F in Friction Diagram: Explaining Opposite Force to Normal

In summary: The diagrams in the question are taken from the official mechanical engineers test of Israel's technological workforce training institute. The one I posted was from a 2006 test. The solution manual is of an Ukrainian-Israeli physics teacher, and the only known solution manual for these tests. Although, I'm considering to make my own solution manual though the rate this is going.
  • #36


Note that I have left out the internal tension forces, but I only drew the external forces on the pulley combined with rope and weight.
Reminds me of the method of joints in trusses, that's pretty much the diagram I'd have to do if I'd look at a certain joint.


I accept your drawing.
Note that they do not cancel each other out completely.
The force at the pulley is transferred to the axis of the pulley, which gives it a different working line.
So in the sum of the horizontal forces they cancel each other out, but in the sum of moments they do not.

I've done sum of all moments on (smachim) A and B, and they perfectly cancel each other out. I can't imagine a point on the structure where they don't cancel each other out even with the sum of all moments.


Well, it's not designed in reverse.
If you try to use this contraption, you would lift the weight the mass m by pulling the rope at E. You would pull with a force T, that is half of the actual weight.
However, as a result the weight exerted on the frame is greater than the actual weight.

I understand now, it did take some time and thought to accept it, but it actually makes sense. Mechanics is beautiful :)
 
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  • #37


Femme_physics said:
I've done sum of all moments on (smachim) A and B, and they perfectly cancel each other out. I can't imagine a point on the structure where they don't cancel each other out even with the sum of all moments.

What then is your sum of all moments?

Femme_physics said:
I understand now, it did take some time and thought to accept it, but it actually makes sense. Mechanics is beautiful :)

Yes it is :)
 
  • #38
  • #39


What then is your sum of all moments?
If I consider point B,

Sigma Mb = 0 ; T x (60+150+250) + T x (60) -T x (60) +Ax x 200 = 0

See, those two T's in the diagram cancel out even with the sum of all moments.

Every so often, in most subjects, a real gem of a textbook is written.
Unfortunately if it is also an older book copies/reprints become like gold dust.

So it is with mechanics - the text extract on the lifting tongs came from this book.

http://www.abebooks.co.uk/servlet/Se...asic+mechanics

The book also covers dynamics and mechanisms in a really accessible, but modern, way.

Hm...well, I'm about to finish statics and start dynamics, so I'm not sure how much it's worth to invest in a book now. You say it also has dynamics, but I want to start studying dynamics with fresh emptiness in my head, actually. If you know what I mean :)
 
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  • #40


Femme_physics said:
If I consider point B,

Sigma Mb = 0 ; T x (60+150+250) + T x (60) -T x (60) +Ax x 200 = 0

See, those two T's in the diagram cancel out even with the sum of all moments.

Yes, this works too, so I'll stop bothering you about it. :)

(Please note that your first term should have a minus sign though. ;()

[edit]And congratulations! :smile: You typed your first formula ever instead of scanning it! :smile: [/edit]
 
  • #41


And congratulations! You typed your first formula ever instead of scanning it!

LOL thank you :)

(Please note that your first term should have a minus sign though. ;()
Argh, yes, I was being careless because I was rushing to prove a point

Thanks :)
 
  • #42


Femme_physics said:
LOL thank you :)

Who needs a scanner if you can't get to it! ;)
 

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