2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Use $x=x_0+\dfrac{1}{2}at^2$ where $F=ma$. $\endgroup$ – Omar Nagib Jan 22 '16 at 12:31
  • $\begingroup$ You need to define velocity vector also. Acceleration a=m/f in same direction as force. The position vector is simply double integral of a. The initial conditions are defined by initial velocity and position. $\endgroup$ – user82412 Jan 22 '16 at 12:45
  • 1
    $\begingroup$ BTW: It is recommended you typeset sines and cosines with \sin and \cos. The difference is $sin(x)$ vs. $\sin(x)$. I already edited your post. $\endgroup$ – John Alexiou Jan 22 '16 at 13:39
  • $\begingroup$ Are you sure the force is on the same direction as the position? Isn't the more general case $\vec{F} = ( F \cos \psi. F \sin \psi)$. Also the velocity of the particle needs to be specified in order to establish the final position. $\endgroup$ – John Alexiou Jan 22 '16 at 13:41
1
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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}

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.