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Calculate magnitude of spring force

  1. Jun 29, 2011 #1
    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. jcsd
  3. Jun 29, 2011 #2
    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??
     
  4. Jun 29, 2011 #3

    ehild

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    '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
     
  5. Jun 29, 2011 #4
    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!
     
  6. Jun 29, 2011 #5
    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!!
     
  7. Jun 29, 2011 #6
    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
     
  8. Jun 29, 2011 #7

    ehild

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    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
     
  9. Jun 29, 2011 #8
    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?
     
  10. Jun 29, 2011 #9

    ehild

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    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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