How to apply a force vector to a position vector?

Question

A particle is present in the two dimensional space. It's position is given by the vector:
$$x = \begin{pmatrix} |x| \cos(\theta) \\ |x| \sin(\theta) \end{pmatrix}$$ It experiences a force of:
$$ F = \begin{pmatrix} |F| \cos(\theta) \\ |F| \sin(\theta) \end{pmatrix}$$ How can I calculate the position of the particle after applying this force?

Answer 1

First of all you have to consider that applying a force to a particle can make it move, it depends on the boundary conditions like the presence of friction.

According to the data you have provided you are in the very simple case of just the particle and the force acting on it so assuming it has mass $ m $ (this data is needed) hence the particle will move so it’s future position will be a function of time.

The final position vector will then be

$$ x(t) = \begin{pmatrix} \left ( |x| + \frac{|F| }{2m} t^{2} \right ) \cos(\theta) \\ \left ( |x| + \frac{|F| }{2m} t^{2} \right ) \sin(\theta) \end{pmatrix} $$

Answer 2

Apply: \begin{equation} F = ma \end{equation} to your particle to get a pair of acceleration vectors. You can then use the acceleration to work out the particle position at a given time. \begin{equation} x = x1 + at^2/2 \end{equation} If you meant impulse instead of force then the process is similar but you equate the impulse to the total change in momentum and then: \begin{equation} impulse = m\Delta v \end{equation} \begin{equation} x = x1 + vt \end{equation}

Answer 3

You start off by drawing a free body diagram, showing the x and y components of the force. Then you write a force balance for the x direction and the y direction to get the accelerations in these directions. Then you can determine the kinematics of the motion, once you know the acceleration components.

Answer 4

How can I calculate the position of the particle after applying this force?

By using Newton's second law.

If the mass of the particle is $m$ and it is constant; then we have: $$\vec F=F\cos \theta \;\vec i+F\sin \theta\;\vec j=m\vec a=m(a_x\;\vec i+a_y\;\vec j)$$ $$\Longrightarrow\;a_x=\frac{F\cos\theta}m\quad\textrm{and}\quad a_y=\frac{F\sin\theta}m$$ On the other hand, we have: $$\frac{\mathrm d^2}{\mathrm dt^2}\vec r(t)=\frac{\mathrm d^2}{\mathrm dt^2} (r\cos\theta\;\vec i+r\sin\theta\;\vec j)=\vec a(t)=a_x\;\vec i+a_y\;\vec j$$ So $$a_x=\frac{\mathrm d^2}{\mathrm dt^2}(r\cos\theta)\quad\textrm{and}\quad a_y=\frac{\mathrm d^2}{\mathrm dt^2}(r\sin\theta)$$ Now, we can determine the position vector of the particle by integrating equations above.