Editing Net force (section)

===Rigid bodies===
[[File:Free body acceleration.JPG|thumb|300px|How a force accelerates a body.]]
In the example shown in the diagram opposite, a single force <math>\mathbf F </math> acts at the application point '''H''' on a free rigid body. The body has the mass <math>m </math> and its center of mass is the point '''C'''. In the constant mass approximation, the force causes changes in the body motion described by the following expressions:

: <math> \mathbf a = {\mathbf F \over m} </math>&nbsp;&nbsp;&nbsp;is the center of mass acceleration; and

: <math> \mathbf \alpha = {\mathbf \tau \over I} </math>&nbsp;&nbsp;&nbsp;is the [[angular acceleration]] of the body.

In the second expression, <math>\mathbf \tau </math> is the [[torque]] or moment of force, whereas <math>I </math> is the [[moment of inertia]] of the body. A torque caused by a force <math>\mathbf F </math> is a vector quantity defined with respect to some reference point:

:<math> \mathbf \tau = \mathbf r \times \mathbf F </math>&nbsp;&nbsp;&nbsp;is the torque vector, and

:<math> \ \tau = Fk </math>&nbsp;&nbsp;&nbsp;is the amount of torque.

The vector <math>\mathbf r </math> is the [[position vector]] of the force application point, and in this example it is drawn from the center of mass as the reference point of (see diagram). The straight line segment <math>k </math> is the lever arm of the force <math>\mathbf F</math> with respect to the center of mass. As the illustration suggests, the torque does not change (the same lever arm) if the application point is moved along the line of the application of the force (dotted black line). More formally, this follows from the properties of the vector product, and shows that rotational effect of the force depends only on the position of its line of application, and not on the particular choice of the point of application along that line.

The torque vector is perpendicular to the plane defined by the force and the vector <math>\mathbf r</math>, and in this example, it is directed towards the observer; the angular acceleration vector has the same direction. The [[right-hand rule]] relates this direction to the clockwise or counterclockwise rotation in the plane of the drawing.

The moment of inertia <math>I</math> is calculated with respect to the axis through the center of mass that is parallel with the torque. If the body shown in the illustration is a homogeneous disc, this moment of inertia is <math>I = m r^2/2</math>. If the disc has the mass 0,5&nbsp;kg and the radius 0,8 m, the moment of inertia is 0,16 kgm<sup>2</sup>. If the amount of force is 2 N, and the lever arm 0,6 m, the amount of torque is 1,2 Nm. At the instant shown, the force gives to the disc the angular acceleration α = {{math|''τ''}}/I = 7,5 rad/s<sup>2</sup>, and to its center of mass it gives the linear acceleration ''a''&nbsp;= ''F''/''m''&nbsp;= 4&nbsp;m/s<sup>2</sup>.