# Help! Force of a Spring for Small displacements

1. Mar 15, 2009

Help! Force of a Spring for "Small" displacements

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

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:

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

Fsp 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.

2. Mar 15, 2009

### gabbagabbahey

Re: Help! Force of a Spring for "Small" displacements

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?

3. Mar 15, 2009

Re: Help! Force of a Spring for "Small" displacements

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.

4. Mar 15, 2009

### gabbagabbahey

Re: Help! Force of a Spring for "Small" displacements

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 [Broken]

Last edited by a moderator: May 4, 2017
5. Mar 16, 2009