Simple but maybe tricky for me

  • Thread starter chawki
  • Start date
In summary, the conversation discusses a problem involving a figure with an upright rigid rod and a flexible rope attached to a weight. The question is asking for the difference in length between two sections, CF and CB, when the rod is translated to a horizontal plane. The solution involves applying Pythagoras' law and taking the square root to find the difference in length.
  • #1
chawki
506
0

Homework Statement


In the figure between AB is an upright rigid rod and flexible access to B from a rope which passes around the points C and D are located in the rollers for spot through the point E, where the rope is attached to weight.


Homework Equations


How much in section E, this weight will increase when the rod AB be translated into a horizontal plane


The Attempt at a Solution


I really don't get it..any hints?
 

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  • #2
What are we trying to find here? tension in rope?

I don't see how the weight will increase by moving the rod down. In fact, it may drop seeing the elevation might change if u want to consider the minute gravity drop.
 
  • #3
hi chawki! :smile:

i'm confused :confused: …

is this just a geometry problem, to find how much higher M goes if the rope goes straight from C to the bottom of that arc, instead of only from C to B?
 
  • #4
tiny-tim said:
hi chawki! :smile:

i'm confused :confused: …

is this just a geometry problem, to find how much higher M goes if the rope goes straight from C to the bottom of that arc, instead of only from C to B?

Hello Tiny-tim :shy:
Actually it will go from B to the bottom, and yes it's probably to find how much higher that mass will go
 
  • #5
then that's just subtracting one side of the triangle from another, isn't it?
 
  • #6
If its the new heigth of the mass.

Lets call the tip of the rod when it's down F. the answer would be FC - BC

u can find the length of FC by using pythaegaros, u got AC and AF(AB)
 
  • #7
Ok guys I'm lost...what triangle?
 
  • #8
using the tip of the rod as F, the triangle would be FCA when the rod is horizontal.
 
  • #9
Like this ?
 

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  • #10
Yep and the amount E rises will simply be the difference between the tip of the rod and point C in both scenarios.
 
  • #11
Ok, if i get it right, the answer would be like this:
(0.5+x)2=2.52+22
we solve the equation x2+x-10=0
Delta=41
x1=-3.70 (it can't be because it's a negative value)
x2=2.70
so x=2.7m.
 

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  • #12
am i right?
 
  • #13
chawki said:
In the figure between AB is an upright rigid rod and flexible access to B from a rope which passes around the points C and D are located in the rollers for spot through the point E, where the rope is attached to weight.
…
How much in section E, this weight will increase when the rod AB be translated into a horizontal plane

The question seems to be asking for the difference between the lengths CF and CB.

That's x = √(2.52+22) - 0.5

That's the same as your equation…

chawki said:
Ok, if i get it right, the answer would be like this:
(0.5+x)2=2.52+22
we solve the equation x2+x-10=0 …

… except for some reason you've turned it into a quadratic equation ……

why?? :confused:​
 
  • #14
i just applied pythagore law
 
  • #15
chawki said:
i just applied pythagore law

yes, your first line was Pythagoras' law, but then you should just have taken the square root (that gets you my equation), not made a quadratic out of it :wink:
 
  • #16
but what do you mean ?
the point is the find x, isn't ?
 
  • #17
yes, but quickly …
tiny-tim said:
That's x = √(2.52+22) - 0.5

:wink:​
 

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