1
$\begingroup$

If there are a multitude of forces acting on an object in different directions, how do we find the TOTAL force?

I know we add up the $ x $- and $ y $-components of the forces individually, but how do we find the total force? Do we take the $ x $- and $ y $-components and apply the Pythagorean Theorem to obtain its magnitude?

Force diagram

$\endgroup$
1
$\begingroup$

Force is a vector which itself has components along different axes, but if you just want the magnitude of the vector, then yes, you use the Pythagorean theorem. There's a good page on vector addition here if you want more info.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Quite simply: yes.

When you add up the x and y components, you result in a vector whose x and y components have already been broken down. The total force is equal to the magnitude of that vector.

As a simple example, if you have $10N$ at $0^o$ and $10N$ at $45^o$ you get a resulting vector $\vec{F}$ whose components are $F_x = 17.07N$ and $F_y = 7.07N$. The Force is then $|\vec{F}| = 18.48N$.

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

Force is a vector, so it has components in both the $x$ and $y$ directions in your specific example. You have to add the components. You find the components by using trig.

$$F_x = F_{x,A}+F_{x,B} + F_{x,C}$$ $$F_y = F_{y,A}+F_{y,B} + F_{y,C}$$

Then the total force is ${\boldsymbol F} = F_x \hat{x} + F_y \hat{y}$

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

The total force (vector) is the sum of all the forces:$$\vec{F_T}=\sum\limits_i\vec{F_i}$$

Mathematically:$$||\vec{F_T}||=\sqrt{\sum\limits_i||\vec{F_i}||^2}\tag{1}$$

If you have doubt on that you can prove $(1)$ using the mathematical induction.

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

Yes. Add up the like components and then use the Pythagorean Theorem.

| 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.