• Seismicmaniac
I will get back to you on that.In summary, if the middle bar is pulled with a force of 100lbs, the block will move.

#### Seismicmaniac

I have a rectangular bar in between two rectangular bars ( a sandwich ). The outer two bars are attached to springs on the outer sides and press against the middle bar with a force of N ( due to springs ) thus acting as support for the middle bar. The middle bar has a free translation motion ( under the effect of friction due to contact in outer two bars). The question is that what should be the value of ' N ' if the middle bar is pulled with a force of 100 Lbs ( for example ) ? Please help out ...

Last edited:

I think you would first have to determine the cofficent of friction between the two bars.

The formula is FF=FN*CF, where FF is the force of friction (100lbs in this example); FN is the normal force made by the springs and CF is the coefficent of friction. I don't know of a way to get CF except to solve for it with known forces (which you don't know)

It might be simpler to look up who manufactured the springs to get the spring constant. Once you know that, just use a ruler to measure the spring deflection; multiply that number by your spring constant and you should have the force N.

Hey Harry thanks a lot for the reply and sorry for being so late.

I figured out what to do and I think it was simple. I will tell you what I did and you comment on it.

Also, These things are to be considered :

- I don know the spring constant because I need to design that as well. I just know the force being applied on A.

-I know the co efficient of friction of the materials in contact ( since I am the one to decide the material , I chose Cast Iron for now and the static COF for it is 1.1 )

So I took the free body diagram of the middle block ( Say A ) . The forces acting on tje block are N1 and N2 ( normal forces) due to top and bottom block and the two frictional forces F1 and F2 in direction opposite to the applied load F. Equating the forces, I got following values :

Applied load F = 2.697 e 5 lbs ( the load is large )

Static COF = 1.1

N1 = N2 = 121772.72 lbs

F1 = F2 = 133950 lbs

here , F = F1 + F2

which means that the block is still at rest.

Now , If I want the blocks to move I need to decrease the values of N1 and N2 so that there is impending motion.

Which means F1 + F2 < F

which finally gives us, N1 < 121772.72 lbs

So, any normal force less that the above mentioned value should give me an impending motion and one the motion starts the frictional forces F1 and F2 can be calculated with that particular N1 and N2 and the kinematic COF.