Newton's second law

Newton's second law: the net force on an object equals its mass times its acceleration. In symbols, $\vec{F}_{\text{net}} = m\vec{a}$. The first law is the special case where the net force is zero, so the acceleration is too; the second law covers everything else.

Read it as a recipe for acceleration, $\vec{a} = \vec{F}_{\text{net}} / m$. Add up every force on the object, divide the sum by the mass, and you have the acceleration: how fast the velocity is changing and in which direction. The acceleration points along the net force, never along the velocity.

It is a vector equation, so it holds one axis at a time. Pick axes, and along each one the components balance the same way: $\sum F_x = m a_x$ and $\sum F_y = m a_y$. Axes that line up with the motion, along an incline or the direction of travel, usually make the problem easy.

Two inputs go in, and both are places people slip. The forces come from a free-body diagram, not from intuition, and the mass is the object's inertia in kilograms, not its weight in newtons. The next sections take those two in turn.

Mass and weight are different quantities with different units. Mass measures inertia in kilograms: how strongly the object resists being accelerated. It does not change if you carry the object to the Moon. Weight is a force, the pull of gravity, equal to $W = mg$ and measured in newtons. On Earth, with $g = 10$ m/s$^2$, a $2$ kg object weighs $20$ N.

The second law takes mass, not weight: $\vec{F}_{\text{net}} = m\vec{a}$ has the mass on the right, not the weight. This matters the moment a problem hands you a weight. A value in newtons is a force; divide it by $g$ to get a mass in kilograms before it can stand in for $m$.

The trap is quiet because the arithmetic still produces a number. Divide a force in newtons by a "mass" that is really a weight, label the result m/s$^2$, and nothing on the page flags the slip. The fix is to track units: newtons mean a force, kilograms mean a mass, linked by a factor of $g$ and never equal.

The same factor explains the Moon. An astronaut whose mass is $80$ kg has that mass everywhere; what changes off Earth is the weight, because $g$ is different. Mass belongs to the object; weight depends on where it is.

Setup. Two blocks hang from a light string over an ideal pulley: $m_1 = 3$ kg and $m_2 = 5$ kg, with $g = 10$ m/s$^2$. Find the acceleration of the system and the tension in the string.

One equation per block. The string ties both blocks to a single acceleration $a$. Take the heavier block $m_2$ as descending. For $m_2$: $m_2 g - T = m_2 a$. For $m_1$, rising: $T - m_1 g = m_1 a$. The tension $T$ is the same on both ends of an ideal string over an ideal pulley.

Add to cancel the tension. Adding the two equations removes $T$: $(m_2 - m_1) g = (m_1 + m_2) a$, so $a = \dfrac{(m_2 - m_1) g}{m_1 + m_2} = \dfrac{(2)(10)}{8} = 2.5$ m/s$^2$.

Back-substitute for the tension. From the $m_1$ equation, $T = m_1 (g + a) = 3(10 + 2.5) = 37.5$ N.

The trap to avoid. The tension is not $m_2 g = 50$ N and not $m_1 g = 30$ N. It lands between the two weights, at $37.5$ N. If the string carried the full weight of either block, that block would not accelerate. The whole point of solving the two equations is that the tension is its own unknown, smaller than the heavy weight and larger than the light one.

The normal force is not a formula to memorize. It is whatever value the surface supplies to keep the object from passing through it, and you find it the same way you find any unknown force: draw the free-body diagram and apply the second law along the axis perpendicular to the surface.

On flat ground with nothing else pushing vertically, that calculation does give $N = mg$. That is one case, not a law, and it is why the identity feels universal. Change the setup and the surface responds:

Added vertical force. Press down on the object with a force $F$ and the surface pushes back harder: $N = mg + F$. Pull up with $F$ and it pushes back less: $N = mg - F$.

Accelerating elevator. Accelerating upward at $a$, the floor pushes harder: $N = m(g + a)$. Accelerating downward, it pushes less: $N = m(g - a)$.

Incline. Tilt the surface to an angle $\theta$ and only part of gravity presses into it: $N = mg\cos\theta$, smaller than the full weight.

Vertical wall. Press a book against a wall and the wall's normal force is horizontal, balancing the push, with nothing to do with $mg$ at all.

The rule that survives every case is the constraint, not the formula: the normal force is whatever it takes to keep the object on the surface, and that value comes out of the FBD.

One limit bounds every one of these: a surface can push but never pull, so $N$ can never be negative. If the perpendicular balance would require $N < 0$, the object has left the surface and $N = 0$ instead.

Setup. A $5$ kg block sits on a frictionless incline angled at $37°$ ($\sin 37° = 0.6$, $\cos 37° = 0.8$, $g = 10$ m/s$^2$). Find the normal force and the acceleration down the slope.

Choose tilted axes. Put the $x$-axis along the incline (down-slope positive) and the $y$-axis perpendicular to it. Gravity $mg = 50$ N points straight down, so it splits into a component $mg\sin\theta$ along the slope and $mg\cos\theta$ into the surface. The normal force is along $+y$.

Perpendicular axis sets the normal force. The block does not accelerate into or off the surface, so $\sum F_y = 0$: $N - mg\cos\theta = 0$, giving $N = mg\cos\theta = 50(0.8) = 40$ N. The normal force is smaller than the $50$ N weight because the surface is tilted.

Along-slope axis sets the acceleration. The only force along the slope is gravity's component: $\sum F_x = mg\sin\theta = ma$, so $a = g\sin\theta = 10(0.6) = 6$ m/s$^2$ down the incline.

The trap to avoid. Neither answer is the flat-ground value. The normal force is $40$ N, not the $50$ N weight, and the acceleration is $6$ m/s$^2$, not the full $g$. Both follow from resolving gravity onto the tilted axes; reusing $N = mg$ or $a = g$ here is the mistake.