How do u find the resultant of 3 forces?

Answer 1

Sometimes it's easy, sometimes it's more difficult, depending on what you're given.

Case I. If the forces are already resolved into components, e.g., F1 = 3i - 2j + 6k (whether in 2 or 3 dimensions), then all you do is add the i, j, and k components together to get the resultant force.

And by the way, if you have vector in component form, you can easily get the magnitude by applying the Pythagorean Theorem twice, so the magnitude of F1 above is sqrt(9+4+36) = 7.

Case II. The problem is a bit more difficult when the three forces are all in the x-y plane, but expressed in polar coordinates in the form (r, theta). Then you need to resolve all three vectors into component form using x = r cos theta and y = r sin theta. Once they're resolved, you can add them as in Case I.

Example: Add F1 = (6, 60 degrees) and F2 = (8, 135 degrees).

F1x = 6 cos 60 = 6(1/2) = 3
F1y = 6 sin 60 = 6 (sqrt 3) / 2 = 3 sqrt 3 = 5.196
F2x = 8 cos 135 = -8(sqrt 2) / 2 = -4 sqrt 2 = -5.657
F2y = 8 sin 135 = 8(sqrt 2) / 2 = 5.657

Then F1 = 3i + 5.196j and F2 = -5.657i + 5.657j

and your resultant is F = -2.657i + 10.853j.

Converting that to polar form, the magnitude is 11.174 by Pythagoras, and the tan theta = -10.853/2.657 = -4.08468, so theta = 103.76 degrees (second quadrant), and F = (11.174, 103.76 degrees).

With three vectors in the x-y plane, you just extend this procedure.

Case III, Part 1. Worst case is where you're in three dimensions using spherical coordinates. Your vectors will take the form F = (r, theta, phi) where r is the magnitude, theta is the angle in the x-y plane (0 to 360) measured from the x-axis, and phi is the angle (0 to 180) measured downward from the z-axis.

To set this up, use the "right-hand rule": From the origin, the x-axis extends down and to the left; the y-axis goes to the right; and the z-axis goes up.

Here's one example I'll work backwards, using the vector <3, -2, 6> that I made up in Case I above. (It's a good one to use because by accident, 3^2 + 2^2 + 6^2 = 7^2.)

Start in the x-y plane with <3, -2>. The resultant vector (rho, theta) is (sqrt 13, -33.69) because tan theta = -2/3.

Use Pythagoras to get the magnitude of the resultant r^2 = rho^2 + z^2 = 13 + 6^2 = 49, so r = 7.

Finally, we get the angle phi from the relation tan phi = rho/z = (sqrt 13) / 6 = 0.60093, so phi = 31.00 degrees.

This vector F will be expressed in spherical coordinates as
F = (r, theta, phi) = (7, -33.7, 31.0)

Case III, Part 2. Having worked this backward, let's do it the way you'd normally have to. Given F = (7, -33.7, 31.0) in spherical coordinates, resolve it into x, y, and z components.

Use these formulas:

z = r cos phi
rho = r sin phi (rho is the projection of r onto the x-y plane)
x = rho cos theta = r sin phi cos theta
y = rho sin theta = r sin phi sin theta

Then we have
z = 7 cos 31.0 = 6
rho = 7 sin 31 = 3.6
(or you can get it by Pythagoras sqrt(r^2 - rho^2) = sqrt(49-36))
x = 3.6 cos (-33.7) = 3
y = 3.6 sin (-33.7) = -2

So the force F = 3i - 2j + 6k = <3, -2, 6> in rectangular components.

And having resolved this, and any other vectors you may have, you can proceed to get 3-D resultants using the methods of Case I above.

Answer 2

you have to decompose every individual force into its x, y, z components.
then add the 3 x components, the y's and the z's
the new 3 x,y, and z components that results, are the components of the resultant.

Answer 3

OK

add two of them. add the third one to the result. sort of a mini fourier transform

Rob