# Newton's laws - why does a plank resting on a friction-less pivot remain in equilibrium when a mass is moved outwards from the centre?

I've been doing some past-papers in order to revise for an upcoming test, and I can't seem to get my head around this question:

A student stands in the middle of a balanced plank which sits on rollers on top of a column. There is zero friction between the plank and the top of the column due to the excellent quality of the rollers. If the student walks (glides) to the right, smoothly and without bouncing state and explain, using Newton’s Laws, what happens to the plank, and the balancing of the system on top of the column.

In my mind, as the person moves to the right, their weight acts on the plank and causes there to be an anticlockwise moment about the pivot. This would then lead to the plank sliding off the rollers as it rotates. However, according to the mark-scheme this isn't the case, with their reasoning being:

The plank will remain balanced. As the student walks, by Newton’s 1st Law (or can argue from 2nd) there is no resultant external force acting and so the centre of mass remains in the same place (above the top of the column)

I don't understand how this is the case, as surely the weight of the student is an external force? Can anybody help me out with what I'm missing here?

• Note the plural: rollers. Not a single roller/pivot... – DJohnM Oct 22 '17 at 15:24
• hint: When you walk you move forward by pushing backwards on the ground – Martin Beckett Oct 22 '17 at 16:04

Consider the situation:

There are at least two parallel rollers on top of a pillar.

A plank is balanced on these rollers; that is, the centre of gravity of the plank is located somewhere between the two outermost rollers.

A student stands on the centre of the plank, so the students centre of mass is located between the those two rollers, as well.

Considering only horizontal motion, the student starts walking to the right, he/she does this by pushing the plank to the left. By Newton's Third, the plank pushes on the student to the right with exactly the same force (but in opposite direction). (as commented by @Martin Beckett)

So the student accelerates to the right, while the plank accelerates to the left.

Since the forces are equal, the accelerations are inversely proportional to the masses: of the plank and of the student.

Since the forces exist and change simultaneously, the velocities and displacements of the student and plank are also inversely proportional to the respective masses.

So, the centre of mass of the student/plank system remains stationary, in balance between the two rollers.

Of course, this could all be summed up by saying; Walking forces are internal to the plank/student system; no movement of system centre of gravity. Hence, no tipping...

The center of mass of the student plus plank will move, because a force is applied to the rollers to make them rotate so there is a reaction force on the student+plank system which will cause its center of mass to move slowly in the direction the student begins walking. (Student moves left, plank moves right applying rightwards force on the rollers, which yields a leftward reaction force.) The solution in the mark-scheme implicitly assumes massless rollers. But if the point is to recognize that the center of mass remains fixed, a better question would be to assume the plank slides frictionlessly on the horizontal top surface of the column.

There are no external horizontal forces so the centre of mass of the plank and man remains in the same horizontal position. If the man moves to the right, the plank moves to the left in such a way that the centre of mass remains in the same place :
$m_1x_1+m_2x_2=0$

If the man moves right this causes a clockwise moment. But the plank also moves left causing an anticlockwise moment. The same equation applies, because we usually calculate the centre of mass by balancing moments.

As long as the centre of mass remains between two rollers, the system is balanced.