sysprog said:
What about the leftward tension of the top ##m## block? Doesn't that act on the hanging ##m## block via the cord?
No. That is not how tension and pulleys work.
What matters for the effect of tension on the side block is the direction of the cord where it contacts the side block. One can look at the tension in the last millimeter of [ideal] cord and the direction of the last millimeter of [ideal] cord and determine the resulting force applied at its point of attachment.
The cord between top block and pulley is horizontal. It exerts a horizontal force at its attachment point on the top block.
The cord between pulley and bottom block is vertical. It exerts a vertical force at its attachment point on the side block.
The interaction with the pulley is more interesting. The cord exerts a uniform radial force along a 90 degree arc of the ideal pulley's surface. The net of this is a force whose magnitude is ##\sqrt{2}## times tension at an angle 45 degrees below the horizontal.
Of course, Newton's third law still applies. The cord is subject to a force from its two attachment points and from the pulley. Since it is an ideal cord and since this means that it is massless, the forces on the cord must sum to zero. And indeed they do. We have a leftward force of magnitude T plus a downward force of magnitude T plus an up-and-to-the-rightward force of magnitude ##\sqrt{2}T##. They sum to zero. Life is still good and the laws of physics still hold.
The side block is not subject to any magical horizontal force transmitted through a vertical section of cord. That would be a shear force. Ideal strings, threads, ropes and cords do not do shear.
Edit to add...
I've never done the coursework on continuum mechanics where this stuff might be discussed, but
@Dale and
@Doc Al have let enough leak through over the years that I have a basic grasp.
In first year physics one treats tension as a force that has particular behaviors. It is always parallel to the cord. It is uniform over the length of an ideal cord. It is transmitted unchanged in magnitude over ideal pulleys.
It turns out that there is a deeper and mathematically correct way to think about tension instead. It is not a force at all. It is a particular condition of
stress in the cord. It is given by the
stress tensor.
Expressed in component form, a tensor is a matrix. For stress within a three-dimensional body it is a 3 by 3 matrix. Its component values are force components per unit area at a point within the volume of the body. There are three directions for the force components. And there are three orthogonal planes on which a directed area can be measured. So there are nine components total.
For a cord under uniform pure tension along its length and a coordinate system aligned nicely with the cord, a pure tension of ##T## results in a stress tensor of
$$\left[ {\begin{array}{ccc}
-T & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array} } \right]$$For an ideal cord, this tensor value will be present at every point within the volume of the cord. [As it passes over an ideal pulley, the stress will change. For a pulley aligned with the coordinate system and a 90 degree turn, a different diagonal element will then have the ##-T## term].
If you perform a tensor multiplication of the stress tensor times a directed area, you get a vector for the force transmitted by the material across that area. For a tensor that corresponds to tension in a cord, it will turn out that no matter how you slice across the cord,
the product will be a vector force parallel to the cord at the point of the cut.
That is the mathematics that goes behind the statement that the direction of the last millimeter of cord is what matters.
More generally, one can use the tensor formalism to do beautiful things like computing the power transmitted through a shaft or belt as a surface integral of local stress times incremental area times local velocity over a chosen slice.
