Calculate magnitude of spring force

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Homework Help Overview

The problem involves two masses on a frictionless surface connected by a spring, with a force applied to one of the masses. The objective is to determine the magnitude of the spring force in relation to the applied force and the masses involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss drawing free body diagrams for both masses to analyze the forces acting on them. There are attempts to establish motion equations based on Newton's second law, and questions arise regarding the correct application of forces and accelerations.

Discussion Status

Participants are actively engaging with the problem, offering guidance on drawing free body diagrams and formulating equations of motion. There is a recognition of the need to clarify the forces acting on each mass and how they relate to the acceleration of the system. Multiple interpretations of the spring force and its relationship to the applied force are being explored.

Contextual Notes

Some participants express confusion over the relationships between forces and accelerations, and there are discussions about the assumptions made regarding the system's behavior under the applied force. The complexity of the problem is acknowledged, with participants seeking to clarify their understanding of the dynamics involved.

PhySci83
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Homework Statement


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.

Homework Equations


F = ma
Fs = -k(xf-xi)


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.
 
Physics news on Phys.org
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??
 
'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
 
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!
 
PhySci83 said:
Fnet,m1,x= a
Fnet,m1,y= 0


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

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!
 
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
 
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
 
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?
 
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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