# Force acting on center of mass

Pictured above is a biker (traveling in a curve) whose center of mass has two forces acting on it. These two forces - a horizontal centrifugal force F(centripetal)* and a vertical gravitational force mg - yield a resultant force which pulls the biker's center of mass diagonally towards the ground.

Here is my question: why does the biker [who I've pictured as a black stripe] stay in place if (and only if) he positions himself according to the resultant force as shown in the image?

My instructor told me that the biker would begin to tilt left or right for any other configuration, and that kind of makes intuitive sense, but I'd rather not base my knowledge on iffy intuition alone.

*I have mistakenly named the centrifugal force as centripetal in the image

EDIT: Could it be that the position of the biker as shown in the image is the only one in which a "conversion" from tilting to the left to tilting to the right occurs (so that the system is balanced, in a sense)?

• Is the cyclist in contact with the ground, which would exert another force? Mar 15, 2014 at 18:20
• "Centripetal force" is a designation of what some force is doing, not a source of a real force, like gravity or the reaction of the road. Mar 15, 2014 at 18:24

For a static configuration you need two conditions fulfilled:

• The sum of all forces must be zero and
• the sum of all torques must be zero.

First, the stable case.

In your case all the forces added together are zero: The normal force is pointing up with the same magnitude as the gravitational force; the force of friction points to the left with the same magnitude as the centrifugal force. So $\vec{F}_{resultant}-(\vec{F}_{normal}+\vec{F}_{friction})=\vec{0}$.

The net torque is given as: $$| \vec{\tau}_{res}|=|\vec r||\vec F_{res}|sin(\theta),$$ where

• $\vec{\tau}_{res}$ is the torque resulting from the resultant force,
• $\vec r$ is the vector from the pivot-point (here the point where the wheels touch the ground) to the point where the force is applied,
• $\vec F_{res}$ is resultant force and
• $\theta$ is the angle between those vectors.

When the biker positions himself according to the resultant force, $\vec r$ and $\vec F_{res}$ coincide (with different signs) so $\theta$ between those two is zero: since $\vec{\tau}=0$ in this case, the biker doesn't fall.

Now, the tilted case.

The forces are still the same, so the sum of all forces is still zero.

The net torque, however, has changed and is now different from zero: the angle between the resultant force and $\vec r$ (the angle between the blue and green line in the hideous picture I added) isn't zero anymore.

So $sin(\theta)\neq 0$ and therefore $\vec{\tau}_{res}\neq0$.

This means that the biker will tilt in one direction: to the right if the resultant force points to the right of the pivot point or else to the left.

• Thank you very much for this, I think I got the picture. Mar 15, 2014 at 22:52

There are two more forces acting on the biker, namely the friction with the ground and the normal force from the ground. I suggest you derive the resultant force from those two and look at the direction of that resultant force.