Equilibrium and displacement of a Particle

1. May 28, 2013

rico22

1. The problem statement, all variables and given/known data

The springs AB and BC have stiffness k and an unstretched length of l. Determine the
displacement d of the cord from the wall when a force F is applied to the cord. See Picture attached.

Given:
l = 3 m
k = 600 N/m
F = 200 N

2. Relevant equations
Fspring= ks

s= l - l0

ƩF=0

3. The attempt at a solution
The force of the spring is 600(s) and I know that s = length of spring stretched - length unstretched. So for the sum of forces in the x direction I get 2(600)(s)[d/√(d2+1.52)] = 200... I divide both sides by 2 and get (600)(s)[d/√(d2+1.52)] = 100...

from here I know that I can get s from equation above thus it becomes 600[√(d2+1.52) - 1.5][d/√(d2+1.52)] = 100... but I don't know what else to do from here... any replies would be greatly appreciated.

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2. May 29, 2013

rico22

anyone?

3. May 29, 2013

jldibble

Do you know what the answer is supposed to be? When I work it out, I get d=0.16667m.

I think you are addressing the forces in the x direction incorrectly. If you look at the force exerted on the spring, you can think about it as stretching the spring a distance d in the x direction while also stretching the spring l/2 in the y direction.

When you look at it like this, then the force due to one spring in the x direction is Fs=(k)(d). And since you have two springs the forces in the x direction should look like this:

200=2(k)(d)

Solve for d.

I don't know why you're using the term [d/√(d2+1.52)]

(someone please correct me if I'm way off on this)

Last edited: May 29, 2013
4. May 29, 2013

butan1ol

jldibble is right

I think you got your idea wrong when you said the force of the spring is 600.

5. May 29, 2013

rico22

the force of the spring is k(s)... k=600 is given, and I know s = l - l0; therefore Fspring=600(l-l0).

l0 is also given and equals 1.5m so the equation becomes 600(l - 1.5). Looking at the triangle the two springs create with the wall we can deduce that l (length of the stretched spring) is going to be equal to √(d2+1.52) if we split it into 2 right triangles and from the Pythagorean theorem.

So putting all this together Fspring=600[√(d2+1.52) - 1.5].

Now if we were to break it down into its x and y components:
ƩFx=0: Fspring[d/√(d2+1.52) + Fspring[d/√(d2+1.52) - 200 = 0

the √(d2+1.52) would be the equivalent of cosθ.

ƩFy=0: Fspring[1.5/√(d2+1.52) - Fspring[1.5/√(d2+1.52).

My issue is when I solve for d.... I guess I might have to do trial and error.

6. May 29, 2013

rico22

sorry it should be s=*lstretched-l0...

7. May 29, 2013

rico22

8. May 29, 2013

milesyoung

9. May 29, 2013

rico22

right I know how to set up the problem... but once I try to solve for d the algebra just throws me off.

10. May 30, 2013

Staff: Mentor

The original post said that the unstretched length of a spring is l (lower case L). But l is given as 3m. Why are you using l/2 for the unstretched length?

11. May 30, 2013

milesyoung

I think it's a typo (or just very poorly worded) since the attached drawing, which seems to have been copied directly from the text, shows their combined unstretched length as l and it's consistent with the solution given.

I doubt you're supposed to solve it analytically. Approximate the solution by making an inital guess and iterate.

12. May 30, 2013

Staff: Mentor

I'd have to side with it being a typo, since the diagram does not actually imply that the unstretched length is l/2 (although it is tempting to believe so for convenience sake!). The initial configuration could have the unstretched springs form an equilateral triangle with the wall.

13. May 30, 2013

milesyoung

Yes, you're right. I assumed too much based on the appearance of the sketch.

14. Jun 8, 2013

rico22

yeah sorry i worded the problem wrong...still, thank you for all the replies...ill do a better job next time.

15. Jun 9, 2013

Staff: Mentor

original length of each spring = l0/2

extended length of each spring = $\sqrt{d^2+(l_0/2)^2}$

tensile force in each spring = $k(\sqrt{d^2+(l_0/2)^2}-(l_0/2))$

horizontal component of tensile force in each spring = $k(\sqrt{d^2+(l_0/2)^2}-(l_0/2))\frac{d}{\sqrt{d^2+(l_0/2)^2}}=kd(1-\frac{(l_0/2)}{\sqrt{d^2+(l_0/2)^2}})$

sum of horizontal components of tensile force from springs = $2kd(1-\frac{(l_0/2)}{\sqrt{d^2+(l_0/2)^2}})$

$F=2kd(1-\frac{(l_0/2)}{\sqrt{d^2+(l_0/2)^2}})=2kd(1-\frac{1}{\sqrt{1+(\frac{2d}{l_0})^2}} )$

The solution to this equation for d is 0.997 m.

Last edited: Jun 9, 2013