equations, can you eliminate the friction forces?
That is, replace them by the formulas you should have just added?
 
  • #4
I like Serena
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Don't be discouraged! :wink:

Let's just see what happens if we move on.

First, though, you adjusted the distance of the force F, but you have the wrong amount: you wrote 2/h, but it should be h/2.

Then, can you go on eliminating unnecessary variables?
(I'm thinking of eliminating first NA, and then NB.)

Then you should be left with one equation with only F and e.
When we have that, we'll see if there is something else you can do! :smile:
 
Last edited:
  • #6
I like Serena
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You forgot to replace 2/h by h/2.

And can you simplify the expression?
(Get rid of the round thingies, calculate the numbers as far as possible, and match up similar parts?)
 
  • #8
I like Serena
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You didn't get rid of the round thingies properly.
(Do you know the "rules" to get rid of round thingies? :wink:)

Edit: just noticed, your distance for F is still not right.
 
  • #10
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As I said, your distance for F is still wrong (in a previous scan).
It should be (e + h/2) and not (e - h/2).

After you have corrected that, you should divide your equation by F.
(Remember the Wheatstone Bridge, pardon me, "Electrobob!" where you did something similar? :wink:)
 
  • #11
Femme_physics
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Can you scoop me up the answer since I'm about to turn in and the test is tomorrow morning? The test is with open material so I can print it in case there'll be anything similar. I know it's a lot to ask I'll understand if you refuse!
 
  • #12
I like Serena
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We're not supposed to give out full solutions! :surprised

That is, unless the OP has shown significant work. :approve:

Ok.

Your last equation should be (with corrected distance for F):
[tex]-F \cdot (e + \frac h 2) + \frac F {0.4} \cdot b + \frac F {0.4} \cdot 0.2 \cdot h = 0[/tex]

Simplifying:
[tex]-F \cdot e - F \cdot \frac h 2 + \frac F {0.4} \cdot b + F \cdot \frac h 2 = 0[/tex]

Canceling equal and opposite terms:
[tex]-F \cdot e + \frac F {0.4} \cdot b = 0[/tex]

Dividing by F:
[tex]-e + \frac b {0.4} = 0[/tex]

Since the problem asked for e, solve for e:
[tex]e = \frac b {0.4}[/tex]

Tadaaaa! :smile:

(Give me a kiss? :blushing:)
 
  • #13
Femme_physics
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Wait, manual says "e" equals 120mm. Is that a mistake? But then again, I now see I wrote "mistake" in this page of the solutions heh. See!!! That's what confused me. Gah! I thought this was a full numeral solution, which is why I figured it was impossible. Now I get it! :D


Thanks!! *smooches*
 
  • #14
I like Serena
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Aha, we missed that b was given as 48 mm!
Fill it in and you get your e=120 mm. :smile:

Still, the clue is that F cancels when you work it out.
This should make some intuitive sense.

If you push twice at hard, all the forces (normal forces and friction forces) will also be twice as big.
For the situation where the system is in equilibrium, it does not matter how big F is!


*smooches* :blushing:
 
  • #15
Femme_physics
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I don't see how F cancels out, frankly, but I don't really have time to work it out

Thanks
 

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