Springs in Two Dimensions

1. Feb 25, 2009

ch010308

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 two-spring 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!

Attached Files:

• MPE_sp2_0.jpg
File size:
6.5 KB
Views:
365
2. Feb 25, 2009

Staff: Mentor

Sure. So find the potential energy of each spring separately, then add to get the total.

3. Feb 25, 2009

makeAwish

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. Feb 25, 2009

makeAwish

are my directions correct??
is it supposed to be negative??

5. Feb 26, 2009

zSuperkz

I think it should be -2kxi-2kyj

6. Feb 26, 2009

Phrak

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

7. Feb 26, 2009

makeAwish

why is it x instead of L?

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

8. Feb 26, 2009

Staff: Mentor

Spell out the details of your calculation, paying attention to signs.

9. Feb 26, 2009

makeAwish

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

10. Feb 26, 2009

Staff: Mentor

The only way to tell where you went wrong is for you to show the details of your calculation.

11. Feb 26, 2009

Dr.D

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((L-X)^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. Feb 26, 2009

ch010308

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

13. Feb 27, 2009

electricblue

please correct me if i am wrong here..
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. Feb 27, 2009

Staff: Mentor

That's the hard way. You must have found the stretch in each spring as a function of x & y. What's the potential energy of a stretched spring? Just find the potential energy of each spring, then add to get the total.

15. Feb 27, 2009

Dr.D

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. Mar 13, 2009

makeAwish

my resultant force,
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. Mar 13, 2009

Staff: Mentor

The problem is with the sign of your x-components. The springs pull in opposite directions along the x-axis, so the x-components of their forces will have opposite signs.

18. Mar 13, 2009

makeAwish

yeah, i saw my mistake! haha. thank you!!! :)

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?

19. Mar 14, 2009

Staff: Mentor

Sure, if you use the correct angle. Don't confuse θ with θ + 180°. And you still need to use the correct direction of the force vector.

20. Mar 14, 2009

makeAwish

hmm. anyway in this qns, the angles are all less than 90 degrees, so their cos and sin will still be positive right?

is it like everytime we deal with such qns, no matter the cos and sin angle will turn out positive or negative, we also need to include the direction of the force vector?