Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to understand to find my homework problem. Its a simple concept of finding force. I wonder if you have two formulas then is the total Force = force from x + force from y. Or $\sqrt{forceX^2+forceY^2}$. And why. It would make sense to me for it to be the sum of two forces if its not so why not??

I have the actual example: 3.00-kg object is moving in a plane, with its x and y coordinates given by $x=5t^2 - 1$ and $y = 3t^3+2$, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at $t = 2.00 s$

so I either get 66 by addition of the forces. Or 46.861 by distance formula.. Which one is correct and why. (feel free to use the example i provided or just explain the concept without it. I just wonder which way of finding force is correct and why)

share|cite|improve this question
Force is a vector. Vectors are added component-wise. Vector magnitude is the square root of the dot product of the vector with itself. – PhysGrad Oct 24 '12 at 7:07
up vote 4 down vote accepted

You're asking quite a basic question, and I'm not sure how well it can be answered without effectively reproducing large parts of your Physics textbook, which isn't really what this site is intended for. However I'll have a go.


Force is a vector, so it has magnitude and direction. In the diagram above I've drawn three forces, $F$, that all have the same magnitude but point in different directions. You can tell the forces all have the same magnitude because they all end on the same circle (drawn dotted) with it's centre at the origin.

When the force is lined up along the $x$ or $y$ axis it's easy to see it's magnitude i.e. how long the arrow is, because it's just $F_x$ or $F_y$ respectively. When the force is at some random angle between the $x$ and $y$ axes it's harder to see what it's magnitude is. Hopefully it should be obvious from the diagram that $F_x$, $F_y$ and $F$ form a right angled triangle:

Right angled triangle

Using Pythagorus' theorem the length of the long side, $F$, is given by:

$$ F = \sqrt{F_x^2 + F_y^2} $$

As PhysGrad says, what you're really doing is a vector addition of all the components of the force, and then a dot product of the force with itself to calculate it's magnitude. However this is possibly a conceptual step too far at the moment.

Regarding your problem, I can't just answer it as site policy forbids us doing people's homework for them, but I can explain how you do it.

Remember Newton's law says that $\vec{F} = m\vec{a}$. You're told the mass of the object, so if you can work out it's acceleration you can calculate the force being applied. We're told that:

$$ x = 5t^2 - 1 $$

so this gives us the position on the $x$ axis as a function of time. To get the velocity along the $x$ axis we differentiate wrt time:

$$ v_x = \frac{dx}{dt} = 10t $$

and to get the acceleration along the $x$ axis differentiate again wrt time:

$$ a_x = \frac{d^2x}{dt^2} = 10 $$

And you can do the same for the y axis to calculate $a_y$. Now you can calculate $F_x$ and $F_y$ by just multiplying by the mass, and finally the vector sum of $F_x$ and $F_y$ gives you the value of the force $F$. I won't give you the answer, but it isn't 66 or 46.861.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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