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Springs in Two Dimensionsby ch010308
Tags: springs in 2d 
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#1
Feb2509, 06:44 AM

P: 10

1. The problem statement, all variables and given/known data
The ends of two identical springs are connected. Their unstretched lengths l are negligibly small and each has spring constant k. After being connected, both springs are stretched an amount L and their free ends are anchored at y=0 and x= (plus minus)L as shown (Intro 1 figure) . The point where the springs are connected to each other is now pulled to the position (x,y). Assume that (x,y) lies in the first quadrant. a) What is the potential energy of the twospring system after the point of connection has been moved to position (x,y)? Keep in mind that the unstretched length of each spring l is much less than L and can be ignored. Express the potential in terms of k, x, y, and L. b) Find the force F on the junction point, the point where the two springs are attached to each other. Express F as a vector in terms of the unit vectors x and y. 2. Relevant equations F=kx 3. The attempt at a solution I don't quite understand how to approach this problem. Do we treat the 2 springs as a single system or 2 springs? Wouldn't the springs stretch by different lengths if the center point is pulled in (x,y) direction? Any help would be great! Thanks! 


#2
Feb2509, 07:02 AM

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P: 41,311




#3
Feb2509, 07:55 PM

P: 128

I have this problem as well. I can only solve part one.
For second part, my force for the string on the left (lets say force1) = kx1(cos angle1)i + kx1(sin angle1)j and my force for the string on the right (lets say force2) = kx2(cos angle2)i + kx2(sin angle2)j then sub in the cos angles and sin angles, my total force is 2kLi + 2kyj note: my i and j are the unit vectors. but the answer is wrong. it is independent of L.. Pls help!! Thanks! 


#4
Feb2509, 08:11 PM

P: 128

Springs in Two Dimensions
are my directions correct??
is it supposed to be negative?? 


#5
Feb2609, 03:23 AM

P: 1

I think it should be 2kxi2kyj



#6
Feb2609, 04:43 AM

P: 4,513

What in the blue blazed does a negligable unstretched length mean? I guess the springs in the drawing are of the non ltype.



#7
Feb2609, 08:29 AM

P: 128

i can understand that it is negative, but i cant figure out the x.. 


#8
Feb2609, 08:39 AM

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P: 41,311




#9
Feb2609, 10:06 AM

P: 128

yeah now i have 2xLi  2kyj
but how to get rid of L? 


#10
Feb2609, 10:25 AM

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#11
Feb2609, 10:54 AM

P: 619

The force in the spring depends on the elongation of the spring. After the connection point moves to (x,y), the spring on the left has length
LL = sqrt((L+x)^2+y^2) and the spring on the right has length LR = sqrt((LX)^2+y^2) The original, free length of each spring was zero (not very realistic, but that is what the problem said) Therefore the force in the spring on the left is FL = k*LL and on the right FR = k*LR, speaking in terms of magnitudes only. From this point it should be easy to complete both parts of the problem. 


#12
Feb2609, 08:26 PM

P: 10

I can solved the question already! Thank you all for helping!



#13
Feb2709, 10:55 AM

P: 6

so i found the force acting on the left spring and right spring.. to find U, it means i have to sum the total work done by both springs? in this case, i should integrate FL and FR over the displacement right? but what should be limits of this integration(i am confused due to x and y components of the question)? 


#14
Feb2709, 11:45 AM

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#15
Feb2709, 12:28 PM

P: 619

The limits of integration for the FL integration are 0 to LL and for the FR integration 0 to LR. Remember that the variable of integration is the stretch, not x or y directly.



#16
Mar1309, 10:53 AM

P: 128

Fr = F1 + F2 = [(kx1 cos angle 1) + (kx2 cos angle 2)] i  [(kx1 sin angle 1 + kx2 sin angle 2)] j =  (kL + kx + kL  kx) i  (ky + ky) j = 2kL i  2ky j where i and j are unit vectors.. 


#17
Mar1309, 01:27 PM

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#18
Mar1309, 09:54 PM

P: 128

but can i ask sth? cos initially i tot the (cos angle) and (sin angle) in the eqn will take care of the directions, so we just put all positive. it doesnt work tat way? 


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