Newton laws for a material point

A material point is a model of a body whose size can be neglected: its position is given by the radius vector $\vec r(t)$, while changes of motion are described by the velocity $\vec v = d\vec r/dt$ and the acceleration $\vec a = d\vec v/dt$. Newton's first law states that there exist inertial reference frames in which a point with $\sum \vec F = 0$ moves uniformly along a straight line, that is, $\vec v = \text{const}$. The second law links the cause of motion to its change: $\sum \vec F = m\vec a,$ so the acceleration points along the net force and its magnitude is inversely proportional to the mass. The third law describes the interaction of two points: the forces are equal in magnitude, opposite in direction, and act on different bodies, $\vec F_{12} = -\vec F_{21}$. That is why third-law forces are not added at one point and do not cancel each other's motion. All three laws are formulated for inertial reference frames and hold for speeds much smaller than the speed of light.

A material point of mass $m = 2{,}0$ kg moves along a straight line. The only force acting on it is $\vec F = 6{,}0$ N directed along the $x$ axis. Find the acceleration of the point.

Two forces act on a material point of mass $m = 2{,}0$ kg: $\vec F_1 = 3{,}0$ N along the $x$ axis and $\vec F_2 = 4{,}0$ N along the $y$ axis. Find the magnitude of the acceleration and the angle between the acceleration and the $x$ axis.

A material point of mass $m = 0{,}50$ kg moves along the $x$ axis with velocity $v_x(t) = 4{,}0 + 6{,}0t$ (in SI units). Find the net force acting on the point and determine whether it is constant.

Run an experiment in a simulation (for example, PhET 'Forces and Motion: Basics'), where you can set the body mass and the applied force and observe the acceleration. Procedure:

1. Set the mass $m = 1{,}0$ kg and apply a horizontal force $F = 2$ N; record the measured acceleration in a table.
2. Double the force at the same mass and check the prediction of the second law: the acceleration should double.
3. Return to $F = 2$ N but double the mass; check that the acceleration halves.
4. Plot $a$ versus $F$ at constant mass and verify that the points lie on a straight line with slope $1/m$.
5. Turn on the interaction of two bodies pushing each other apart in the simulation and compare the forces: they should be equal in magnitude and opposite in direction.

A simulation is used because directly measuring forces on fast-moving bodies at home is unsafe and inaccurate. Record your results in a mass — force — acceleration table and check them against the formula $\sum \vec F = m\vec a$.

Show that the acceleration vector of a material point always points along the net force, and that its magnitude is proportional to the force magnitude and inversely proportional to the mass; also show that third-law forces act on different points.

A sketch and a graph. Sketch: a point $M$ on the $x$ axis with arrows $\vec F_1$ and $\vec F_2$ drawn from one point, their sum $\sum \vec F$ shown as a dashed arrow, and a vector $\vec a$ drawn from $M$ parallel to $\sum \vec F$. Below, two separate points $A$ and $B$ are joined by arrows $\vec F_{AB}$ and $\vec F_{BA}$ of equal length and opposite direction. Graph: the straight line $a(F)$ at a fixed mass $m$; its slope equals $1/m$, and for a larger mass the line is shallower.

A player kicks a stationary ball of mass $m = 0{,}43$ kg on grass. $\Delta t = 0{,}015$ s after contact begins the ball leaves at $v = 27$ m/s. The ball's acceleration during the kick is $a = \frac{27}{0{,}015} = 1800$ m/s², and the average kicking force is $F = ma = 0{,}43 \cdot 1800 \approx 770$ N.

Newton's second law in the form $\vec F = m\vec a$ is at work: the mass and the change of velocity are known, and the force follows from them. Newton's third law shows up in the boot–ball pair: the ball acts on the boot with a force of $770$ N directed opposite to the kick. The first law is visible after the kick: the ball flies straight and almost uniformly until air drag becomes noticeable.

A car of mass $M = 1200$ kg travelling at $20$ m/s comes to rest in $4$ s. The acceleration points opposite to the motion and equals $a = \frac{20}{4} = 5$ m/s². This acceleration is produced by friction from the road, of magnitude $F = Ma = 1200 \cdot 5 = 6000$ N. A passenger of mass $60$ kg is held by the belt with a force of $60 \cdot 5 = 300$ N.

Newton's second law $F = ma$ links the deceleration to the net force: the acceleration points backwards, so the net force points backwards too. Newton's third law describes the wheel–road pair: the road acts on the wheel with $6000$ N backwards, and the wheel acts on the road with $6000$ N forwards. The first law explains why a belt is needed: without it the passenger would keep moving at $20$ m/s until some force stopped him.

A phone of mass $m = 0{,}17$ kg lies on a shelf in the cabin. The car accelerates from $0$ to $15$ m/s in $5$ s, that is with $a = 3$ m/s². For the phone to move together with the car, the shelf must exert a friction force $F = ma = 0{,}17 \cdot 3 = 0{,}51$ N on it. If the maximum static friction is only $0{,}3$ N, the phone gets an acceleration of just $0{,}3/0{,}17 \approx 1{,}8$ m/s² and slides backwards relative to the shelf.

Newton's first law: a body keeps its velocity while the net force is zero, so in the frame of the ground the phone simply fails to keep up with the car. Newton's second law names the reason: it is the net force that sets the acceleration, and if static friction caps that force, it caps the acceleration too — $a \le \mu N/m$.