Leaning Column Free Body Diagram

[dstretan]

This is a self-study homework problem. I am trying to verify the "falling column physics" associated with leaning forward with our center of mass in front of our feet.

Homework Statement

Describe the free body diagram for a column (tree, pencil, person) leaning from vertical. Assume there are no forces at the ground inhibiting/resisting the forward fall. Also assume there is no force of air on the column as it moves forward.

Homework Equations

- Newton's first three laws.

The Attempt at a Solution

See the attached diagrams.

Vertical Column Preface:
- The force of gravity acts downward on the column. This is depicted as F(g) acting on the column's center of mass (COM).
- A normal force F(n) opposes F(g).

Leaning Column:
- The force of gravity acts downward on the column. This is depicted as F(g) acting on the column's center of mass (COM).
- This force F(g) has two components: F(c) acting down the column and F(p) acting perpendicular to the column as in a moment arm.
- Since the contact point with the ground is stationary, a normal force F(n) and a frictional force F(f) make up a force F(nf) equal and opposite to F(c).
- F(p) remains and acts on the COM which results in COM acceleration.
- F(p) has a F(p-x) and F(p-y) component, with movement x>y relative to f(h/l). Meaning the higher the center of mass the more x movement forward for y height decrease.
- The forces change as the position of the column moves forward relative to the contact point with the ground.

This might seem like a rather trivial physics problem but there are a lot of people who suggest that gravity cannot move a column/person forward. That is not my experience in terms of walking/running, and it would seem this simple diagram confirms the "feeling" of being moved forward by a series of micro-falls and micro-catches. The COM loses a small height with each fall, which it regains with each step.
- Moving downhill would mean it does not need to regain height; and probably not lean as much.
- Moving uphill would mean it would need to regain and add height; and possibly lean more.
- The person accelerates when falling - and decelerates otherwise (when in the air if we consider air resistance; and upon landing if we consider landing position and tension). The average velocity across the running gait cycle determines the runner's pace.

This answer might also explain how this "robot" is possible:
- http://www.wired.co.uk/news/archive/2011-10/26/bluebiped-robolegs

Similar examples might be a person walking on stilts or a unicyclist.

Your input would be greatly appreciated.
Thanks,
David.

 

[ehild]

The acceleration of the CM is determined by the resultant of external forces: The forward acceleration is proportional to the horizontal component of the external forces. Gravity and normal force are vertical, the only horizontal force is the force of friction. Without it, the CM would accelerate vertically downward.
What you wrote about the way of walking is true. By leaning your body, you get a horizontal force from the ground which pushes you forward.

ehild

 

[dstretan]

ehild,
Thank you for your response.

I don't follow you. The force of friction acts on the base of the column which is stationary (velocity=0) as the column's COM falls forward (velocity > 0). If the base of the column is not moving, how can friction F(f) be moving the column forward?

Thanks again for your input.
David.

 

[PhanthomJay]

ehild,
Thank you for your response.

I don't follow you. The force of friction acts on the base of the column which is stationary (velocity=0) as the column's COM falls forward (velocity > 0). If the base of the column is not moving, how can friction F(f) be moving the column forward?

Thanks again for your input.
David.

You are mixing 2 problems into one, the first being that of a leaning person or object, like a pencil, and the 2nd being that of a human or robot in the rather complex process of walking. For the latter case, it is friction from the ground that propels the person forward; for the first case, no such forward translational motion exists unless static friction is overcome, in which case translational motion will be backwards. Better check that direction of the friction force in your 2nd FBD.

 

[ehild]

ehild,
Thank you for your response.

I don't follow you. The force of friction acts on the base of the column which is stationary (velocity=0) as the column's COM falls forward (velocity > 0). If the base of the column is not moving, how can friction F(f) be moving the column forward?

Thanks again for your input.
David.

It is static friction, a force that prevents the relative motion of the surfaces touching each other. Without friction, the top end of the rod would move forward and downward, while the other end touching the floor would slide backward, and the CM would move vertically downward. Static friction opposes the backward motion along the floor, so it has to act forward at the lower end of the rod.

ehild

 

[dstretan]

Thank you for your responses.

Let's forget the case of walking/running and back up to the case of a leaning pencil. You have a brand new pencil and stand it on end; if on a level surface and vertical it will stay there. See position A in the attached diagram.
- You take your finger and move the top of the pencil to position B just right of vertical.
- You let go of the pencil and it moves through position C. The eraser at Point E does not move (at least not horizontally) consistent with *static* friction I think.

Given that point E does not move, that would imply zero net horizontal and vertical forces on point E. I believe this would be consistent with "static friction forward" opposing the rearward component of F(c).

Q. How does point E's zero net force move the column's COM?
- That would seem to break Newton's Second Law, F=ma. With no net force on the column at C, it would seem other force(s) would need to act on the pencil at other point(s) to move it from B to C.

I understand that the *static* friction is necessary at point E for the pencil to move from B to C, but I do not see how *static* friction moves the pencil.

Thanks again.

 

[PhanthomJay]

Static friction force at E does not move the pencil. The torque from the gravity force rotates it, and the tangential component of the gravity force moves the COM circumferentially in a circle. If there is no friction, or if the static friction force is overcome and kinetic friction takes over, the COM still moves and E moves leftward since the centripetal force component at E will be greater than any kinetic friction force.

 

[ehild]

- You let go of the pencil and it moves through position C. The eraser at Point E does not move (at least not horizontally) consistent with *static* friction I think.

Given that point E does not move, that would imply zero net horizontal and vertical forces on point E. I believe this would be consistent with "static friction forward" opposing the rearward component of F(c).

The force must act on an object, at a point. Force on a point has no meaning.
If you meant the net force acting on a tiny piece of the pencil situated at point E, you need to include also the internal forces. That piece with mass dm will not move till the forces from the floor and the force from the pencil (those which keep the pencil together) and force of gravity sum up to zero.

Q. How does point E's zero net force move the column's COM?

The forces which act on the tiny piece at E will move that piece or keep it stationary. These forces include the forces of interaction from the surrounding pieces. According to Newton's third law, the piece at E will exert force on the other pieces around. These pieces interact with their surroundings and this will result some net force on the pencil.

- That would seem to break Newton's Second Law, F=ma. With no net force on the column at C, it would seem other force(s) would need to act on the pencil at other point(s) to move it from B to C.

It is not true that the external forces acting on the pencil at point E sum up to zero. Zero was the resultant of all forces-external and internal- on a tiny piece at E.

I understand that the *static* friction is necessary at point E for the pencil to move from B to C, but I do not see how *static* friction moves the pencil.

The pencil is a rigid body, it keeps size and shape during motion. There must be internal forces acting on each piece to ensure this.

You certainly was shown how a system of point masses moves under the effect of external forces and internal ones. Writing up the equation of motion of each point mass, and adding up the equations, the internal forces cancelled, and you got a single equation for a point called Centre of Mass. The acceleration of the CM multiplied with the total mass is equal to the resultant of all external forces.
The rigid body can also rotate about a point. The rotation is governed by the resultant torque of the external forces.
The CM of a rigid body accelerates according to the net external force on the object. The angular acceleration of rotation about the CM or other point corresponds to the net torque of the external forces.

The net force on the pencil include gravity and normal force, both vertical, and the horizontal force of friction. When the pencil falls forward, its CM moves along a circle, it has both forward and vertical acceleration. So the resultant force has a horizontal component, pointing forward. The force of friction is the single horizontal force: It has to point forward when the pencil falls.

There is also the torque of the external forces, which rotate the pencil around point E. There are three equations for the motion of the pencil which describe its rotation about point E still the force of friction provides the necessary force to this. You can determine these forces, and you would get a forward force of friction.

I hope you made experiments with your pencil. If you repeat it with a pen without eraser at the end, you will see that the end of the pen starts to slide backward at some leaning angle, when the force of static friction is not enough to keep it stationary, that is, to keep the CM on its circular path.

ehild