Solving Spring Problem: Find F? Angle for 1 mm Compression

[Pellefant]

Homework Statement

I am seeking f?

Homework Equations

See picture

The Attempt at a Solution

See picture

http://img14.imageshack.us/img14/1778/industri1.jpg

......

Also some private thinking if the angel would be 45 degree then i would use 50% of F?, because if F? puch the rod 1 mm down; the spring would only be compressed 0,5 mm; am i rigth?

 

[Pellefant]

Forgot it shall be multiplyed with 2 ...

So is it ok?

 

[Redbelly98]

Not quite.

Your calculation of F_n is really for F_friction, since you are multiplying by the friction coefficient 0.15.

Also, you need to think about:

In what direction is the friction force acting on the vertical part?

In what direction does the normal force act on the vertical part? It looks like you did not account for this in your calculation.

 

[ashishsinghal]

Can you improve the image so that the situation is clearer. What are the cylindrical things.
What are pairs of squares on them.
Or at least describe the situation. The image is really repelling...

 

[Pellefant]

Not quite.

Your calculation of F_n is really for F_friction, since you are multiplying by the friction coefficient 0.15.

Also, you need to think about:

In what direction is the friction force acting on the vertical part?

In what direction does the normal force act on the vertical part? It looks like you did not account for this in your calculation.

About the friction force; it is equilibrium on the x direction; because the two friction forces in the right and left case are countering each other, in the x-direction.

I am just interested in the force in the y-direction; so the only force who is interesting is the friction force in the y-direction.

The force from the spring is in the x-direction so it does only indirectly act on the y-direction; by the friction force that is.

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I forgot; there r gearing too; with 30 degree we get the following relationship:

y-> 0,5
x-> 0,866

So i shall multiply F? with 0,866/05 = 1,73

Are there anything i have missed, remember i am only looking for F?

Can you improve the image so that the situation is clearer. What are the cylindrical things.
What are pairs of squares on them.
Or at least describe the situation. The image is really repelling...

The squares are bearings

 

[SammyS]

As usual, a Free Body Diagram will be a great help here. I encourage you to draw at least one for the vertical plunger - or whatever it is.

There is more than frictional force acting in the vertical direction! There is a component of the normal force that is vertical on each of the 2 contact surfaces. These will tend to push the plunger up.

The frictional force also has a vertical component - possibly positive or negative.

BTW: You really haven't given us your problem. Are you solving for the force that must be exerted on the plunger to keep it from moving upward, or are you solving for the range of the force that must be applied to the plunger to keep the system in equilibrium?

The Normal force on the plunger due to the spring on the right is:

$$\left|\vec{F}_{N}\right|=4,5\cdot\cos(30^\circ)$$ in a direction 30° above the horizontal.

So F_N has a component in the y-direction (vertical direction) of:

$$\left(F_{N}\right)_y=4,5\cdot\cos(30^\circ)\sin(30^\circ)$$

The frictional force associated with the spring on the left is in a direction which is either +120° or ‒60° from the positive x-axis. So the vertical component of this frictional force is:

$$\pm\mu\left|\vec{F}_{N}\right|\sin(60^\circ)$$

 

[Pellefant]

I will look deeper into your reply in this week hopefully tomorrow. You say that there r more who r acting on it then the friction force; that is what has confused me.

Because the force of the springs acts in the x direction, so who can it create a force in the y direction; let me illustrate:

The equations you showed me

Fn= 4,5*cos(30)

And you r rigth

Fny= cos(30)*sin(30)*4,5

But the force who go along the surface will neutralise that force, which is:

F1= 4,5*sin(30)

Se picture:
http://img291.imageshack.us/img291/8417/testgi.jpg

F1y=4,5*sin(30)*cos(30)

So Fny+F1y=0

So we still won't get a force in the y-direction, except for the friction force; or am I wrong?


What i am after in this stage, is to verify that the only force who acts in the y-direction is the friction force; under the conditions I have given.

Thank you for your feedback, i need it :)

 

[SammyS]

Hello Pellefant.

Of course, the normal force the "plunger" exerts on the spring is equal & opposite the normal force the spring exerts on the plunger - that's just Newton's 3rd Law- a force and it's reaction force. Those forces do NOT cancel each other out because each is exerted on entirely different objects. ALL forces appear in pairs this way, even the frictional forces in this problem.

Don't be fooled by my user-name. I've been solving problems like this for over 40 years.

If you draw a Free Body Diagram for each spring and one for the plunger, you should see the angled surface between each spring and the plunger will allow each spring to exert a vertical force on the plunger. The countering force on each spring comes from the horizontal surface supporting each spring.

 

[SammyS]

Free Body Diagram for the "plunger".

I'll do the left or right spring when I get a chance.

 

[Pellefant]

Hi i think i get it now; The force F1 i wrote about do no not act on the plunger, that so explains it; i can also see that on you free body diagram and understand it by intution. It was nice to get that problem out of the way :).

That changes the problem definition-> i will no see it was a dynamic problem, for just the moment in the picture, which represent the worst case senario: So where i correct when i wrote:I forgot; there r gearing too; with 30 degree we get the following relationship:

y-> 0,5
x-> 0,866

So i shall multiply F? with 0,866/05 = 1,73

 

[Pellefant]

Am i rigth or not? i should add this is no homework it is for a real welding fixture; i can explain by pm if someone r interested. The problem definition is relevant as defined.

My question is as stated above:

...
There r gearing too; with 30 degree we get the following relationship:

y-> 0,5
x-> 0,866

So i shall multiply F? with 0,866/05 = 1,73
...

I just wonder if the princip is correct?

 

[SammyS]

For the statics problem as you presented it, "gearing" is not an issue. It would be if you solved it using "virtual-work".

If this is a "real-world" problem, it's going to be difficult to execute this as a "real-world" device. The bearings will be hard to design. Friction will show up in many more places than what you have allowed for.

Concerning your post #10:

That changes the problem definition-> i will no see it was a dynamic problem, for just the moment in the picture, which represent the worst case senario: So where i correct when i wrote:

So, what did you write? I see no solution given for the initial problem you posted. What is the range of values you get for F ?

 

[Pellefant]

I can represent the problem when i come back; you can get springs where the force only vary from 1 till 2%; and where it is sliding i will use a surface finish on 0,3; but i have not found a table who tells the relationship between surface finish and the coefficient for friction, but my guess is that it will be 0,15. Tool steel will be used ..