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