- #1
fisico30
- 374
- 0
Hello Forum,
If a bike is negotiation a curve, the static friction force will supply the centripetal force F_c.
If we go too fast and the friction is not enough we skid along the tangent.
There are two forces: the weight W=mg pointing down, the normal (contact) force N pointing up and equal to the weight W, and the static friction 0<=f_s<= mu_s*(N)= m_s*(mg).
If we go fast, we automatically lean into the curve. Why?
Some free body diagrams show the normal force being inclined w.r.t the pavement and pointing along the rider axis. How can that be? The normal force is called normal because it is perpendicular to the floor.
Does the leaning increase the normal force somehow? If we increase N we automatically increase f_s and supply a larger centripetal force. Maybe there is "camber thrust": the normal force gets inclined because there is a deformation of the tires...
Or does the leaning simply help creating a torque about point of contact due to the weight W that counterbalance the "centrifugal" force torque (centrifugal force does not exist really, only in noninertial frames)... Otherwise, how come the bike does not fall if the CM is outside the base of support (wheels)?
Thanks,
fisico30
If a bike is negotiation a curve, the static friction force will supply the centripetal force F_c.
If we go too fast and the friction is not enough we skid along the tangent.
There are two forces: the weight W=mg pointing down, the normal (contact) force N pointing up and equal to the weight W, and the static friction 0<=f_s<= mu_s*(N)= m_s*(mg).
If we go fast, we automatically lean into the curve. Why?
Some free body diagrams show the normal force being inclined w.r.t the pavement and pointing along the rider axis. How can that be? The normal force is called normal because it is perpendicular to the floor.
Does the leaning increase the normal force somehow? If we increase N we automatically increase f_s and supply a larger centripetal force. Maybe there is "camber thrust": the normal force gets inclined because there is a deformation of the tires...
Or does the leaning simply help creating a torque about point of contact due to the weight W that counterbalance the "centrifugal" force torque (centrifugal force does not exist really, only in noninertial frames)... Otherwise, how come the bike does not fall if the CM is outside the base of support (wheels)?
Thanks,
fisico30