User:Physikaficionado/Tangential force Andy

Tangential and normal force along a cycloid

The tangential force acts tangentially to the trajectory of a moving body. This means that it acts along the direction in which the object is currently moving. The change in direction is insignificant for it. If no other forces are acting, this can lead to a change in speed $\Delta v$ in the direction of the force.

If several forces act on the acceleration of a body in a plane, the resultant can be divided into the two perpendicular components of the tangential force ${\vec {F}}_{||}$ and the normal force ${\vec {F}}_{\bot }$. The tangential component only changes the amount of velocity and not the direction. The normal force only changes the direction of movement depending on the speed of the body and the shape of the path.

Examples

Free fall in a homogeneous gravitational field

The free fall in a homogeneous gravitational field describes the movement of a body that is only influenced by the constant gravitational force, without air resistance or other forces acting. In the homogeneous gravitational field, the gravitational force is the same everywhere and acts in the same direction. The tangential force ${\vec {F}}_{\|}$ runs parallel to the gravitational force ${\vec {F}}_{\text{G}}$.

${\vec {F}}_{\|}={\vec {F}}_{\text{G}}=m{\vec {g}}$

The body therefore experiences a constant acceleration $g$. Without friction and air resistance, the distance traveled $s$ is proportional to the square of the falling time $t$: $s\sim t^{2}$.

Slope downforce on the inclined plane

Tangential force along the inclined plane

In a homogeneous gravitational field, the tangential force ${\vec {F}}_{\|}$ accelerates a body $P$ of mass $m$ downwards on the inclined plane without the effect of friction:

${\vec {F}}_{\|}={\vec {F}}_{G}+{\vec {F}}_{Z}\quad {\text{mit}}\quad |{\vec {F}}_{\|}|=|{\vec {F}}_{G}|\sin \alpha =mg\sin \alpha$

The acceleration $g\sin \alpha$ due to the downhill force ${\vec {F}}_{\|}$ is smaller by the factor $\sin \alpha$ than in free fall.

The constraining force ${\vec {F}}_{Z}$ exerted by the inclined plane on the body $P$ is equal in magnitude to the normal component of the weight force $|{\vec {F}}_{G\bot }|=mg\sin \alpha$, but acts in the opposite direction ${\vec {F}}_{Z}=-{\vec {F}}_{G\bot }$. There is a balance of forces perpendicular to the plane, so that the body is not accelerated in this direction.

Frictional force between solid bodies

An example of a tangential force is the force that pushes us into the seat when we accelerate in a car. The tangential velocity changes unless other forces prevent this. In the real case with friction, the acceleration force $F_{\|}$ must exceed a threshold value for a body to start moving. This acceleration force must always be greater than the frictional force $F_{\text{R}}$, which as a tangential force with the unit vector ${\vec {e}}_{\|}$ is always directed in the opposite direction to the direction of movement. With the normal force $F_{\bot }$ of the body perpendicular to the support, experience has shown that the frictional force is

${\vec {F}}_{\text{R}}=-\mu \,F_{\bot }{\vec {e}}_{\|}$

The coefficient of friction $\mu$ increases from rolling friction to sliding friction to static friction.