How do we add vectors numerically?

[fawk3s]

Recently I started thinking about vectors and I ran into some confusion. So I would need a bit of help.
There is no task or any kind of homework question, but I figured this would be a good place to post the question.

[PLAIN]http://img5.imageshack.us/img5/175/vectorsm.jpg

So say we have these 2 vectors which we add up, vectors a and b. The sum vector is c.
How do we actually add their numerical values up? Are we supposed to approach it like triangles, with Pythagoran theorem or angles? Because when they are drawn on a scale, it would make sense.
If that's the case, that's where I get confused. If we find the sum vector c by Pythagoran theorem, it gives us a value which is smaller than the actual sum of the 2 vectors. Its like the resultant force on the object these forces are applied on is smaller than the actual forces, even though the forces are not opposite.
If we were to apply a force in the direction which vector c is pointed to, the component forces would be the same as in the first case. But the sum of these component forces is bigger than the actual force which we applied on the object. How could that be?

Obviously my logic fails somewhere, so please point it out.

 

[zorro]

So say we have these 2 vectors which we add up, vectors a and b. The sum vector is c.
How do we actually add their numerical values up? Are we supposed to approach it like triangles, with Pythagoran theorem or angles? Because when they are drawn on a scale, it would make sense.

You can use Pythagoras' theorem to find the resultant only if a and b (both vectors) are perpendicular to each other. You have to use the parallelogram law of vector addition to find out the resultant of these two if they are at an angle other than 90 degrees.

R² = A² + B² + 2ABcosθ

Where R,A,B stand for the magnitude of the vectors R,A,B and θ is the angle between A and B.
Notice that the above equation changes to Pythagoras' result if θ is 90.

 

[gneill]

So say we have these 2 vectors which we add up, vectors a and b. The sum vector is c.
How do we actually add their numerical values up? Are we supposed to approach it like triangles, with Pythagoran theorem or angles? Because when they are drawn on a scale, it would make sense.

Yes, that's correct (for vectors at right-angles as you've drawn).

If that's the case, that's where I get confused. If we find the sum vector c by Pythagoran theorem, it gives us a value which is smaller than the actual sum of the 2 vectors. Its like the resultant force on the object these forces are applied on is smaller than the actual forces, even though the forces are not opposite.

Can you give a numerical example of this?

 

[fawk3s]

You can use Pythagoras' theorem to find the resultant only if a and b (both vectors) are perpendicular to each other. You have to use the parallelogram law of vector addition to find out the resultant of these two if they are at an angle other than 90 degrees.

R² = A² + B² + 2ABcosθ

Where R,A,B stand for the magnitude of the vectors R,A,B and θ is the angle between A and B.
Notice that the above equation changes to Pythagoras' result if θ is 90.

Yes, sorry, I figured I'd use vectors which make up a right triangle, for simplification.

Can you give a numerical example of this?

Well, say the horisontal vector is 3 Newtons. The vertical is 4 N.
By Pythagoran theorem we get that the resultant vector of the two is 5 N.

Say I push a box with 3 N in one direction, and with 4 N in a direction 90 degrees from the first direction. The resultant force on the box is supposed to be 5 N. So what happens to the 2 N from the 7 N I am applying on that box?

 

[gneill]

Well, say the horisontal vector is 3 Newtons. The vertical is 4 N.
By Pythagoran theorem we get that the resultant vector of the two is 5 N.

Say I push a box with 3 N in one direction, and with 4 N in a direction 90 degrees from the first direction. The resultant force on the box is supposed to be 5 N. So what happens to the 2 N from the 7 N I am applying on that box?

Suppose that, instead of forces, these were distances. You walk 3 miles north and 4 miles east. The resultant displacement from your origin is 5 miles. What happened to the extra 2 miles from the seven miles total that you walked? They're still in the individual components, the "legs" of the triangle, but the resultant reflects the net effect taking direction into account.

The same is true of the addition of forces. You can always recover the contributions of the components if you have the magnitude of the resultant and the angle it makes with the axes.

When you study the results of forces acting on bodies you find that the individual components retain their expected effects independently in their given directions (such as acceleration and the resulting velocities), and that the resultant vector of these effects is the same as if the force components were combined first and then applied. The benefit is that you don't have to keep reorienting your axes for making calculations, and when there are multiple forces being applied in different directions, you can simply sum up the their same-direction components to achieve the net result in those directions.

 

[fawk3s]

I think I get it now. I was thinking basically the same as you guys, but for some reason it seemed unfamiliar and didnt quite nail it.

Thanks !

 

[Andrew Mason]

Say I push a box with 3 N in one direction, and with 4 N in a direction 90 degrees from the first direction. The resultant force on the box is supposed to be 5 N. So what happens to the 2 N from the 7 N I am applying on that box?

Analyse the problem from the direction of the resultant force. Resolve each of the applied forces into components in the direction of the resultant force and perpendicular to that direction. You will see that the components in the direction of the resultant force add to 5 N and the components in the perpendicular are equal and opposite, so they cancel each other out.

AM

 

[fawk3s]

Analyse the problem from the direction of the resultant force. Resolve each of the applied forces into components in the direction of the resultant force and perpendicular to that direction. You will see that the components in the direction of the resultant force add to 5 N and the components in the perpendicular are equal and opposite, so they cancel each other out.

AM

Thanks! This is a really good answer.
I actually did resolve the component forces like that, but I didnt realize that the forces perpendicular to the resultant cancel each other out. I don't know how I missed that!