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Let us consider three bodies of equal mass connected to each other with 2 ideal strings of length l. The three bodies are placed in a straight line

In this arrangement there is 1 body connected to 2 strings, and two bodies at the end connected to only 1 string. (2 strings in total). The body in the middle, let it be B. Now let B be given a constant acceleration perpendicular to the strings. . Neglect friction. My questions are as follows ::

Does constant acceleration being imparted to B equal constant force?

Are the tensions, in both the strings equal?

Does the acceleration of Center of Mass of the system remain constant?

Will the directions of the velocities of the other 2 blocks be along the string?

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  • $\begingroup$ Can you clarify the direction of acceleration? Because it the body is pushed perpendicularly to the springs, then the system will not be placed in a straight line anymore. $\endgroup$
    – QuantumBrick
    Commented Oct 29, 2014 at 17:00
  • $\begingroup$ Yes it is pushed perpendicularly to the strings. The system will not remain in a straight line $\endgroup$
    – Sashurocks
    Commented Oct 29, 2014 at 17:12
  • $\begingroup$ Is this a homework problem? $\endgroup$
    – garyp
    Commented Oct 29, 2014 at 17:21
  • $\begingroup$ At time=0 there is no tension in the springs since deflection is unchanged. $\endgroup$
    – John Alexiou
    Commented Oct 29, 2014 at 18:43
  • $\begingroup$ This question would make more sense with massless rigid rods connecting the masses. $\endgroup$
    – John Alexiou
    Commented Oct 29, 2014 at 19:59

2 Answers 2

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As mass B accelerates, the other two "stay behind". The string will no longer be at 90 degrees to the motion, and tension in the strings will pull back on B. So in order to maintain constant acceleration on B, the force needs to increase. The tension in the two strings is the same, by symmetry. As the force on B (and thus, the system) changes, the center of mass will not accelerate with constant acceleration (it moves as though all external forces act on it).

As for the motion of the other two: they will start out by moving inwards, then will follow the mass B. The vector of their acceleration is always along the string, but their velocity along the string is given by the angle of the string and the velocity of B (which you know because of the constant acceleration).

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  • $\begingroup$ I have 2 questions: what causes the occurence of tension. This may be basic but with no friction what force will the other 2 bodies exert on the strings. Apart from that are you telling that there exists a relationship between velocity of B and that of the bodies along the string $\endgroup$
    – Sashurocks
    Commented Oct 29, 2014 at 17:26
  • $\begingroup$ since the bodies have mass there must be a force to accelerate them, otherwise they won't move. The only "medium" for this force is the string. $\endgroup$
    – Joseph Hankel
    Commented Oct 29, 2014 at 17:46
  • $\begingroup$ When B moves away, the other masses must follow (or the string will break). This means they too must accelerate. It is their inertia that gives rise to the tension (which will increase until it results in an acceleration $a=F/m$ sufficient to "keep up" with B). $\endgroup$
    – Floris
    Commented Oct 29, 2014 at 18:25
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Let's assume that the balls are identical, and the strings are identical, and are elastic (obey Hooke's Law) and massless.

You have stated that you've somehow arranged for B to have a constant acceleration. This means that the net force on B is constant. There are three forces on B: one due to the string connected to A, one due to the string connected to C, and one due to whatever is causing B to move in the first place: let's call it the applied force. The vector sum of these three forces is constant, but the magnitudes of the applied force and the tension forces, if considered individually, will change with time. That this is the case can be seen by considering that the strings will stretch, and the forces they apply will change according to Hooke's Law.

Because the system is perfectly symmetric, the tensions will be identical at any point in time, but the tensions will change with time.

The acceleration of the center of mass will not remain constant, as the force applied to the system (the applied force) is not constant.

The accelerations of the other two blocks will be in the direction of the string, but the velocities will not.

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