Lesson 1 · 20 min

Newton's Second Law

Kinematics describes motion without asking what causes it. Kinetics asks: which forces act on a body, and what motion do they produce? One law answers both directions of that question. The resultant force on a particle equals its mass times its acceleration, \(\sum\Fvec = m\avec\), and every problem in this module is that equation, written carefully for a well-drawn free-body diagram.

Learning objectives

A push on smooth ice

A crate rests on smooth ice. You push it horizontally with a steady force. Before you look at Figure 1.1, make a prediction.

Figure 1.1 A crate pushed by a constant horizontal force \(\colP{P}\). The chart under the crate plots its velocity against time: a straight line, whose slope is the acceleration. The bars in the panel add the horizontal forces and land on \(\colMA{ma}\). Change the push, the mass or the friction: the acceleration follows \(a = \sum F_x / m\) at every setting. Forces are drawn as in a free-body diagram: the weight \(\colW{W}\) ends at the center of gravity, the push and the ice's normal force \(N\) end on the faces they push, and friction \(\colF{F_f}\) starts on the bottom face.

Double the push and the velocity line gets twice as steep. Double the mass instead and it gets half as steep. Add friction and the slope drops by \(\mu_k g\); make the push smaller than the largest static friction and nothing moves at all: the forces balance, and the acceleration is zero.

Newton's laws of motion

  1. First law. A particle at rest stays at rest, and a particle in motion keeps moving in a straight line at constant speed, unless an unbalanced force acts on it.
  2. Second law. A particle acted on by an unbalanced force accelerates in the direction of the force, at a rate proportional to the force.
  3. Third law. The forces two particles exert on each other are equal in size, opposite in direction and collinear.

The equation of motion

\[ \textstyle\sum\Fvec = m\colMA{\avec} \]

The resultant of all the forces on a particle equals its mass times its acceleration. Both sides are vectors, so the equation holds component by component.

Mass, weight and units

Mass \(m\) is a property of the body: its resistance to acceleration. It is the same everywhere. Weight is the force of gravity on the body, and it depends on where the body is. Near the Earth's surface:

Weight

\[ \colW{W = mg}, \qquad g = 9.81\ \text{m/s}^2 \]

In SI units, mass is in kilograms, acceleration in meters per second squared, and force in newtons. The newton is defined by the second law: \(1\ \text{N}\) gives \(1\ \text{kg}\) an acceleration of \(1\ \text{m/s}^2\). So a \(70\ \text{kg}\) person weighs \(70(9.81) = 686.7\ \text{N}\) on Earth, and about \(113\ \text{N}\) on the Moon, where \(g = 1.62\ \text{m/s}^2\), with the same mass.

A bathroom scale measures the force it pushes up with, not the mass. That is why it reads differently in an accelerating elevator (Lesson 2).

Example 1.1 — The force to accelerate a car

A \(1200\ \text{kg}\) car accelerates uniformly from rest to \(100\ \text{km/h}\) in \(8\ \text{s}\) on a level road. What net horizontal force acts on it?

Show solution

Kinematics first: \(v = 100/3.6 = 27.78\ \text{m/s}\), so \(a = 27.78/8 = 3.472\ \text{m/s}^2\). Then kinetics:

\[ \textstyle\sum F_x = ma = (1200)(3.472) = 4167\ \text{N} \]

This is the net force: the push of the road on the driven wheels less the air drag and rolling resistance. It is about \(35\%\) of the car's weight, \(11.77\ \text{kN}\).

Three kinds of kinetics problem

What is given, and what \(\sum\Fvec = m\avec\) finds
GivenFindExample
The forcesThe acceleration, then the motion by kinematicsHow fast is a crate moving after it slides \(3\ \text{m}\) down a rough ramp?
The motionThe forces that cause itWhat tension does an elevator cable carry while the car speeds up?
Some of eachThe unknown forces and the acceleration togetherTwo blocks joined by a cord: find their acceleration and the tension.

In every case the work is the same: a free-body diagram showing every force, a kinetic diagram showing \(m\avec\), and one equation for each direction. Lesson 2 sets out the procedure.

Color tells you which force is which

Every figure, readout and highlighted formula in this module uses the same colors.

Cord tensions are drawn in gray, starting where the cord is tied.

The road ahead

Check your understanding

Key takeaways