Forces with vectors (Mechanics)

Forces as a vector

A force applied to a particle is also a vector as it has both a magnitude and a direction. We can use the formula $\mathbf{F} = m \mathbf{a}$, where $m$ is the mass of the particle, to solve problems about force.

If a particle is resting in equilibrium then the resultant of all the forces acting on it is zero. Therefore the sum of the vectors of the forces must be equal to the zero vector.

==Example using $\mathbf{F} = m \mathbf{a}$==

Example

A constant force $\mathbf{F} \mathrm{N}$ acts on a particle of mass $3 \mathrm{kg}$ for $7 \mathrm{seconds}$. The particle is initially at rest, and $7 \mathrm{seconds}$ later it has velocity $(11 \mathbf{i} + 22 \mathbf{j})\mathrm{ms^{-1} }$. Find $\mathbf{F}$.

Solution

We can use $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ to find the acceleration of the particle, and from this find $\mathbf{F}$ using $\mathbf{F} = m \mathbf{a}$. As it is initially at rest we have that $\mathbf{u}=\mathbf{0}$. \begin{align} \mathbf{v} & = \mathbf{u} + \mathbf{a} t, \\ (11 \mathbf{i} + 22 \mathbf{j}) & = \mathbf{0} + 7 \mathbf{a}, \\ \mathbf{a} & = \frac{1}{7} \left(11 \mathbf{i} + 22 \mathbf{j} \right), \\ \mathbf{F} & = m \mathbf{a}, \\ & = 3 \times \frac{1}{7} \left(11 \mathbf{i} + 22 \mathbf{j} \right), \\ & = \left(\frac{363}{7} \mathbf{i} + \frac{726}{7} \mathbf{j}\right) \mathrm{N}. \end{align}

Example using forces in equilibrium

Example

Suppose that the forces $3\mathbf{i} +5 \mathbf{j}$, $-4\mathbf{i} + 2\mathbf{j}$, $6\mathbf{i} - 3\mathbf{j}$ and $p\mathbf{i} + q\mathbf{j}$ act on a particle which is in equilibrium. What are the values $p$ and $q$?

Solution

If the particle is in equilibrium then the resultant force will be zero. \begin{align} (3\mathbf{i} +5 \mathbf{j}) + (-4\mathbf{i} + 2\mathbf{j}) + (6\mathbf{i} - 3\mathbf{j}) + (p\mathbf{i} + q\mathbf{j}) & = \mathbf{0}, \\ (3 -4 + 6 + p) \mathbf{i} + (5 + 2 -3 + q) \mathbf{j} & = \mathbf{0}. \end{align} Therefore for the $\mathbf{i}$ component we have that \begin{align} 5 + p & = 0, \\ p & = -5. \end{align} Similarly for the $\mathbf{j}$ component we have that \begin{align} 4 + q & = 0, \\ q & = -4. \end{align}