Help Force of a Spring for Small displacements

[Saladsamurai]

Help! Force of a Spring for "Small" displacements

Homework Statement

Okay, here is the diagram,

For "small horizontal displacements" of the mass:

For the force exerted by each spring (TOTAL, not resolved) they are using:

F_sp=kxcos(45) and then resolving it onto the x-axis they use

F_{sp x}=kxcos25*cos45

I do not understanding this at all.

Why is the unresolved force using cos*45 ?

I am thinking it has to do with the fact that we are assuming that we are making
only "small horizontal displacements" of the mass.

 

[gabbagabbahey]

Why is the unresolved force using cos*45 ?

I am thinking it has to do with the fact that we are assuming that we are making
only "small horizontal displacements" of the mass.

The key word is horizontal as-in 'along the x-axis'. The springs exert a force in response to displacements along their lengths. What component of the horizontal displacement points along the length of each spring?

 

[Saladsamurai]

Well, apparently x*cos(45)...

But I have drawn it out like 4 times and cannot seem to see it. Guess you cannot really help me with that part though.

 

[gabbagabbahey]

Well, let's look at the spring in the first quadrant: the angle between the horizontal displacement (call the displacement \dec{d}=x\hat{i}), and the unit vector along th length of the spring (call it \hat{n}) is 45 deg. To get the component of d along n you take the dot product of the two vectors, and \vec{d}\cdot\hat{n}=||\vec{d}||*||\hat{n}||\cos(45)=(x)*(1)\cos(45)

Alternatively you can just use trig:

http://img19.imageshack.us/img19/1719/springa.th.jpg

 

[Saladsamurai]

Yes, I was using the wrong triangle like the jerk that I am. Thank you for that G... That offers a lot of insight into my problem-solving strategies.

Thanks again for your time,
Casey