Relationship of acceleration in two-pulley system

AI Thread Summary
The discussion centers on determining the relationship between the accelerations a1 and a2 in a two-pulley system. Participants express confusion about how the movements of masses m1 and m2 relate to each other, particularly whether m1 moves twice as far as m2 or vice versa. Through various attempts, some participants suggest that a2 = 2a1, while others arrive at a1 = 2a2. The conversation highlights the importance of understanding the constraints imposed by the lengths of the strings in the system. Ultimately, the correct relationship is clarified as a1 = 2a2.
jake jungle
Messages
12
Reaction score
0

Homework Statement


upload_2015-2-4_20-24-52.png

What is the relationship between a1 and a2
A.a1=a2
B.a1=2a2
C.a2=2a1
D.a1=a2√2
E.a2=a1√2
F.There is no simple relationship between the two acceleration

Homework Equations


a = F/m

The Attempt at a Solution


I am quite confused with the second pulley , so the weight of m2 is divided to 2 tension through the second pulley , but i can't find what is the magnitude of the upper tension ,therefore we can't find the lower tension or the acceleration either.
So i choose F as my answer , however i am not sure . Is my answer right ?.
 
Last edited:
Physics news on Phys.org
Conserve string.
 
DEvens said:
Conserve string.

sorry i don't get it . can you be more specific , and which answer do you choose :D
 
Really? That's not enough?

Ok. When mass m1 moves a distance of 1 unit, how far must the mass m2 move?
 
That's where i was confused.
If m2 goes a distance of x. Will m1 goes a distance of x/2 or 2x?
 
jake jungle said:
That's where i was confused.
If m2 goes a distance of x. Will m1 goes a distance of x/2 or 2x?
Let the lengths of the horizontal portions be x (on the left), y (top right), z (lower right).
What equations relate them?
 
I have the same problem too
haruspex said:
Let the lengths of the horizontal portions be x (on the left), y (top right), z (lower right).
What equations relate them?

I think that it is Δx = Δy + Δz
Since Δy is the same as Δz, so Δx = 2Δy = 2Δz
So, a2 = 2a1, right ?
 
Try again.
 
BvU said:
Try again.
Hmm..
I tried and got the same answer, a2 = 2a1
Please explain the correct answer to me

EDIT :
Tried again and get a1 = 2a2
 
Last edited:
  • #10
terryds said:
Tried again and get a1 = 2a2
Good.
 
  • Like
Likes terryds
  • #11
terryds said:
Hmm..
I tried and got the same answer, a2 = 2a1
Please explain the correct answer to me

EDIT :
Tried again and get a1 = 2a2

sr i didnt understand , since S = 1/2 at^2
then if Δx=2Δy .
then ax should be equal 2 ay ?
then a2 = 2a1 ?
 
  • #12
jake jungle said:
sr i didnt understand , since S = 1/2 at^2
That's irrelevant to how the lengths x, y and z relate.
jake jungle said:
then if Δx=2Δy .
It isn't.
What combinations of lengths x, y and z are known to be constant?
 

Similar threads

Back
Top