Free Body Diagrams

Back in 1988, I once learned a dab of Korean, which was pretty easy to pick up. But in the meantime I have forgotten enough that I can't answer definitively.

I don't think either diagram is wrong, but A addresses the dynamic case of movement, whereas B is a static one-shot. Thus A seems more inclusive. (For what little this may be worth.)

If JDTardis sees this question, he may well be able to answer it better.

Neither A nor B is a Free Body Diagram. They are both force vector diagrams. If the system is in equilibrium, the point mass needs a horizontal force to the left preventing it from falling to the ground.

The axial force in the rod and spring is F_L which is numerically equal to W/cosθ where W is the weight of the point mass and θ is the angle between the rod and a vertical line.

In moving through angle θ, the point of contact has moved from the bottom of the rod to the black dot shown on the diagram. The line of action of the force must be from the black dot to the center of mass, so the line of action is at an angle φ from the vertical. The force Fc is equal to W/cosφ. Force Fθ is needed to complete the force polygon.

Because the resultant force is not colinear with the rod and spring, there is a moment in the rod varying from Fθ*d at the bottom of the rod to zero at the point mass. Dimension d is the length of rod plus spring.

Thanks,

Why the line of action of the force must be from the black dot to the center of

mass ?

Notice the forces shown in red in the attached Free Body Diagram. Gravity force, W and horizontal force H act at the center of mass.

At the point of contact between the arc and the floor, i.e. the 'black dot', we also have a vertical and horizontal force.

For equilibrium in the system, the sum of horizontal forces, vertical forces and moments must all be zero.

To satisfy the first two conditions, H and W at the bottom must be equal to H and W at the top.

To satisfy the third condition, the moment of all forces must be zero at all points in the system. In order to have a zero moment at the center of mass, W*x - H*y = 0 or, alternatively, H = W*x/y.

The resultant force top and bottom is easily seen to be co-linear with a line drawn through the two points.

I forgot to define x and y in the above. They are the horizontal and vertical distance respectively between the two points.

The two forces W, separated by distance x, form a clockwise couple W*x. The two forces H, separated by distance y, form a counter clockwise couple H*y. For equilibrium, the two couples must be equal in magnitude and opposite in direction.

Thank you very much ^^

Why do I need to think that "moments must all be zero" ?

Is it related that black dot (contact point) is now moving ?

If the system is in equilibrium, the sum of horizontal forces, sum of vertical forces and sum of moments must be zero. Those are three necessary conditions for equilibrium.

If the sum of all moments acting on a body is not zero, the body will accelerate rotationally which means it is not in equilibrium.

If in the Free Body Diagram, the only forces acting on the system were the W forces, we would have satisfied the first two conditions of equilibrium but not the third because, in effect we would be applying a torque of W*x to the system. This would cause it to accelerate rotationally.

The black dot in the diagram is significant only because it is the point of application of force from the floor to the system. If the angle theta were changed, the black dot would move and the angle phi would change.

If you remove the curved rocker from the bottom of the rod and allow the system to be supported at a point, the black dot stays in the same place during system rotation. The angle phi would be the same as the angle theta. In that case, the rod and spring would carry axial load and there would be no bending moment in the rod.

Thanks !

What I mean is... theta, phi are changing. in this case (dynamic motion), why do I need to assume that system is in equilibrium?, including moment equilibrium condition..

I think I am missing an important point.. what is it ?!?!

thank you very much for your explanation.

I was not aware that the system is in dynamic motion. Did you explain that in your question? If so, I missed it. The laws of statics do not apply to a system in motion.

And if that is the case, what prevents the mass from falling to the floor, perhaps bouncing once or twice, than resting flat on the floor?

Are you concerned about the instantaneous forces when the system has fallen through a particular angle? If so, I would have to think about that some more.

Actually that is a part of simple walking model.

Center of Mass is now on swing motion (falling) with curved foot.

- Just mass-spring inverted pendulum with an arch is now falling to the ground. befere collision, the other inverted pendulum will show up for an periodic walking motion. -

Pendulum (with initial angular velocity) is now on Rolling without slip.

Statics is a subject I know well. Kinematics is not familiar to me, but let me have a try anyway.

The only forces acting on the system are the weight of the mass, W and the reaction at the point of contact between curved arch and floor. That point may be considered an instantaneous hinge.

Force W is resolved into two components, Wcosφ directed toward the contact point at the floor and Wsinφ directed normal to that. The normal component is not in equilibrium and causes an acceleration of the mass in the same direction.

At the contact point on the floor, thrust will be Wcosφ (Fc on the diagram) directed toward the center of mass. The thrust may be resolved into components Wcos²φ vertically and W*cosφ*sinφ horizontally.

Axial force (F_L), shear (F_θ) and bending moment in the rod and spring may be determined from the magnitude and direction of the thrust.

sorry for being late, I really appreciate that you have given some good comments !

I've been very busy on some stuffs. sorry for late response.

AND I am still thinking about it..

[-fact-] W is an weight of Center of Mass, ground reaction force Fz sometimes shows lager value than W.

Do you mean that the direction of resultant force of ground reaction force must be lying on the line of action (on figure) ?

Yes, the ground reaction force must be lying on the line of action, i.e. a line between the point of contact on the floor and the center of mass. That is because there is effectively a hinge at each end. A member hinged at each end is incapable of transferring anything other than an axial load.

As for the suggestion that the ground reaction force F_Z (did you mean F_C?) can be larger than W, I don't see how that is possible. When the leg is vertical, the ground reaction force is precisely equal to W. As the leg moves beyond vertical, the ground force reduces by cosφ which is less than 1.0.

But there is more to the problem than we see by this model alone. When a person walks, he transfers his weight from one leg to the other. When one foot leaves the floor, the entire weight is carried by the other leg. While both legs make contact with the floor, they share the weight such that W_L + W_R = W but the portion carried by each leg is constantly changing as theta changes.