Problem: two bodies linked via a spring

In summary, the problem involves two bodies, A and B, connected by a spring with a natural length l_0 and elastic constant k. At a specific time t_i, an external force F is applied to body A in the direction of B. The problem requires calculating the initial acceleration of the center of mass of the system, as well as the initial acceleration of each of the two bodies. The respective accelerations are also required when the spring is compressed by a length x, as well as identifying the pairs of action-reaction forces at that moment. The center of mass is calculated to be at \frac{l_0}{m_1+m_2} from body A, and the acceleration of the center of mass is given by
  • #1
fluidistic
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Homework Statement


Two bodies [tex]A[/tex] and [tex]B[/tex] of mass [tex]m_1[/tex] and [tex]m_2[/tex] are connected via a spring of natural longitud [tex]l_0[/tex] and elastic constant [tex]k[/tex]. Both bodies are free of net force until at an instant [tex]t_i[/tex] something applies a constant force [tex]F[/tex] to the body [tex]A[/tex], in the direction of [tex]B[/tex]. (see the diagram)
a)Calculate the initial acceleration of the center of mass of the system
b)Calculate the initial acceleration of each of the 2 bodies
c)Calculate the respective accelerations in the instant in which the spring is compressed by a length x.
d)Indicate all the pars of action-reaction forces in the moment in which the spring is compressed by a length x.


Homework Equations


[tex]F_{spring}=k \varedelta x[/tex]
[tex]\sum \vec{F}=m\vec{a}[/tex].



The Attempt at a Solution


a)Just for fun I calculated the center of mass to be at [tex]\frac{l_0}{m_1+m_2}[/tex] if the origin is situated at body [tex]A[/tex] in instant [tex]t_i[/tex].
I think I've read somewhere that if an external force is applied, then it will modify the acceleration of the center of mass of the system following Newton's second law.
So [tex]\vec{a}_{CM}=\frac{F}{m_1+m_2}i[/tex]. Am I right?
b)Using Newton's second law, [tex]\vec{a}=\frac{F}{m_1+m_2}i[/tex] for the body [tex]A[/tex].
And here start my problems. I'm not sure how to find the acceleration of the body [tex]B[/tex] at [tex]t_i[/tex]. I'm tempted to go against my intuition and say that it will be the same as the center of mass of the system, but I don't think it's possible. Please help me going further this.
 

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  • #2
fluidistic said:
a)Just for fun I calculated the center of mass to be at [tex]\frac{l_0}{m_1+m_2}[/tex] if the origin is situated at body [tex]A[/tex] in instant [tex]t_i[/tex].
I think I've read somewhere that if an external force is applied, then it will modify the acceleration of the center of mass of the system following Newton's second law.
So [tex]\vec{a}_{CM}=\frac{F}{m_1+m_2}i[/tex]. Am I right?
Yes, you are right.
b)Using Newton's second law, [tex]\vec{a}=\frac{F}{m_1+m_2}i[/tex] for the body [tex]A[/tex].
Not exactly. What's the net force on body A? The mass of body A?
And here start my problems. I'm not sure how to find the acceleration of the body [tex]B[/tex] at [tex]t_i[/tex]. I'm tempted to go against my intuition and say that it will be the same as the center of mass of the system, but I don't think it's possible.
The way to find the acceleration of any system (which could be one mass or both masses) is to apply Newton's 2nd law to that system. Start by identifying the forces acting on the system of interest at the moment of interest.
 
  • #3
Oops! I did it right but get confused when typing here.
b) For the body [tex]A[/tex], [tex]\vec{a}=\frac{F}{m_1}i[/tex].
So for the body [tex]B[/tex], I have to draw a F.B.D. in order to identify the forces acting on it at [tex]t_i[/tex].
I'm not sure... but at [tex]t_i[/tex] the spring isn't compressed, right? Or it is, by a differential x? If so, then [tex]\vec{a}=\frac{k \cdot \Delta x}{m_2}[/tex] and I think I could even work out [tex]\Delta x[/tex] in function of [tex]F[/tex]. Am I in the right direction?
 
  • #4
fluidistic said:
b) For the body [tex]A[/tex], [tex]\vec{a}=\frac{F}{m_1}i[/tex].
Good.
So for the body [tex]B[/tex], I have to draw a F.B.D. in order to identify the forces acting on it at [tex]t_i[/tex].
Right.
I'm not sure... but at [tex]t_i[/tex] the spring isn't compressed, right?
Right. At that moment, the spring has not had a chance to compress.
Or it is, by a differential x? If so, then [tex]\vec{a}=\frac{k \cdot \Delta x}{m_2}[/tex] and I think I could even work out [tex]\Delta x[/tex] in function of [tex]F[/tex]. Am I in the right direction?
Let [tex]\Delta x = 0[/tex]. :wink:

(You're doing fine.)
 
  • #5
Thank you very much Doc Al. I've completed the problem now.
I understand better now that it's possible for a body to have an acceleration but with a velocity equal to 0. It happens in springs and pendulums for example.
In this problem, the fact that A has no velocity in [tex]t_i[/tex] is the responsible of the no-compression of the spring and so that B cannot suffer any force thus any acceleration.
 
  • #6
Good.

It's certainly possible for something to have an acceleration but (at least for the moment) zero velocity. Consider a ball tossed straight up. When it reaches the top, its velocity is zero, yet its acceleration is always g downward.

In this problem, the trick is realizing that the force exerted by a spring is simply given by its degree of stretch or compression. No stretch or compression, no force.
 

1. What is a "spring" in the context of two bodies linked together?

A spring is an elastic object that can store mechanical energy and exert a force when stretched or compressed. In the context of two bodies linked together, a spring acts as a connecting element between the two bodies, allowing them to move in response to applied forces.

2. How does the spring affect the motion of the two bodies?

The spring affects the motion of the two bodies by exerting a restorative force that opposes any changes in the distance between the bodies. This force is proportional to the displacement of the bodies from their equilibrium position and follows Hooke's Law.

3. What is the equilibrium position in a system of two bodies linked via a spring?

The equilibrium position is the point at which the spring is neither stretched nor compressed and the two bodies are at rest. This position represents the state of minimum potential energy in the system.

4. Can the spring in a system of two bodies be stretched or compressed indefinitely?

No, the spring in a system of two bodies has a maximum limit to its stretching or compression. This limit is known as the spring's elastic limit and if exceeded, the spring will permanently deform or break.

5. What factors affect the strength of the spring's restorative force?

The strength of the spring's restorative force is affected by the spring constant, which is a measure of the stiffness of the spring, and the displacement of the two bodies from their equilibrium position. The greater the spring constant or displacement, the stronger the restorative force will be.

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