Conservation of String Exercise

Click For Summary
The discussion focuses on applying the conservation of string principle to a problem involving a symmetric sub-pulley. Participants emphasize that the string's length remains constant, which leads to equations relating the vertical positions of the masses and the pulley. They suggest differentiating these equations to derive relationships for velocities and accelerations. A key point is that the acceleration of the left mass is the negative average of the accelerations of the right two masses, highlighting the interconnectedness of the system. Overall, the conversation aims to clarify the application of these concepts in solving the problem effectively.
mancity
Messages
26
Reaction score
2
Homework Statement
Explain why the acceleration of the left mass equals negative the average of the accelerations of the right two masses.
Relevant Equations
Conservation of string
I'm not quite sure how to apply conservation of string to this problem, so guidance would be appreciated. Normally as long as there isn't a "sub-pulley" I can do the problem fairly easily but this one tricks me up. Thanks
 

Attachments

  • 2.png
    2.png
    2.6 KB · Views: 68
Physics news on Phys.org
mancity said:
Normally as long as there isn't a "sub-pulley" I can do the problem fairly easily but this one tricks me up. Thanks
The sub-pulley is symmetric, correct? That is, there is no difference between its right and left hand sides? They are mirror images of each other?
 
jbriggs444 said:
The sub-pulley is symmetric, correct? That is, there is no difference between its right and left hand sides? They are mirror images of each other?
I take the masses as unknown.

mancity said:
I'm not quite sure how to apply conservation of string to this problem,
Write equations relating string (section) lengths to heights of masses, throwing in constants as necessary. Differentiate twice.
 
mancity said:
Homework Statement: Explain why the acceleration of the left mass equals negative the average of the accelerations of the right two masses.
Relevant Equations: Conservation of string

I'm not quite sure how to apply conservation of string to this problem, so guidance would be appreciated. Normally as long as there isn't a "sub-pulley" I can do the problem fairly easily but this one tricks me up. Thanks
'Conservation of string' is an unusual way to state that the length of the string is constant. So when you set up a set of equations for your exercise, one of them is a relationship between
##y_2##, the vertical position of the middle mass,
##y_3##, idem rightmost mass
##y_5##, the vertical position of the center of the pulley on the right:
##y_5-y_2+y_5-y_3=C##

Differentiation wrt time gives an equation for the vertical velocities; a second differentiation yields another for the accelerations.

##\ ##
 
mancity said:
Homework Statement: Explain why the acceleration of the left mass equals negative the average of the accelerations of the right two masses.
Relevant Equations: Conservation of string

I'm not quite sure how to apply conservation of string to this problem, so guidance would be appreciated. Normally as long as there isn't a "sub-pulley" I can do the problem fairly easily but this one tricks me up. Thanks
Relative to the bottom pulley, the average acceleration of the bottom two masses is zero. The acceleration of the upper mass is minus the acceleration of the bottom pulley.
 
Chestermiller said:
Relative to the bottom pulley, the average acceleration of the bottom two masses is zero. The acceleration of the upper mass is minus the acceleration of the bottom pulley.
My impression is that the OP has been instructed to use conservation of string length to obtain the result. That suggests to me applying your framework to positions and then differentiating.
 
  • Like
Likes Chestermiller
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 9 ·
Replies
9
Views
1K
Replies
2
Views
1K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 12 ·
Replies
12
Views
994
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
989
  • · Replies 6 ·
Replies
6
Views
2K