# Thread: how can get anything from this?....

1. ## how can get anything from this?....

Three forces AB and C act at, and away from, the orgin of a three-dimensional coordinate system. Force acts along the x axis and has a magnitude of 3 N; force B acts along the y axis and has a magnitude of 5 N; force C acts along the z axis and has a magnitude of 2 N. Evaluate the magnitude of the resultant force and specify the angle to the xy plane at which it acts. The resultant force and its projection both lie in a plane at an angle X to the xz plane. Find this angle between these two planes.

thanx

x

2. Originally Posted by steph21
Three forces AB and C act at, and away from, the orgin of a three-dimensional coordinate system. Force acts along the x axis and has a magnitude of 3 N; force B acts along the y axis and has a magnitude of 5 N; force C acts along the z axis and has a magnitude of 2 N. Evaluate the magnitude of the resultant force and specify the angle to the xy plane at which it acts. The resultant force and its projection both lie in a plane at an angle X to the xz plane. Find this angle between these two planes.

thanx

x
Hello steph21,

Lets write each of your vectors in component notation.
$F_1=3 \vec i \\\ F_2=5 \vec j \\\ F_3=2 \vec k$

Now our resultant force is the sum of all the forces.

$F_r=3\vec i+ 5 \vec j +2 \vec k$

The magnitude of the force is the vector's length

$|F_r|=\sqrt{3^2+5^2+2^2}=\sqrt{38}$

To find the angle between the vector and the xy plane lets draw a right triangle. Lets start by drawing a line from the origin to the tip of the vector, and from the tip of the vector drop a perpendicular into the xy plane.

We know both of these distances. The length of the vector is its magnitude and the perpendicular is the z component of the vector. We can find the angle by using the $\sin^{-1}\left( \frac{2}{\sqrt{38}}\right) \approx .33 rad \approx 18.9^\circ$

For the last part try to find a different triangle. Good luck.