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Two different known masses at rest on a frictionless surface are connected by a rope. A known force F is applied horizontally to mass 2 so that the boxes begin to accelerate in that direction. The acceleration of both masses together is given by a=F/(m1+m2) (I think they can be considered the same object because of the rope and the lack of friction). The only horizontal force acting on mass 1 is the pull coming from the tension in the rope. Therefore the tension T is given by T=m1*F/(m1+m2) (again, the acceleration is the same because of the rope and the lack of friction, I think).

Intuitively, this makes absolutely no sense to me, because the tension in the rope ends up being less than the force acting on mass 2. What's going on? Are you supposed to not count both masses in the calculation of acceleration? If so, why?

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It is best to draw free body diagrams for the two masses.

enter image description here

$F$ is the applied force and $T$ the tension in the massless and inextensible rope joining the two masses.
There is no friction and both masses have the same acceleration $a$.

Applying Newton's second law for each of the masses:

$T = m_1\;a$ and $F-T= m_2\; a \Rightarrow F = (m_1+m_2)\;a$ so $F>T$

You can think of it as the force $F$ is accelerating both masses whereas the force $T$ only has to accelerate mass $m_2$.

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    $\begingroup$ Note to novices who read this: the procedure outlined in this answer is much better than the procedure that considers the two masses as part of a single system. That commonly applied short cut works in very simple cases, but as complexity is added it leads to confusion, ambiguity, and wrong answers. Don't do that. Do this, even when the system is simple; it's a good habit. In this method we have Newton's second law for each object, and a set of constraint equations, here $a_1=a_2$. Clean bookeeping. $\endgroup$
    – garyp
    Commented Oct 10, 2016 at 16:24
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Since the system is frictionless, the force applied to m2 must cause an acceleration in the direction of the force. Both blocks accelerate together, same value. The reason tension in the rope exists is due to the inertia of m1 resisting the pull from m1. The greater the force, the greater the acceleration, the greater the tension.

But the tension can not be equal or greater than the applied force. If you remove the force, the two blocks will now continue to move at constant velocity and the tension will disappear.

Here's another way to see this. Take your first equation and rearrange for force:

F = (m1 + m2) * a

Now, remembering the acceleration of m1 is also a, solve for the tension using only m2 as your free body:

T = m1 * a

Easy to see how F must be greater than T.

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In your example the only quantity that should be equivalent between both masses is the acceleration. This is intuitive because the blocks are not moving closer or farther apart as they are pulled. That means by the Second Law, the net force on each block varies by its own mass. This is the steady state solution for your scenario.

Your intuition is not wrong per se, however. You are imagining initial conditions when the external force is first applied, in which case tension will equal the external force but starts dropping as soon as the mass being dragged starts accelerating.

Imagine an ant tied by rope to an iron block on which a force is applied. There would be very high force on the block but almost no tension in the rope, yet both block and ant accelerate equally.

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  • $\begingroup$ The tension on each block is also the same. $\endgroup$
    – bpedit
    Commented Oct 10, 2016 at 3:26

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