# Equation of Motion

Newton’s second law of motion states that  “the relation between an object’s mass m, its acceleration a, and the applied force F is F=ma, where acceleration and force are vectors, and the direction of the force vector is the in the same direction as the acceleration vector.”  This means that if there is more than one force pulling or pushing on an object,  than the resultant force of those forces can be used to find the acceleration of the object.  Refer to the equations below.

(Eq 1) $∑F=ma=F_R$

(Eq 2) $F_R=\sqrt{F_x^2+F_y^2+F_z^2}$

The above equations can be visually expressed in the following image.

Graphically the magnitude of each force and its direction can be accounted for by drawing a free body diagram of the forces.  The summation of these forces will result in a resultant force that will equal the mass times the acceleration of the particle.  If the summation of the forces equals 0 than the acceleration of particle P will be zero.  If the acceleration is at 0 the particle will either remain stationary or move along a straight path at a constant velocity.  This is considered static equilibrium which satisfies Newton’s first law of motion.

In order to effectively use the equation of motion an acceleration needs to be measured at a Newtonian or inertial frame of reference. A Newtonian frame of reference is based off a coordinate system that is not able to rotate and is fixed or is translating at a constant acceleration so that there is zero acceleration.  By following this rule it can be assured that two different observers in two different location will be able to measure the same resultant forces caused by that acceleration.

Example

Determine the acceleration of the box in the image below.  The box weights 250 Newtons, and the coefficient of kinetic friction is 0.2.  There is a force of 300 Newtons pulling on the box at a 25 degree angle.  It can be assumed that the surface the boxing is being pulled across is perfectly flat.

Solution

Step 1:  Draw a free body diagram.

Step 2:  Based off the free body diagram define the equations of motion.

$F_y+F_c-W=0$

$F_x-F_{fiction}+ma=0$

where $F_{friction}=0.2F_c$ and $m=\frac{W}{g}$

Step 3:  Find the acceleration a

$F_c=W-F_y$$=250-300sin(25)=123.2 N a=\frac{F_{friction}-F_x}{m}$$=\frac{24.64-300cos(25)}{-25.48}=9.7\frac{m}{s^2}$