0
$\begingroup$

I guess the title says it all, it doesn't seem intuitive that we can consider something as abstract as forces to be decomposable vectors. An example would be a mass sliding off an inclined plane, we consider the tangential component of the weight to the plane. What guarantees us in this case that our results will be correct?

$\endgroup$
7
  • $\begingroup$ Decomposing is just for mathematical ease. Newton's second law says $F = ma$. To compute $F = F_{net}$, you add up all of the forces. When mathematically adding up the forces, you decompose into components. $\endgroup$
    – mathworker21
    Jul 18, 2019 at 4:03
  • 1
    $\begingroup$ Forces can be represented as vectors, and thus vector mathematics is applicable. What precisely don't you understand? $\endgroup$
    – David G. Stork
    Jul 18, 2019 at 4:05
  • $\begingroup$ It is a theorem in linear algebra that every element of a vector space can be written uniquely as a linear combination of the basis vectors. $\endgroup$
    – M. Nestor
    Jul 18, 2019 at 4:06
  • $\begingroup$ @M.Nestor I know, I am asking on why is it okay to treat forces as vectors. $\endgroup$
    – GDGDJKJ
    Jul 18, 2019 at 4:12
  • 2
    $\begingroup$ "I know, I am asking on why is it okay to treat forces as vectors." Because they have magnitude and direction and Newton's laws say you can add them. $\endgroup$
    – David G. Stork
    Jul 18, 2019 at 4:33

4 Answers 4

3
$\begingroup$

It is clear that displacement is a vector. Since displacement is a vector, so is its derivative with respect to time, which we call velocity. Momentum is velocity times mass, which is a scalar - so momentum is also a vector.

Newton's second law (justified by experimental evidence) says that force is equal to the rate of change of momentum with respect to time. Since momentum is a vector, then force is also a vector. And this is what guarantees that we get the same answer whether we regard a net force as a single item or as the superposition of several components.

$\endgroup$
1
  • $\begingroup$ this is a satisfying answer, thank you. $\endgroup$
    – GDGDJKJ
    Jul 18, 2019 at 16:18
1
$\begingroup$

It is okay to do so because we have experimentally verified that force is indeed a vector (in the sense that it satisfies the triangle rule of addition). For instance (try this), pushing an object with a force of $1 $ N east and a force of $1 $N north simultaneously has the same effect (in terms of the acceleration) as pushing it with a force of $\sqrt 2 \approx 1.4$N north of east.

Another way to think of this: if you pulled (with a force $F$) on a toy train (mass $m$) along a straight railway (at an angle of $\theta$) and the train was somehow restricted to move along the railway and you measured the acceleration of the train, you would get $\frac{F \cos \theta}{m}$. If the railway is at right angles to the original direction, the acceleration is $\frac{F \sin \theta}{m}$. Now, if the railway is in the direction that you're pulling it in, then the acceleration is $\frac{F}{m}$.

$\endgroup$
0
$\begingroup$

It is okay to decompose forces because if you hit a baseball with a bat, most likely the force that the bat makes with the ball will not be entirely in one direction. So you decompose the forces to see where exactly those forces are being applied (x or y or z) direction.

$\endgroup$
1
  • 3
    $\begingroup$ I think this is why it is useful to do that, not why you can do that. $\endgroup$ Jul 18, 2019 at 9:22
0
$\begingroup$

You can decompose forces because of the superposition principle. According to Newton's laws of motion, a force of $a\vec{v} + b \vec{w}$ has the exact same effect as two forces $a\vec{v}$ and $b\vec{w}$ acting independently

$\endgroup$
4
  • 3
    $\begingroup$ This is tautological, then the question becomes:"why do you accept superposition principle?" $\endgroup$ Jul 18, 2019 at 9:23
  • $\begingroup$ Well, at some point you will have to stop the math and go make some experiments... $\endgroup$
    – David
    Jul 18, 2019 at 10:04
  • 1
    $\begingroup$ Right! That's exactly what I meant! $\endgroup$ Jul 18, 2019 at 10:06
  • $\begingroup$ So, your expected answer was.. "because experiments have proved so"? $\endgroup$
    – David
    Jul 18, 2019 at 10:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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