16.2 Newton's second law and F = ma — Learn

Prerequisites: Newton's first law and resultant forces (Section 16.1); rearranging a formula; the equations of constant acceleration (Section 15.4).

By the end of this section you can: state Newton's second law; use $F = ma$ along the line of motion; find an acceleration, a resultant force or a mass; decide the direction of the acceleration; and read a simple modelling assumption from a question.

1. Newton's second law

Newton's second law links the resultant force on a body to the way its velocity changes. For a body of constant mass $m$ , moving in a straight line with acceleration $a$ , and with resultant force $F$ along that line:

F = ma.

The force is measured in newtons, the mass in kilograms, and the acceleration in metres per second squared. One newton is the force that gives a mass of $1$ kg an acceleration of $1$ m s $^{-2}$ .

The law has three consequences worth stating plainly.

The acceleration is in the same direction as the resultant force.
A zero resultant force gives a zero acceleration, which is Newton's first law again.
For a fixed force, a larger mass gives a smaller acceleration.

The symbol $F$ in $F = ma$ is the resultant force, not one of the forces. Add all the forces along the line first, then use the total.

2. Using F = ma

Choose the direction of motion as positive, then write down the resultant force along that direction. A force that opposes the motion is negative.

Three rearrangements are useful.

a = \frac{F}{m}, \qquad m = \frac{F}{a}, \qquad F = ma.

Example 1. A crate of mass $45$ kg is pushed along a horizontal floor. The two workers push with horizontal forces of $120$ N and $95$ N in the same direction. A constant resistance of $65$ N opposes the motion. Find the acceleration of the crate.

Take the direction of motion as positive. The resultant force along the floor is

120 + 95 - 65 = 150 \text{ N}.

Now use $F = ma$ with $F = 150$ N and $m = 45$ kg:

a = \frac{150}{45} = \frac{10}{3} \approx 3.33 \text{ m s}^{-2}.

The crate accelerates at about $3.33$ m s $^{-2}$ in the direction of the push.

Example 2. A car of mass $1200$ kg travels along a straight road. The driving force is $2500$ N and the total resistance is $700$ N. Find the acceleration.

Take the direction of motion as positive. The resultant force is

2500 - 700 = 1800 \text{ N}.

Then

a = \frac{1800}{1200} = 1.5 \text{ m s}^{-2}.

Example 3. A body of mass $8$ kg has an acceleration of $2.5$ m s $^{-2}$ . Find the resultant force on it.

Use $F = ma$ directly:

F = 8 \times 2.5 = 20 \text{ N}.

The force is in the direction of the acceleration.

Example 4. A trolley of mass $6$ kg accelerates at $1.5$ m s $^{-2}$ against a resistance of $5$ N. Find the driving force.

First find the resultant force needed:

F = ma = 6 \times 1.5 = 9 \text{ N}.

The driving force must give this resultant and also overcome the resistance, so

\text{driving force} = 9 + 5 = 14 \text{ N}.

3. Reading a modelling question

Mechanics questions use a short list of modelling words. Each one removes a complication.

Particle means the body is treated as a single point, so its size and any turning effect are ignored.
Smooth means there is no friction.
Rough means friction is present, and a coefficient of friction is given.
Light means a string or rod has negligible mass.
Inextensible means a string does not stretch, so two bodies joined by it have accelerations of the same size.
Air resistance is negligible means drag is left out of the calculation.

Watch out. Three slips are common.

Using one force instead of the resultant. In Example 1, $120$ N is not the force in $F = ma$ . The resultant is $150$ N, after the second push and the resistance. Use the total along the chosen direction.

Forgetting the direction. If the acceleration is negative, the body is slowing down. Write the sign down with the number and say what it means. Re-check: a car braking has a forward driving force and a larger backward braking force, so the resultant points backwards.

Mixing up mass and weight. The $m$ in $F = ma$ is the mass in kilograms, not the weight in newtons. A mass of $45$ kg has a weight of about $441$ N. Put the mass in the formula, and Section 16.3 explains the difference.

Progress check

Return to the chapter problem. The two workers push the $45$ kg crate along the horizontal floor with forces of $120$ N and $95$ N. A resistance of $65$ N opposes the motion.

Find the resultant force along the floor.
Find the acceleration of the crate.
The resistance doubles to $130$ N while the push stays the same. Find the new acceleration, and say what happens to the motion.

You should now be able to

state Newton's second law and write it as $F = ma$ ;
use the resultant force along the chosen direction;
find an acceleration, a resultant force or a mass;
say which way the acceleration points;
read the modelling words particle, smooth, rough, light, inextensible and air resistance is negligible.