Vector composition of forces

Force: magnitude, direction and sense

If you push a trolley forwards with $50\ \text{N}$ it moves, but if you apply the same $50\ \text{N}$ downwards you just press it against the floor: why is the numerical value not enough to describe what a force does? In Leaving Cert Physics you will use the symbol $F_{net}$ for the resultant force.

Forces in the same direction and graphical addition: the parallelogram

When two or more forces act along the same line, the resultant is obtained with an algebraic sum, giving a positive sign to one sense and a negative sign to the opposite one. When they are not aligned, we need the graphical construction of the parallelogram.

Resolving a force into components

To add forces at any angle to each other, the general method is the analytical one: resolve each force into its horizontal component ($F_x$) and vertical component ($F_y$).

Weight on a slope

On a slope at angle $\theta$ to the horizontal, it is handy to resolve the weight $mg$ into a component parallel to the slope and one perpendicular to it. The angle between the weight and the perpendicular to the slope is also $\theta$, so: $$mg\sin\theta \text{ (parallel)} \qquad mg\cos\theta \text{ (perpendicular)}$$ The parallel component tends to make the object slide down; the perpendicular one presses it against the surface (the normal reaction $R$ balances it).

Resultant of several forces using components

The components method is essential when there are three or more forces at different angles, or when the question already gives the components and asks you to reconstruct the magnitude and the angle.

Finding the angle: inverse trig and angles beyond 90°

The angle of the resultant can be found with any inverse ratio: $\sin\alpha = R_y/R$, $\cos\alpha = R_x/R$ or $\tan\alpha = R_y/R_x$. Use the absolute values of the components to get an acute angle $\alpha$ from the horizontal, then use the signs of $R_x$ and $R_y$ to decide the quadrant: if $R_x<0$ and $R_y>0$ the force points up and to the left, so its angle from the positive $x$-axis is $180^\circ-\alpha$. For forces at angles above $90^\circ$, the formulas $F_x=F\cos\theta$ and $F_y=F\sin\theta$ still work if $\theta$ is measured from the positive $x$-axis: the calculator gives the negative signs automatically (e.g. $\cos 120^\circ=-0.5$).

Equilibrium of a particle

In a tug-of-war at a school sports day the central flag stays still if both teams pull with the same force, but if several cables pull in different directions, what exact mathematical condition predicts whether or not the object will start to move?

Verifying the addition of forces in the lab

In the lab you can verify the law of addition of vectors with a force board (force table). Hang masses from two cords over pulleys, so the tensions are $T = mg$. In 1D, two forces along the same line add or subtract: check that the reading of one newton balance matches $F_1 \pm F_2$. In 2D, place two forces at $90^\circ$ and use a third force to bring the ring to the centre of the board: this third force is the equilibrant, equal in magnitude and opposite in sense to the resultant.

Back to the start: (a different case from the exercise). If two tugboats in Dublin Port each pull with $3000\ \text{N}$ and the angle between their cables is $60^\circ$ ($30^\circ$ to the line of travel), the useful forward force is not $6000\ \text{N}$ but $2 \cdot 3000 \cdot \cos 30^\circ \approx 5200\ \text{N}$.

Summary