# Calculate magnitude of spring force

1. Jun 29, 2011

### PhySci83

1. The problem statement, all variables and given/known data
A 2.0-kg mass (m1) and a 3.0-kg mass (m2) are on a horizontal, frictionless surface, connected by a massless spring with spring constant k = 140 N/m. A 15-N force (F) is app0lied to the larger mass, m2, as shown in the figure:

|__m1__|-/-/-/-/-|____m2____| ----------> F

Show that the magnitude of the spring force (spring tension) is given by:

Fs= F(1/(1+(m2/m1))) <---Sorry, do not know how to make that look more asthetically pleasing.

2. Relevant equations
F = ma
Fs = -k(xf-xi)

3. The attempt at a solution

Since I am looking for the Fs, I attempted to find the net force of the system with the direction of the force being the positive x direction and the N force of the masses being the positive y.

So for the F going to the right, I said F = (m1+m2)*a; a = F/(m1+m2).

For the F of the spring, I said Fs = -k(xf-xi) and rearranged the equation so that delta_x = magnitude_F/K; delta_x=(ma)/K; then I plug in the value for a that I solved prior.

The problem is, I keep going around in circles with this thing. I am pretty far removed from basic HS math, but if the F wasn't pulled out of the equation, would it be:

Fs = F/(F + [(F*m2/F*m1)])?

Any help that would point me in the right direction would be greatfully appreciated. It seems like I have to solve for the various variables and plug them into the original equation:

Fs = -k(xf-xi)

but getting rid of K and delta_x are troubling me.

2. Jun 29, 2011

### Mandeep Deka

1. Draw the free body diagrams of both the blocks
2. Write down the motion equations of both the bodies.
What do u find then??

3. Jun 29, 2011

### ehild

'Fs = F/(F + [(F*m2/F*m1)])'

Are not there too many F-s in your formula? F*m2/Fm1=m2*m1. If you meant F*m2/(F*m1), it is m2/m1=1.5, you can not add newtons to it!

Draw the free body diagram, and see what force acts on m1 if it moves with the acceleration a=F/(m1+m2)

ehild

4. Jun 29, 2011

### PhySci83

OK, so I think I have the FBDs, but this was the part I was struggling the most with.

So for m1 I have:

|
| FN
[__m1__] -------> a
|
| Fg

Fnet,m1,x= a
Fnet,m1,y= 0

Am I missing the force of the spring on this? Isn't there an added force in the positive x direction?

For m2:

| FN
|
<--Fs[__m2__]--------> a
|
| Fg

Fnet,m2,x= a - Fs
Fnet,m2,y= 0

Is this all for the FBDs or am I missing something? Sorry for the drawings as I am at work and do not have access to a scanner.

THANKS FOR ALL THE HELP!

5. Jun 29, 2011

### Mandeep Deka

My friend, how can you equate force with acceleration!!!
These are two different physical quantities!!

Don't you think you are missing a quantity called mass, coz what newton's 2nd law says is:

Force = Mass x Acceleration!!

6. Jun 29, 2011

### PhySci83

Whoops! What I meant was:

Fnet,m1,x= (m1 + m2)*a
Fnet,m1,y= 0

Fnet,m2,x= (m1+m2)*a - Fs
Fnet,m2,y= 0

Then F_net = Fnet,m2,x + Fnet,m1,x

7. Jun 29, 2011

### ehild

Do not forget that ∑F =ma. The resultant of all forces acting on m1 is equal m1*a. What horizontal force(s) act on m1? What is its acceleration?

ehild

8. Jun 29, 2011

### PhySci83

Are the forces that act on m1 the force of system accelerating from the force acted upon it plus the force of the spring? So a FBD would look like:

|
| FN --->Fs
[__m1__] -------> a
|
| Fg

F,x,m1= (m1*a) + Fs
F,x,m2=(m1+m2)*a -Fs

Then doesn't Fs cancel out?

acceleration of m1 is equal to the acceleration of the system which is a = F/(m1+m2)?

Sorry, I have tried this every way and still cannot come up with something that even looks correct. Are my above assumptions correct or am I still missing something?

9. Jun 29, 2011

### ehild

I show you a proper free body diagram. There is only one horizontal force exerted on m1: the spring force, Fs. And there are two horizontal forces exerted on m2: F=15 N and -Fs. The magnitude of the spring force is the same for both bodies, only the direction is opposite. The bodies move with the same acceleration a.

Newton's equations ∑Fi =ma for both objects:

m1*a = F - Fs
m2*a = Fs.

A force exerted on one body does not act on the other one.

If you add up the equations you get

(m1+m2)*a=F,

a=F/(m1+m2). You got this a already.

Use this a in the equation for m2 to get Fs.

ehild

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