High School Feynman problem about a cylinder in a corner

[mainguy]

As always a fairly devious problem from Feynman, it's getting the better of me and I imagine some of you may be able to solve!

Excuse the poor drawing. He writes 'Consider cylinders radius πcm, he cylinders are chopped into thirds and two thirds are connected as shown in fig. The thirds have uniform mass densities, but different total masses - one has 2M the other M. Find the force exerted by the wall on the cylinder'
*Note, the radial line from the center of cylinder across mass M sector is parallel to the x-axis.

So how I solved this was by finding the center of mass, which I found to be 1.256cm from the center and at 23.5 degrees to the horizontal. I then proceeded to calculate the torque of the this center of mass about the pivot point which is directly below the center.
And then by equating this torque to that exerted by the wall (which I found to be Ffriction*pi) I found F.

I got it wrong by a long way, the actual answer is 8.1N and my force was 84N...

 

[jbriggs444]

It seems immediately obvious that it is statically indeterminate. The cylinder can be under a combination of friction and normal forces that would allow it to wedge itself into the corner as tightly as it pleases. [Consider the equivalent problem of a rock climber ascending a chimney with not-quite-parallel sides -- the climber can choose to ascend or descend as he pleases].

 

[mainguy]

So you mean to say the question is unanswerable? I'm not sure what you mean

 

[jbriggs444]

So you mean to say the question is unanswerable? I'm not sure what you mean

Yes. Many solutions are possible. Without further information, it will be impossible to pick a single right answer.

Suppose that a solution exists. To this solution, add a 1 Newton pressure from the cylinder at a 45 degree angle upwards against the wall and downwards against the floor. Increase the normal force from wall on cylinder by $\frac{\sqrt{2}}{2}$ N and the frictional force from the wall by a like amount. Similarly for the normal force and friction at the floor.

[If the coefficient of friction is too small to support a 1 N force without slipping, make the added force small enough so that slipping does not occur. There will always be a force that is small enough, barring the corner case where the cylinder is perfectly balanced already and does not need the wall at all] Edit: I see that the problem statement makes it clear that the cylinder is unbalanced.

By inspection, the cylinder is still under no net force. It does not accelerate away.
By inspection, the cylinder is still under no net torque. It does not twist in place.

If we are to settle on a single solution, we need some additional constraint. Perhaps the cylinder is set gently in place and allowed to settle into the wall. Perhaps the wall is frictionless.

 

[A.T.]

Perhaps the wall is frictionless.

Or the floor?

 

[jbriggs444]

Or the floor?

If the floor has zero friction then equilibrium requires that the wall have zero normal force. The cylinder will sag, push itself away from the wall and slide indefinitely to the right, rocking all the way.

 

[jbriggs444]

Going back to the original post, I see...

And then by equating this torque to that exerted by the wall (which I found to be Ffriction*pi) I found F.

The wall exerts two component forces on the cylinder. One is friction. Given your chosen reference axis, there is another non-zero torque from wall on cylinder.

 

[A.T.]

If the floor has zero friction then equilibrium requires that the wall have zero normal force.

Sticky wall? ;-)