If a force acts upon an object or particle it will accelerate. An object will remain at rest or will continue to move at a constant speed in a straight line unless it is acted upon by a **resultant force.**

Every action has an equal and opposite reaction. So if an object exerts a downward force of magnitude $R$ on a surface then the surface is also exerting an upward force of equal magnitude on the object. This is called the **normal reaction** between the object and the surface, it acts perpendicular to the surface.

The relationship between the resultant force $F$ acting on a particle of mass $m$ which results in an acceleration of $a \mathrm{ms^{-2} }$ is $F=ma$. This can be used to solve problems involving force and acceleration.

**Tension** is the force acting on an object that is being pulled by a string. If an object is being pushed by a rod the force is called the **thrust** or **compression** in the rod.

**Resistance** is experienced when an object travels through air or fluid, this force acts in the opposite direction to the direction of motion.

**Gravity** is the force between any object and the Earth. The weight of an object is $W=mg$, where $m$ is the mass of the object and $g=9.8 \mathrm{ms^{−2} }$ is the acceleration due to gravity.

So important equations are:

\begin{align*} F &= ma,\\ W & = mg. \end{align*}

When there is more than one force acting on an object you resolve the forces using $F=ma$ in the direction of acceleration and in the direction perpendicular to acceleration.

Suppose that there is a block of mass $8 \mathrm{kg}$ attached to a vertical rope. Find the tension in the rope when the block moves downwards with an acceleration of $4 \mathrm{ms^{−2} }$.

We can resolve $F=ma$ in the vertical direction to find the tension, $T$ in the rope. As the block is accelerating downwards we take downwards as the positive direction, so the tension will act in the opposite direction to the weight, $W$ of the block.

\begin{align} F & = ma, \\ W - T & = ma, \\ mg - T & = ma, \\ T & = mg - ma, \\ & = \left(8 \times 9.8\right) - \left(8 \times 4\right), \\ & = 78.4 - 32, \\ & = 46.4 \mathrm{N}. \end{align}

The tension in the rope is $46.4\mathrm{N}$.

Suppose that there is a particle of mass $7 \mathrm{kg}$ which is pulled along a rough horizontal surface by a horizontal force of magnitude $15 \mathrm{N}$ against a constant frictional force of $3 \mathrm{N}$. Find the acceleration of the particle given that it is initially at rest and the magnitude of the normal reaction between the particle and the surface.

We can resolve $F=ma$ in the horizontal direction, taking our positive direction as our direction of the acceleration.

\begin{align} F & = ma, \\ 15 - 3 & = ma, \\ 12 & = 7a, \\ a & = \frac{12}{7} \mathrm{ms^{-2} }. \end{align}

The particle accelerates at $1.714 \mathrm{ms^{−2} }$ (to 3d.p.). To find the normal reaction, $R$, we resolve $F=ma$ vertically. We take the upwards direction as the positive and the acceleration in this direction is zero.

\begin{align} R - mg & = ma, \\ R - \left(7 \times 9.8 \right) & = 7 \times 0, \\ R & = 68.6 \mathrm{N}.\end{align} The normal reaction has magnitude $64.6 \mathrm{N}$.

Try our Numbas test on forces: Forces