# Why doesn't a resting bar always rotate when pushed upon?

I'm trying to clear up some of my understanding about torque. Imagine that we have a uniform, heavy, thin metal bar resting on a table with less than half of it hanging over the edge of the table. My understanding is that because the center of mass is not past the edge of the table, the torque caused by gravity is cancelled out by the normal force of the table.

Say I push down lightly on the part of the bar that is hanging over the edge - because the bar has some heft, it will not rotate. However, if I push down hard enough though, it will rotate around the edge of the table.

1. Why does the bar not move when I push down lightly? There's now additional torque happening in addition to what gravity and the normal force are doing so the net torque should be non-zero. What's cancelling out my torque?

2. Why does the bar eventually rotate when I push down hard? Why is whatever force in question 1 not sufficient to keep cancelling it out? What's the break even point?

• When you push down on the part of the bar hanging over the edge of the table you are effectively moving the center of gravity (COG) of the bar towards the edge of the table. Only if you push down hard enough will you effectively move the COG past the edge. Aug 1, 2023 at 15:04
• The net force holding the bar onto the table is likely non-zero, so a bit of torque only reduces it, not negates it. Your current mental model would have any force on a bar with even just a sliver of it off the end rotate easily. You know that does not happen. Aug 1, 2023 at 15:22

The normal force has two important properties:

1. It can take on any magnitude it needs to to ensure that an object does not accelerate through a surface. Note that the normal force is not necessarily equal to the object's weight.1
2. It is always directed away from the surface, never towards.

When you push down lightly on the cantilevered end of the bar, the normal force from the table does not stay constant; it decreases to compensate (see property #1). This ensures that the net torque on the bar is still zero, and the bar doesn't rotate. So long as the torque you apply is less than the torque from gravity, the normal force can do this, since doing so requires the normal force to push upwards on the bar.

But if you push too hard, and the magnitude of the torque from your finger becomes greater than the torque from gravity, then the normal force would have to pull the bar downwards to keep the bar in place. Because of property #2, above, it can't do this. So the bar will rotate.

As an analogy, consider a block with a string attached to it, sitting on a table. If I pull upwards on the string lightly, the normal force on the table decreases, but the block still doesn't move. If I pull sufficiently forcefully, then the force I'm applying will exceed the weight of the block, the normal force is unable to "hold it down", and so the block lifts off the table.

1 A classic misconception of introductory physics students everywhere. If I had a dollar for every time I saw a student make this mistake... well, I wouldn't be rich, exactly, but I could take my wife out to a really nice dinner.

Suppose the centre of mass of the rod is close to the edge of a table.
The centre of mass of bar is above ground level and when end of the bar is pushed down the bar pivots on the edge and a vertical line through the centre of mass moves closer to the edge for a small push and over the edge if the push is large enough.
Thus for a small push the torque about the edge is in a direction to restore the initial static equilibrium position but for a large push the direction of the torque will make the rod rotate further.

The break-even point is when the vertical line passing through the centre of mass touches the edge.