Equilibrium and displacement of a Particle

[rico22]

Homework Statement

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

Homework Equations

F_spring= ks

s= l - l₀

ƩF=0

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.

 

[rico22]

anyone?

 

[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 F_s=(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)

 

[butan1ol]

jldibble is right

The force of the spring is 600(s)

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

 

[rico22]

the force of the spring is k(s)... k=600 is given, and I know s = l - l₀; therefore F_spring=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 F_spring=600[√(d²+1.5²) - 1.5].

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

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

ƩF_y=0: F_spring[1.5/√(d²+1.5²) - F_spring[1.5/√(d²+1.5²).

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

 

[rico22]

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

 

[rico22]

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

QUOTE]

the correct answer is d= 0.997

 

[milesyoung]

Your solution is fine:
WolframAlpha calculation

Is it solving for d that's the problem?

 

[rico22]

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

 

[gneill]

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

l₀ is also given and equals 1.5m so the equation becomes 600(l - 1.5).

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?

 

[milesyoung]

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?

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.

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

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

 

[gneill]

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'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.

 

[milesyoung]

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.

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

 

[rico22]

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

 

[Chestermiller]

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.