Mobile equillibrium

  • Thread starter Jasbroek
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  • #1
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Homework Statement


physics-1.jpg



Homework Equations


-

The Attempt at a Solution


I've tried to calculate moment counter clockwise,
0.4 * 10 (Fg of the yellow piece) + 0.6* 50 = 34 N cm
this is how far I got...
I don't know how to proceed from here on..
please help ;)

and sorry for my english
and sorry if this is posted in the wrong section ;)
 
Last edited:

Answers and Replies

  • #2
gneill
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Be careful with the "extra" lengths of the horizontal rails (yellow pieces). The unbalanced lengths will act like an extra weight located at the center of mass of that "excess" bit. So, for example, the top rail is 30cm longer on its left than on its right. There's a 30cm "excess" or unbalanced part of the rail on the left end. The center of mass of that unbalanced end will be located 30cm/2 = 15cm from the left end of the rail.
 
  • #3
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wops sorry
my mistake..
The "first" bar isn't 20cm on the right, but 30...
updating picture atm
 
  • #4
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ok
I might have the solution.
please somebody confirm if I'm doing the right thing. ;)

So the Mccw = 34 N cm
Mccw = Mcw = 34 N cm
F=M/distance F= 34/30 = 1.13333333333334 N
so the weight has to be 113 grams.
The stick is 40 grams, so B + C = 73 grams
The sticks centre of gravity acts 5 cm counter clockwise.
so M = 5 x 0.4 = 2 N cm
to cancel this force, weight C has to be at least 20 grams (0.2 x 10 = 2 N cm)
so now, B + C = 53 grams.
B = 1/3 of 53 = 17 2/3
C = 2/3 of 53 = 55 1/3 grams

can any one confirm this?
if so,
thanks a lot ;)
 
  • #5
gneill
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Can you detail how you found the 34 N cm for the Mccw?
 
  • #6
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0.4 * 10 (Fg of the yellow piece) + 0.6* 50 = 34 N cm

this is already in the first post ;)
calculate Fg of weight A (0.06kg * 10 = 0.6N)
and M= F * d
so M = 0.6 * 50 = 30 N cm
for the stick itself
Fg = 0.04 * 10 = 0.4N
0.4 *10 = 4N cm
30+4 = 34N cm
 
Last edited:
  • #7
gneill
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So you are assuming that the acceleration due to gravity is 10 m/s^2. that's fine, it cancels out everywhere anyways.

You are also assuming that the "sticks" are 40 grams on both sides of the support? So the entire top rail would be 80 grams? It's not obvious from your problem statement.

What is the "10" in your second to last line? I presume you're multiplying by some distance in centimeters (although you put the result in Newtons ???). Where does that 10cm come from?
 
  • #8
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the sticks in total ar 40 grams, so the compelete rail is 40 grams.
the 10 are the 10cm, it should be N cm indeed, my mistake.
the 10 cm comes from the centre of gravity.
as the centre of gravity acts from the middle of the stick (80cm /2 = 40cm so 10 cm to the left of the string)
 
  • #9
gneill
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20,907
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Hmm. The counter clockwise torque should be due to all the mass that resides to the left of the string. You should find the center of mass of the portion of the stick that is to the left of the string, as well as the mass of that portion.

Alternatively, you could "wave a hand" and declare that only the excess length of stick makes a difference, since the rest is equally balanced by an identical amount of stick on the other side. Then you'd just have to worry about the 20cm more stick on the left.
 

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